Mixing
Evaluating $int_0^1 log log left(frac1xright) fracdx1+x^2$
Clash Royale CLAN TAG#URR8PPP
up vote
32
down vote
favorite
Show that $displaystyleint_0^1 log log left(frac1xright) fracdx1+x^2 = fracpi2log left(sqrt2pi Gammaleft(frac34right) / Gammaleft(frac14right)right)$
This question was posted as part of this question:
Solve the integral $S_k = (-1)^k int_0^1 (log(sin pi x))^k dx$
I cannot think of a change of variable nor other integrating methods. May be there is a known method that I am missing.
calculus integration
edited Apr 13 '17 at 12:21
Community♦
1
asked Mar 18 '12 at 0:44
Kirthi Raman
6,2702855
add a comment |Â
up vote
32
down vote
favorite
Show that $displaystyleint_0^1 log log left(frac1xright) fracdx1+x^2 = fracpi2log left(sqrt2pi Gammaleft(frac34right) / Gammaleft(frac14right)right)$
This question was posted as part of this question:
Solve the integral $S_k = (-1)^k int_0^1 (log(sin pi x))^k dx$
I cannot think of a change of variable nor other integrating methods. May be there is a known method that I am missing.
calculus integration
edited Apr 13 '17 at 12:21
Community♦
1
asked Mar 18 '12 at 0:44
Kirthi Raman
6,2702855
@PeterT.off: That isn't true at all.
– Eric Naslund
Mar 18 '12 at 12:18
For some variants of this integral see C.J. Malmsten.
– Raymond Manzoni
Mar 9 '14 at 19:30
add a comment |Â
up vote
32
down vote
favorite
up vote
32
down vote
favorite
Show that $displaystyleint_0^1 log log left(frac1xright) fracdx1+x^2 = fracpi2log left(sqrt2pi Gammaleft(frac34right) / Gammaleft(frac14right)right)$
This question was posted as part of this question:
Solve the integral $S_k = (-1)^k int_0^1 (log(sin pi x))^k dx$
I cannot think of a change of variable nor other integrating methods. May be there is a known method that I am missing.
calculus integration
edited Apr 13 '17 at 12:21
Community♦
1
asked Mar 18 '12 at 0:44
Kirthi Raman
6,2702855
Show that $displaystyleint_0^1 log log left(frac1xright) fracdx1+x^2 = fracpi2log left(sqrt2pi Gammaleft(frac34right) / Gammaleft(frac14right)right)$
This question was posted as part of this question:
Solve the integral $S_k = (-1)^k int_0^1 (log(sin pi x))^k dx$
I cannot think of a change of variable nor other integrating methods. May be there is a known method that I am missing.
calculus integration
edited Apr 13 '17 at 12:21
Community♦
1
asked Mar 18 '12 at 0:44
Kirthi Raman
6,2702855
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Mar 18 '12 at 0:44
Kirthi Raman
6,2702855
asked Mar 18 '12 at 0:44
Kirthi Raman
6,2702855
asked Mar 18 '12 at 0:44
Kirthi Raman
6,2702855
6,2702855
@PeterT.off: That isn't true at all.
– Eric Naslund
Mar 18 '12 at 12:18
For some variants of this integral see C.J. Malmsten.
– Raymond Manzoni
Mar 9 '14 at 19:30
add a comment |Â
@PeterT.off: That isn't true at all.
– Eric Naslund
Mar 18 '12 at 12:18
For some variants of this integral see C.J. Malmsten.
– Raymond Manzoni
Mar 9 '14 at 19:30
@PeterT.off: That isn't true at all.
– Eric Naslund
Mar 18 '12 at 12:18
@PeterT.off: That isn't true at all.
– Eric Naslund
Mar 18 '12 at 12:18
For some variants of this integral see C.J. Malmsten.
– Raymond Manzoni
Mar 9 '14 at 19:30
For some variants of this integral see C.J. Malmsten.
– Raymond Manzoni
Mar 9 '14 at 19:30
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
43
down vote
accepted
By the substitution $x = e^-t$, we find that
$$beginalign*
int_0^1 frac(log (1/x))^s1+x^2 ; dx
&= int_0^infty fract^s e^-t1 + e^-2t ; dt \
&= int_0^infty sum_n=0^infty (-1)^n t^s e^-(2n+1)t ; dt \
&= sum_n=0^infty (-1)^n , fracGamma(s+1)(2n+1)^s+1 \
&= Gamma(s+1)L(s+1, chi_4),
endalign*$$
where $L(s, chi_4)$ is the Dirichlet L-function of the unique non-principal character $chi_4$ to the modulus 4. Often it is denoted as $beta(s)$ and called the Dirichlet beta function. Thus differentiating both sides with respect to $s$ and plugging $s = 0$, we obtain a representation formula
$$int_0^1 fraclog log (1/x)1+x^2 ; dx = psi_0(1) beta(1) + beta'(1),$$
and the problem reduces to find the value of $beta(1)$ and $beta'(1)$. Note that $beta(1) = fracpi4$ is just the Gregory series. For $beta'(1)$, we first notice that the following functional equation holds.
$$ beta(s)=left(fracpi2right)^s-1 Gamma(1-s) cos left( fracpi s2 right),beta(1-s). $$
This follows from the general theory of Dirichlet L-functions, and one can consult with any analytic number theory textbook to find its proof. Therefore it is sufficient to calculate $beta'(0)$. For $0 < s$, we have
$$beginalign*
-beta'(s)
&= sum_n=1^infty left[ fraclog(4n+1)(4n+1)^s - fraclog(4n-1)(4n-1)^s right] \
&= sum_n=1^infty frac1(4n)^s left[ log left( frac4n+14n-1 right) - frac12n right] + 2^-2s-1zeta(s+1) \
& qquad + sum_n=1^infty left( frac1(4n+1)^s - frac1(4n)^s right) log (4n+1) \
& qquad + sum_n=1^infty left( frac1(4n)^s - frac1(4n-1)^s right) log (4n-1) \
& =: A(s) + 2^-2s-1zeta(s+1) + B(s) + C(s).
endalign*$$
We first estimate $B(s)$. As $n to infty$, we have
$$ log left( frac4n4n+1 right) = -frac14n + Oleft( frac1n^2 right), quad log left( frac4n4n-1 right) = frac14n + Oleft( frac1n^2 right). $$
Thus when $s to 0$,
$$beginalign*
B(s)
&= sum_n=1^infty frac1(4n)^s left[ expleft( s log left( frac4n4n+1 right) right) - 1 right] left[ log (4n) - log left(frac4n4n+1 right) right] \
&= sum_n=1^infty frac1(4n)^s left[ - fracs4n + O left(fracs^2n^2 right) right] left[ log (4n) + O left(frac1n right) right] \
&= -s 2^-2s-2 sum_n=1^infty frac1n^s+1 log (4n) + O(s) \
&= s 2^-2s-2 left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).
endalign*$$
Similar consideration also shows that
$$ C(s) = s 2^-2s-2 left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).$$
Thus we have
$$ 2^-2s-1zeta(s+1) + B(s) + C(s) = 2^-2s-1 left[ zeta(s+1) + s zeta'(s+1) - s zeta(s+1) log 4 right] + O(s). $$
But since
$$zeta(1+s) = frac1s + gamma + O(s),$$
we have
$$ lim_sdownarrow 0 left( 2^-2s-1zeta(s+1) + B(s) + C(s) right) = fracgamma2 - log 2.$$
For $A(s)$, the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that
$$ lim_sdownarrow 0 A(s) = sum_n=1^infty left[ log left( frac4n+14n-1 right) - frac12n right]. $$
Let $L$ denote this limit. Then by Stirling's formula,
$$beginalign*
e^L
& stackrelNtoinftysim prod_n=1^N left( frac4n+14n-1 right) e^-1/2n
sim frace^-gamma/2sqrtN prod_n=1^N left( fracn+(1/4)n-(1/4) right) \
& sim frace^-gamma/2sqrtN fracGammaleft(frac34right)Gammaleft(frac54right) fracGammaleft(N+frac54right)Gammaleft(N+frac34right)
sim frace^-gamma/2sqrtN fracGammaleft(frac34right)Gammaleft(frac54right) fracleft( fracN + (5/4)e right)^N+frac54left( fracN + (3/4)e right)^N+frac34 \
& sim e^-gamma/2 fracGammaleft(frac34right)Gammaleft(frac54right)
= 4 e^-gamma/2 fracpi sqrt2Gammaleft(frac14right)^2,
endalign*$$
where we have used the Euler's reflection formula in the last line. Combining all the efforts, we obtain
$$-beta'(0) = log (2 pi sqrt2) - 2 log Gammaleft(frac14right) .$$
Now taking logarithmic differntiation to the functional equation, we have
$$ fracbeta'(s)beta(s) = logleft(fracpi2right) - psi_0 (1-s) - fracpi2 tan left( fracpi s2 right) - fracbeta'(1-s)beta(1-s). $$
Taking $s = 0$, we have
$$ fracbeta'(0)beta(0) = logleft(fracpi2right) + gamma - fracbeta'(1)beta(1) quad Longrightarrow quad beta'(1) = beta(1) left[ logleft(fracpi2right) + gamma - fracbeta'(0)beta(0) right]. $$
But again by the functional equation, we have $beta(0) = frac12$. Therefore
$$ beta'(1) = fracpi4 left[ gamma + 2 log 2 + 3 log pi - 4 log Gammaleft(frac14right) right], $$
and hence
$$ int_0^1 fraclog log (1/x)1+x^2 ; dx = fracpi4 left[ 2 log 2 + 3 log pi - 4 log Gammaleft(frac14right) right], $$
which is identical to the proposed answer.
answered Mar 18 '12 at 6:48
Sangchul Lee
85.6k12155253
Excellent answer. I was trying to figure out how to evaluate $beta^'(1)$, but couldn't get very far.
– Eric Naslund
Mar 18 '12 at 12:18
Ahhh, a small typo. Each of the zeta terms should be of the form $zeta(s+1)$, specifically it should be $2^-2s-1zeta(s+1)$ instead of $2^-2s-1zeta(s)$. Otherwise in the line where you take $srightarrow 0$, the pole from $szeta^'(s+1)$ will not cancel out with anything. ($zeta^'(s+1)$ has a double pole)
– Eric Naslund
Mar 18 '12 at 12:29
@EricNaslund, thanks for pointing out typos! I will fix it right now.
– Sangchul Lee
Mar 18 '12 at 12:32
5
@sos440, I just wanted to add, this is one of my favorite proofs I have read so far on Math Stack Exchange.
– Eric Naslund
Apr 14 '12 at 9:33
2
This is nuts, man.
– Pedro Tamaroff♦
Oct 21 '12 at 18:49
 |Â
show 2 more comments
up vote
16
down vote
$newcommand+^dagger
newcommandangles[1]leftlangle #1 rightrangle
newcommandbraces[1]leftlbrace #1 rightrbrace
newcommandbracks[1]leftlbrack #1 rightrbrack
newcommandceil[1],leftlceil #1 rightrceil,
newcommandddrm d
newcommanddowndownarrow
newcommandds[1]displaystyle#1
newcommandequalby[1]#1 atop = atop vphantomhuge A
newcommandexpo[1],rm e^#1,
newcommandfermi,rm f
newcommandfloor[1],leftlfloor #1 rightrfloor,
newcommandhalf1 over 2
newcommandicrm i
newcommandiffLongleftrightarrow
newcommandimpLongrightarrow
newcommandisdiv,left.rightvert,
newcommandket[1]leftvert #1rightrangle
newcommandol[1]overline#1
newcommandpars[1]left( #1 right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandppcal P
newcommandroot[2],sqrt[#1],#2,,
newcommandsech,rm sech
newcommandsgn,rm sgn
newcommandtotald[3]fracrm d^#1 #2rm d #3^#1
newcommandul[1]underline#1
newcommandverts[1]leftvert, #1 ,rightvert$
$dsint_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsrootvphantomlarge A2pi,
Gammapars3/4 over Gammapars1/4: Large ?$
With $dsx to 1/x$:
$$
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
int_infty^1lnparslnparsx,pars-,dd x/x^2 over 1 + 1/x^2
=int_1^inftylnparslnparsx over 1 + x^2,dd x
$$
With $x equiv expotquadiffquad t = lnparsx$:
beginalign
&color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
int_0^inftylnparst over 1 + expo2t,expotdd t
=int_0^inftylnparstexpo-t,1 over 1 + expo-2t,dd t
\[3mm]&=int_0^inftylnparstexpo-t,
sum_ell = 0^inftypars-1^ellexpo-2ell t,dd t
=sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
int_0^inftyt^muexpo-pars2ell + 1t,dd t
\[3mm]&=sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
1 over pars2ell + 1^mu + 1
overbraceint_0^inftyt^muexpo-t,dd t^dsGammaparsmu + 1
endalign
where $Gammaparsz$ is the
Gamma Function.
beginalign
&color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
Gammaparsmu + 1 over pars2ell + 1^mu + 1
\[3mm]&=
lim_mu to 0partialdmubracks%
Gammaparsmu + 1sum_ell = 0^infty
pars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=
lim_mu to 0braces%
Gamma'parsmu + 1sum_ell = 0^infty
pars-1^ellover pars2ell + 1^mu + 1
+Gammaparsmu + 1partialdmubracks%
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=-gammasum_ell = 0^inftypars-1^ellover 2ell + 1
+
lim_mu to 0partialdmu
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1tag1
endalign
In this result, we used $Psipars1 = -gamma$ and $Gammapars1 = 1$ where
$Psiparsz equiv ddlnGammaparsz/dd z$ is the
Digamma Function and $gamma$ is the
Euler-Mascheroni constant.
The first $ell$-sum in the right member of $pars1$ is given by:
beginalign
&sum_ell = 0pars-^ell over 2ell + 1=
sum_ell = 0pars1 over 4ell + 1 - 1 over 4ell + 3
=1 over 8sum_ell = 01 over parsell + 1/4parsell + 3/4
\[3mm]&=-,1 over 4bracksPsipars1 over 4 - Psipars3 over 4
= pi over 4
endalign
where we used the identities:
beginalign
sum_ell = 0^infty1 over parsell + z_0parsell + z_1
&=Psiparsz_0 - Psiparsz_1 over z_0 - z_1tag1.1
\[3mm]Psiparsz - Psipars1 - z &= -picotparspi ztag1.2
endalign
$$
mboxThen,quad
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
-,1 over 4,gammapi + lim_mu to 0partialdmu
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1tag2
$$
Also,
beginalign
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
&=sum_ell = 0^infty1 over bracks2pars2ell + 1^mu + 1
-sum_ell = 0^infty1 over bracks2pars2ell + 1 + 1^mu + 1
\[3mm]&=2^-2mu - 2bracks%
sum_ell = 0^infty1 over parsell + 1/4^mu + 1
-sum_ell = 0^infty1 over parsell + 3/4^mu + 1
\[3mm]&=2^-2mu - 2bracks%
zetaparsmu + 1,1 over 4 - zetaparsmu + 1,3 over 4
endalign
where $dszetaparss,q equiv sum_n = 0^infty1 over parsq + n^s$
with $Reparss > 1$ and $Reparsq > 0$ i s the
Hurwitz Zeta Function or/and Generalizated Zeta Funcion .
So,
beginalign
&lim_mu to 0partialdmusum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=-,1 over 4,lnpars2
underbraceoverbracesum_ell = 0^infty1 over parsell + 3/4parsell + 1/4
^ds2bracksPsipars3/4 - Psipars1/4 = 2pi
_dsmboxSee pars1.1 mboxand pars1.2
+ 1 over 4
overbracepartialdmubracks%
zetaparsmu,1 over 4 - zetaparsmu,3 over 4_mu = 1
^ds-gamma_1pars1/4 + gamma_1pars3/4
endalign
where $gamma_nparsz$ is a
Generalizated Stieltjes Constant
.
With this result, $pars2$ is reduced to:
beginalign
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2
&=-,1 over 4,braces%
pibracks%
gamma + 2lnpars2 + gamma_1pars1 over 4 - gamma_1pars3 over 4
tag3
endalign
The difference $gamma_1pars1/4 - gamma_1pars3/4$ is evaluated with the 1846 Carl Malmsten identity
:
$$
gamma_1parsm over n - gamma_1pars1 - m over n
=-pibracksgamma + lnpars2pi ncotparsmpi over n
+ 2pisum_ell = 1^n - 1
sinpars2pi m over n,elllnparsGammaparsell over n
$$
With $m = 1$ and $n = 4$:
beginalign
&gamma_1pars1 over 4 - gamma_1pars3 over 4
\[3mm]&=-pibracksgamma + lnpars8picotparspi over 4
+ 2pisum_ell = 1^3sinparspi,ell over 2
lnparsGammaparsell over 4
\[3mm]&=-pibracksgamma + 2lnpars2 + lnpars2pi
\[3mm]&phantom=
+ 2pibrackssinparspi over 2lnparsGammapars1 over 4
+ sinparspilnparsGammapars1 over 2
+ sinpars3pi over 2Gammapars3 over 4
\[3mm]&=-pibracksgamma + 2lnpars2 + lnpars2pi
+2pibrackslnparsGammapars1 over 4 - lnparsGammapars3 over 4
\[3mm]&=-pibracksgamma + 2lnpars2 - 2pibracks%
lnparsroot2pi +lnparsGammapars3 over 4 over Gammapars1 over 4
\[3mm]&=-pibracksgamma + 2lnpars2
-2pilnparsroot2pi,Gammapars3 over 4 over Gammapars1 over 4
endalign
By replacing this result in $pars3$, we find:
$$color#00flarge%
int_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsrootvphantomlarge A2pi,
Gammapars3/4 over Gammapars1/4
$$
As an 'extra-bonus' we can use the identity
$dsGammaparsz = pi over Gammapars1 - zsinparspi z$ to 'kill' one of the $Gamma,$'s functions:
$dsGammapars1 over 4 = root2pi over Gammapars3/4$ which yields:
$$
int_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsGamma^,2pars3/4 over rootpi
$$
answered Mar 2 '14 at 0:28
Felix Marin
65.6k7105135
add a comment |Â
up vote
5
down vote
See also V. Adamchik's formula $$int_0^1 fracx^p-11+x^nlog log frac1xdx = fracgamma+log(2n)2n(psi(fracp2n)-psi(fracn+p2n))+frac12n(zeta'(1,fracp2n)-zeta'(1,fracn+p2n))$$ in http://dx.doi.org/10.1145/258726.258736 .
answered Mar 1 '14 at 18:40
R. J. Mathar
6912
add a comment |Â
up vote
4
down vote
The above integral was long-time known as the Vardi's integral. Quite recently, an interesting research work devoted to this integral was published by Iaroslav Blagouchine. It appears that this integral was first evaluated by Carl Malmsten in 1842 (and not by Ilan Vardi in 1988). Blagouchine describes two different methods for its evaluation: the original Malmsten's method and the contour integration method. In the same paper, numerous integrals similar to the above one are also treated, and it is shown that many of them may be evaluated by the contour integration methods.
More recent paper by Blagouchine (mentioned above in the context of Malmsten's formula derived in 1846) is also integersting, but it is less directly related to the integral in question than the first one.
edited Nov 29 '14 at 2:42
Iaroslav Blagouchine
32727
answered Jun 24 '14 at 21:50
Mark
812
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
43
down vote
accepted
By the substitution $x = e^-t$, we find that
$$beginalign*
int_0^1 frac(log (1/x))^s1+x^2 ; dx
&= int_0^infty fract^s e^-t1 + e^-2t ; dt \
&= int_0^infty sum_n=0^infty (-1)^n t^s e^-(2n+1)t ; dt \
&= sum_n=0^infty (-1)^n , fracGamma(s+1)(2n+1)^s+1 \
&= Gamma(s+1)L(s+1, chi_4),
endalign*$$
where $L(s, chi_4)$ is the Dirichlet L-function of the unique non-principal character $chi_4$ to the modulus 4. Often it is denoted as $beta(s)$ and called the Dirichlet beta function. Thus differentiating both sides with respect to $s$ and plugging $s = 0$, we obtain a representation formula
$$int_0^1 fraclog log (1/x)1+x^2 ; dx = psi_0(1) beta(1) + beta'(1),$$
and the problem reduces to find the value of $beta(1)$ and $beta'(1)$. Note that $beta(1) = fracpi4$ is just the Gregory series. For $beta'(1)$, we first notice that the following functional equation holds.
$$ beta(s)=left(fracpi2right)^s-1 Gamma(1-s) cos left( fracpi s2 right),beta(1-s). $$
This follows from the general theory of Dirichlet L-functions, and one can consult with any analytic number theory textbook to find its proof. Therefore it is sufficient to calculate $beta'(0)$. For $0 < s$, we have
$$beginalign*
-beta'(s)
&= sum_n=1^infty left[ fraclog(4n+1)(4n+1)^s - fraclog(4n-1)(4n-1)^s right] \
&= sum_n=1^infty frac1(4n)^s left[ log left( frac4n+14n-1 right) - frac12n right] + 2^-2s-1zeta(s+1) \
& qquad + sum_n=1^infty left( frac1(4n+1)^s - frac1(4n)^s right) log (4n+1) \
& qquad + sum_n=1^infty left( frac1(4n)^s - frac1(4n-1)^s right) log (4n-1) \
& =: A(s) + 2^-2s-1zeta(s+1) + B(s) + C(s).
endalign*$$
We first estimate $B(s)$. As $n to infty$, we have
$$ log left( frac4n4n+1 right) = -frac14n + Oleft( frac1n^2 right), quad log left( frac4n4n-1 right) = frac14n + Oleft( frac1n^2 right). $$
Thus when $s to 0$,
$$beginalign*
B(s)
&= sum_n=1^infty frac1(4n)^s left[ expleft( s log left( frac4n4n+1 right) right) - 1 right] left[ log (4n) - log left(frac4n4n+1 right) right] \
&= sum_n=1^infty frac1(4n)^s left[ - fracs4n + O left(fracs^2n^2 right) right] left[ log (4n) + O left(frac1n right) right] \
&= -s 2^-2s-2 sum_n=1^infty frac1n^s+1 log (4n) + O(s) \
&= s 2^-2s-2 left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).
endalign*$$
Similar consideration also shows that
$$ C(s) = s 2^-2s-2 left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).$$
Thus we have
$$ 2^-2s-1zeta(s+1) + B(s) + C(s) = 2^-2s-1 left[ zeta(s+1) + s zeta'(s+1) - s zeta(s+1) log 4 right] + O(s). $$
But since
$$zeta(1+s) = frac1s + gamma + O(s),$$
we have
$$ lim_sdownarrow 0 left( 2^-2s-1zeta(s+1) + B(s) + C(s) right) = fracgamma2 - log 2.$$
For $A(s)$, the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that
$$ lim_sdownarrow 0 A(s) = sum_n=1^infty left[ log left( frac4n+14n-1 right) - frac12n right]. $$
Let $L$ denote this limit. Then by Stirling's formula,
$$beginalign*
e^L
& stackrelNtoinftysim prod_n=1^N left( frac4n+14n-1 right) e^-1/2n
sim frace^-gamma/2sqrtN prod_n=1^N left( fracn+(1/4)n-(1/4) right) \
& sim frace^-gamma/2sqrtN fracGammaleft(frac34right)Gammaleft(frac54right) fracGammaleft(N+frac54right)Gammaleft(N+frac34right)
sim frace^-gamma/2sqrtN fracGammaleft(frac34right)Gammaleft(frac54right) fracleft( fracN + (5/4)e right)^N+frac54left( fracN + (3/4)e right)^N+frac34 \
& sim e^-gamma/2 fracGammaleft(frac34right)Gammaleft(frac54right)
= 4 e^-gamma/2 fracpi sqrt2Gammaleft(frac14right)^2,
endalign*$$
where we have used the Euler's reflection formula in the last line. Combining all the efforts, we obtain
$$-beta'(0) = log (2 pi sqrt2) - 2 log Gammaleft(frac14right) .$$
Now taking logarithmic differntiation to the functional equation, we have
$$ fracbeta'(s)beta(s) = logleft(fracpi2right) - psi_0 (1-s) - fracpi2 tan left( fracpi s2 right) - fracbeta'(1-s)beta(1-s). $$
Taking $s = 0$, we have
$$ fracbeta'(0)beta(0) = logleft(fracpi2right) + gamma - fracbeta'(1)beta(1) quad Longrightarrow quad beta'(1) = beta(1) left[ logleft(fracpi2right) + gamma - fracbeta'(0)beta(0) right]. $$
But again by the functional equation, we have $beta(0) = frac12$. Therefore
$$ beta'(1) = fracpi4 left[ gamma + 2 log 2 + 3 log pi - 4 log Gammaleft(frac14right) right], $$
and hence
$$ int_0^1 fraclog log (1/x)1+x^2 ; dx = fracpi4 left[ 2 log 2 + 3 log pi - 4 log Gammaleft(frac14right) right], $$
which is identical to the proposed answer.
answered Mar 18 '12 at 6:48
Sangchul Lee
85.6k12155253
Excellent answer. I was trying to figure out how to evaluate $beta^'(1)$, but couldn't get very far.
– Eric Naslund
Mar 18 '12 at 12:18
Ahhh, a small typo. Each of the zeta terms should be of the form $zeta(s+1)$, specifically it should be $2^-2s-1zeta(s+1)$ instead of $2^-2s-1zeta(s)$. Otherwise in the line where you take $srightarrow 0$, the pole from $szeta^'(s+1)$ will not cancel out with anything. ($zeta^'(s+1)$ has a double pole)
– Eric Naslund
Mar 18 '12 at 12:29
@EricNaslund, thanks for pointing out typos! I will fix it right now.
– Sangchul Lee
Mar 18 '12 at 12:32
5
@sos440, I just wanted to add, this is one of my favorite proofs I have read so far on Math Stack Exchange.
– Eric Naslund
Apr 14 '12 at 9:33
2
This is nuts, man.
– Pedro Tamaroff♦
Oct 21 '12 at 18:49
 |Â
show 2 more comments
up vote
43
down vote
accepted
By the substitution $x = e^-t$, we find that
$$beginalign*
int_0^1 frac(log (1/x))^s1+x^2 ; dx
&= int_0^infty fract^s e^-t1 + e^-2t ; dt \
&= int_0^infty sum_n=0^infty (-1)^n t^s e^-(2n+1)t ; dt \
&= sum_n=0^infty (-1)^n , fracGamma(s+1)(2n+1)^s+1 \
&= Gamma(s+1)L(s+1, chi_4),
endalign*$$
where $L(s, chi_4)$ is the Dirichlet L-function of the unique non-principal character $chi_4$ to the modulus 4. Often it is denoted as $beta(s)$ and called the Dirichlet beta function. Thus differentiating both sides with respect to $s$ and plugging $s = 0$, we obtain a representation formula
$$int_0^1 fraclog log (1/x)1+x^2 ; dx = psi_0(1) beta(1) + beta'(1),$$
and the problem reduces to find the value of $beta(1)$ and $beta'(1)$. Note that $beta(1) = fracpi4$ is just the Gregory series. For $beta'(1)$, we first notice that the following functional equation holds.
$$ beta(s)=left(fracpi2right)^s-1 Gamma(1-s) cos left( fracpi s2 right),beta(1-s). $$
This follows from the general theory of Dirichlet L-functions, and one can consult with any analytic number theory textbook to find its proof. Therefore it is sufficient to calculate $beta'(0)$. For $0 < s$, we have
$$beginalign*
-beta'(s)
&= sum_n=1^infty left[ fraclog(4n+1)(4n+1)^s - fraclog(4n-1)(4n-1)^s right] \
&= sum_n=1^infty frac1(4n)^s left[ log left( frac4n+14n-1 right) - frac12n right] + 2^-2s-1zeta(s+1) \
& qquad + sum_n=1^infty left( frac1(4n+1)^s - frac1(4n)^s right) log (4n+1) \
& qquad + sum_n=1^infty left( frac1(4n)^s - frac1(4n-1)^s right) log (4n-1) \
& =: A(s) + 2^-2s-1zeta(s+1) + B(s) + C(s).
endalign*$$
We first estimate $B(s)$. As $n to infty$, we have
$$ log left( frac4n4n+1 right) = -frac14n + Oleft( frac1n^2 right), quad log left( frac4n4n-1 right) = frac14n + Oleft( frac1n^2 right). $$
Thus when $s to 0$,
$$beginalign*
B(s)
&= sum_n=1^infty frac1(4n)^s left[ expleft( s log left( frac4n4n+1 right) right) - 1 right] left[ log (4n) - log left(frac4n4n+1 right) right] \
&= sum_n=1^infty frac1(4n)^s left[ - fracs4n + O left(fracs^2n^2 right) right] left[ log (4n) + O left(frac1n right) right] \
&= -s 2^-2s-2 sum_n=1^infty frac1n^s+1 log (4n) + O(s) \
&= s 2^-2s-2 left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).
endalign*$$
Similar consideration also shows that
$$ C(s) = s 2^-2s-2 left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).$$
Thus we have
$$ 2^-2s-1zeta(s+1) + B(s) + C(s) = 2^-2s-1 left[ zeta(s+1) + s zeta'(s+1) - s zeta(s+1) log 4 right] + O(s). $$
But since
$$zeta(1+s) = frac1s + gamma + O(s),$$
we have
$$ lim_sdownarrow 0 left( 2^-2s-1zeta(s+1) + B(s) + C(s) right) = fracgamma2 - log 2.$$
For $A(s)$, the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that
$$ lim_sdownarrow 0 A(s) = sum_n=1^infty left[ log left( frac4n+14n-1 right) - frac12n right]. $$
Let $L$ denote this limit. Then by Stirling's formula,
$$beginalign*
e^L
& stackrelNtoinftysim prod_n=1^N left( frac4n+14n-1 right) e^-1/2n
sim frace^-gamma/2sqrtN prod_n=1^N left( fracn+(1/4)n-(1/4) right) \
& sim frace^-gamma/2sqrtN fracGammaleft(frac34right)Gammaleft(frac54right) fracGammaleft(N+frac54right)Gammaleft(N+frac34right)
sim frace^-gamma/2sqrtN fracGammaleft(frac34right)Gammaleft(frac54right) fracleft( fracN + (5/4)e right)^N+frac54left( fracN + (3/4)e right)^N+frac34 \
& sim e^-gamma/2 fracGammaleft(frac34right)Gammaleft(frac54right)
= 4 e^-gamma/2 fracpi sqrt2Gammaleft(frac14right)^2,
endalign*$$
where we have used the Euler's reflection formula in the last line. Combining all the efforts, we obtain
$$-beta'(0) = log (2 pi sqrt2) - 2 log Gammaleft(frac14right) .$$
Now taking logarithmic differntiation to the functional equation, we have
$$ fracbeta'(s)beta(s) = logleft(fracpi2right) - psi_0 (1-s) - fracpi2 tan left( fracpi s2 right) - fracbeta'(1-s)beta(1-s). $$
Taking $s = 0$, we have
$$ fracbeta'(0)beta(0) = logleft(fracpi2right) + gamma - fracbeta'(1)beta(1) quad Longrightarrow quad beta'(1) = beta(1) left[ logleft(fracpi2right) + gamma - fracbeta'(0)beta(0) right]. $$
But again by the functional equation, we have $beta(0) = frac12$. Therefore
$$ beta'(1) = fracpi4 left[ gamma + 2 log 2 + 3 log pi - 4 log Gammaleft(frac14right) right], $$
and hence
$$ int_0^1 fraclog log (1/x)1+x^2 ; dx = fracpi4 left[ 2 log 2 + 3 log pi - 4 log Gammaleft(frac14right) right], $$
which is identical to the proposed answer.
answered Mar 18 '12 at 6:48
Sangchul Lee
85.6k12155253
Excellent answer. I was trying to figure out how to evaluate $beta^'(1)$, but couldn't get very far.
– Eric Naslund
Mar 18 '12 at 12:18
Ahhh, a small typo. Each of the zeta terms should be of the form $zeta(s+1)$, specifically it should be $2^-2s-1zeta(s+1)$ instead of $2^-2s-1zeta(s)$. Otherwise in the line where you take $srightarrow 0$, the pole from $szeta^'(s+1)$ will not cancel out with anything. ($zeta^'(s+1)$ has a double pole)
– Eric Naslund
Mar 18 '12 at 12:29
@EricNaslund, thanks for pointing out typos! I will fix it right now.
– Sangchul Lee
Mar 18 '12 at 12:32
5
@sos440, I just wanted to add, this is one of my favorite proofs I have read so far on Math Stack Exchange.
– Eric Naslund
Apr 14 '12 at 9:33
2
This is nuts, man.
– Pedro Tamaroff♦
Oct 21 '12 at 18:49
 |Â
show 2 more comments
up vote
43
down vote
accepted
up vote
43
down vote
accepted
By the substitution $x = e^-t$, we find that
$$beginalign*
int_0^1 frac(log (1/x))^s1+x^2 ; dx
&= int_0^infty fract^s e^-t1 + e^-2t ; dt \
&= int_0^infty sum_n=0^infty (-1)^n t^s e^-(2n+1)t ; dt \
&= sum_n=0^infty (-1)^n , fracGamma(s+1)(2n+1)^s+1 \
&= Gamma(s+1)L(s+1, chi_4),
endalign*$$
where $L(s, chi_4)$ is the Dirichlet L-function of the unique non-principal character $chi_4$ to the modulus 4. Often it is denoted as $beta(s)$ and called the Dirichlet beta function. Thus differentiating both sides with respect to $s$ and plugging $s = 0$, we obtain a representation formula
$$int_0^1 fraclog log (1/x)1+x^2 ; dx = psi_0(1) beta(1) + beta'(1),$$
and the problem reduces to find the value of $beta(1)$ and $beta'(1)$. Note that $beta(1) = fracpi4$ is just the Gregory series. For $beta'(1)$, we first notice that the following functional equation holds.
$$ beta(s)=left(fracpi2right)^s-1 Gamma(1-s) cos left( fracpi s2 right),beta(1-s). $$
This follows from the general theory of Dirichlet L-functions, and one can consult with any analytic number theory textbook to find its proof. Therefore it is sufficient to calculate $beta'(0)$. For $0 < s$, we have
$$beginalign*
-beta'(s)
&= sum_n=1^infty left[ fraclog(4n+1)(4n+1)^s - fraclog(4n-1)(4n-1)^s right] \
&= sum_n=1^infty frac1(4n)^s left[ log left( frac4n+14n-1 right) - frac12n right] + 2^-2s-1zeta(s+1) \
& qquad + sum_n=1^infty left( frac1(4n+1)^s - frac1(4n)^s right) log (4n+1) \
& qquad + sum_n=1^infty left( frac1(4n)^s - frac1(4n-1)^s right) log (4n-1) \
& =: A(s) + 2^-2s-1zeta(s+1) + B(s) + C(s).
endalign*$$
We first estimate $B(s)$. As $n to infty$, we have
$$ log left( frac4n4n+1 right) = -frac14n + Oleft( frac1n^2 right), quad log left( frac4n4n-1 right) = frac14n + Oleft( frac1n^2 right). $$
Thus when $s to 0$,
$$beginalign*
B(s)
&= sum_n=1^infty frac1(4n)^s left[ expleft( s log left( frac4n4n+1 right) right) - 1 right] left[ log (4n) - log left(frac4n4n+1 right) right] \
&= sum_n=1^infty frac1(4n)^s left[ - fracs4n + O left(fracs^2n^2 right) right] left[ log (4n) + O left(frac1n right) right] \
&= -s 2^-2s-2 sum_n=1^infty frac1n^s+1 log (4n) + O(s) \
&= s 2^-2s-2 left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).
endalign*$$
Similar consideration also shows that
$$ C(s) = s 2^-2s-2 left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).$$
Thus we have
$$ 2^-2s-1zeta(s+1) + B(s) + C(s) = 2^-2s-1 left[ zeta(s+1) + s zeta'(s+1) - s zeta(s+1) log 4 right] + O(s). $$
But since
$$zeta(1+s) = frac1s + gamma + O(s),$$
we have
$$ lim_sdownarrow 0 left( 2^-2s-1zeta(s+1) + B(s) + C(s) right) = fracgamma2 - log 2.$$
For $A(s)$, the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that
$$ lim_sdownarrow 0 A(s) = sum_n=1^infty left[ log left( frac4n+14n-1 right) - frac12n right]. $$
Let $L$ denote this limit. Then by Stirling's formula,
$$beginalign*
e^L
& stackrelNtoinftysim prod_n=1^N left( frac4n+14n-1 right) e^-1/2n
sim frace^-gamma/2sqrtN prod_n=1^N left( fracn+(1/4)n-(1/4) right) \
& sim frace^-gamma/2sqrtN fracGammaleft(frac34right)Gammaleft(frac54right) fracGammaleft(N+frac54right)Gammaleft(N+frac34right)
sim frace^-gamma/2sqrtN fracGammaleft(frac34right)Gammaleft(frac54right) fracleft( fracN + (5/4)e right)^N+frac54left( fracN + (3/4)e right)^N+frac34 \
& sim e^-gamma/2 fracGammaleft(frac34right)Gammaleft(frac54right)
= 4 e^-gamma/2 fracpi sqrt2Gammaleft(frac14right)^2,
endalign*$$
where we have used the Euler's reflection formula in the last line. Combining all the efforts, we obtain
$$-beta'(0) = log (2 pi sqrt2) - 2 log Gammaleft(frac14right) .$$
Now taking logarithmic differntiation to the functional equation, we have
$$ fracbeta'(s)beta(s) = logleft(fracpi2right) - psi_0 (1-s) - fracpi2 tan left( fracpi s2 right) - fracbeta'(1-s)beta(1-s). $$
Taking $s = 0$, we have
$$ fracbeta'(0)beta(0) = logleft(fracpi2right) + gamma - fracbeta'(1)beta(1) quad Longrightarrow quad beta'(1) = beta(1) left[ logleft(fracpi2right) + gamma - fracbeta'(0)beta(0) right]. $$
But again by the functional equation, we have $beta(0) = frac12$. Therefore
$$ beta'(1) = fracpi4 left[ gamma + 2 log 2 + 3 log pi - 4 log Gammaleft(frac14right) right], $$
and hence
$$ int_0^1 fraclog log (1/x)1+x^2 ; dx = fracpi4 left[ 2 log 2 + 3 log pi - 4 log Gammaleft(frac14right) right], $$
which is identical to the proposed answer.
answered Mar 18 '12 at 6:48
Sangchul Lee
85.6k12155253
By the substitution $x = e^-t$, we find that
$$beginalign*
int_0^1 frac(log (1/x))^s1+x^2 ; dx
&= int_0^infty fract^s e^-t1 + e^-2t ; dt \
&= int_0^infty sum_n=0^infty (-1)^n t^s e^-(2n+1)t ; dt \
&= sum_n=0^infty (-1)^n , fracGamma(s+1)(2n+1)^s+1 \
&= Gamma(s+1)L(s+1, chi_4),
endalign*$$
where $L(s, chi_4)$ is the Dirichlet L-function of the unique non-principal character $chi_4$ to the modulus 4. Often it is denoted as $beta(s)$ and called the Dirichlet beta function. Thus differentiating both sides with respect to $s$ and plugging $s = 0$, we obtain a representation formula
$$int_0^1 fraclog log (1/x)1+x^2 ; dx = psi_0(1) beta(1) + beta'(1),$$
and the problem reduces to find the value of $beta(1)$ and $beta'(1)$. Note that $beta(1) = fracpi4$ is just the Gregory series. For $beta'(1)$, we first notice that the following functional equation holds.
$$ beta(s)=left(fracpi2right)^s-1 Gamma(1-s) cos left( fracpi s2 right),beta(1-s). $$
This follows from the general theory of Dirichlet L-functions, and one can consult with any analytic number theory textbook to find its proof. Therefore it is sufficient to calculate $beta'(0)$. For $0 < s$, we have
$$beginalign*
-beta'(s)
&= sum_n=1^infty left[ fraclog(4n+1)(4n+1)^s - fraclog(4n-1)(4n-1)^s right] \
&= sum_n=1^infty frac1(4n)^s left[ log left( frac4n+14n-1 right) - frac12n right] + 2^-2s-1zeta(s+1) \
& qquad + sum_n=1^infty left( frac1(4n+1)^s - frac1(4n)^s right) log (4n+1) \
& qquad + sum_n=1^infty left( frac1(4n)^s - frac1(4n-1)^s right) log (4n-1) \
& =: A(s) + 2^-2s-1zeta(s+1) + B(s) + C(s).
endalign*$$
We first estimate $B(s)$. As $n to infty$, we have
$$ log left( frac4n4n+1 right) = -frac14n + Oleft( frac1n^2 right), quad log left( frac4n4n-1 right) = frac14n + Oleft( frac1n^2 right). $$
Thus when $s to 0$,
$$beginalign*
B(s)
&= sum_n=1^infty frac1(4n)^s left[ expleft( s log left( frac4n4n+1 right) right) - 1 right] left[ log (4n) - log left(frac4n4n+1 right) right] \
&= sum_n=1^infty frac1(4n)^s left[ - fracs4n + O left(fracs^2n^2 right) right] left[ log (4n) + O left(frac1n right) right] \
&= -s 2^-2s-2 sum_n=1^infty frac1n^s+1 log (4n) + O(s) \
&= s 2^-2s-2 left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).
endalign*$$
Similar consideration also shows that
$$ C(s) = s 2^-2s-2 left[ zeta'(s+1) - zeta(s+1) log 4 right] + O(s).$$
Thus we have
$$ 2^-2s-1zeta(s+1) + B(s) + C(s) = 2^-2s-1 left[ zeta(s+1) + s zeta'(s+1) - s zeta(s+1) log 4 right] + O(s). $$
But since
$$zeta(1+s) = frac1s + gamma + O(s),$$
we have
$$ lim_sdownarrow 0 left( 2^-2s-1zeta(s+1) + B(s) + C(s) right) = fracgamma2 - log 2.$$
For $A(s)$, the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that
$$ lim_sdownarrow 0 A(s) = sum_n=1^infty left[ log left( frac4n+14n-1 right) - frac12n right]. $$
Let $L$ denote this limit. Then by Stirling's formula,
$$beginalign*
e^L
& stackrelNtoinftysim prod_n=1^N left( frac4n+14n-1 right) e^-1/2n
sim frace^-gamma/2sqrtN prod_n=1^N left( fracn+(1/4)n-(1/4) right) \
& sim frace^-gamma/2sqrtN fracGammaleft(frac34right)Gammaleft(frac54right) fracGammaleft(N+frac54right)Gammaleft(N+frac34right)
sim frace^-gamma/2sqrtN fracGammaleft(frac34right)Gammaleft(frac54right) fracleft( fracN + (5/4)e right)^N+frac54left( fracN + (3/4)e right)^N+frac34 \
& sim e^-gamma/2 fracGammaleft(frac34right)Gammaleft(frac54right)
= 4 e^-gamma/2 fracpi sqrt2Gammaleft(frac14right)^2,
endalign*$$
where we have used the Euler's reflection formula in the last line. Combining all the efforts, we obtain
$$-beta'(0) = log (2 pi sqrt2) - 2 log Gammaleft(frac14right) .$$
Now taking logarithmic differntiation to the functional equation, we have
$$ fracbeta'(s)beta(s) = logleft(fracpi2right) - psi_0 (1-s) - fracpi2 tan left( fracpi s2 right) - fracbeta'(1-s)beta(1-s). $$
Taking $s = 0$, we have
$$ fracbeta'(0)beta(0) = logleft(fracpi2right) + gamma - fracbeta'(1)beta(1) quad Longrightarrow quad beta'(1) = beta(1) left[ logleft(fracpi2right) + gamma - fracbeta'(0)beta(0) right]. $$
But again by the functional equation, we have $beta(0) = frac12$. Therefore
$$ beta'(1) = fracpi4 left[ gamma + 2 log 2 + 3 log pi - 4 log Gammaleft(frac14right) right], $$
and hence
$$ int_0^1 fraclog log (1/x)1+x^2 ; dx = fracpi4 left[ 2 log 2 + 3 log pi - 4 log Gammaleft(frac14right) right], $$
which is identical to the proposed answer.
answered Mar 18 '12 at 6:48
Sangchul Lee
85.6k12155253
edited Mar 18 '12 at 13:33
answered Mar 18 '12 at 6:48
Sangchul Lee
85.6k12155253
answered Mar 18 '12 at 6:48
Sangchul Lee
85.6k12155253
answered Mar 18 '12 at 6:48
Sangchul Lee
85.6k12155253
85.6k12155253
Excellent answer. I was trying to figure out how to evaluate $beta^'(1)$, but couldn't get very far.
– Eric Naslund
Mar 18 '12 at 12:18
Ahhh, a small typo. Each of the zeta terms should be of the form $zeta(s+1)$, specifically it should be $2^-2s-1zeta(s+1)$ instead of $2^-2s-1zeta(s)$. Otherwise in the line where you take $srightarrow 0$, the pole from $szeta^'(s+1)$ will not cancel out with anything. ($zeta^'(s+1)$ has a double pole)
– Eric Naslund
Mar 18 '12 at 12:29
@EricNaslund, thanks for pointing out typos! I will fix it right now.
– Sangchul Lee
Mar 18 '12 at 12:32
5
@sos440, I just wanted to add, this is one of my favorite proofs I have read so far on Math Stack Exchange.
– Eric Naslund
Apr 14 '12 at 9:33
2
This is nuts, man.
– Pedro Tamaroff♦
Oct 21 '12 at 18:49
 |Â
show 2 more comments
Excellent answer. I was trying to figure out how to evaluate $beta^'(1)$, but couldn't get very far.
– Eric Naslund
Mar 18 '12 at 12:18
Ahhh, a small typo. Each of the zeta terms should be of the form $zeta(s+1)$, specifically it should be $2^-2s-1zeta(s+1)$ instead of $2^-2s-1zeta(s)$. Otherwise in the line where you take $srightarrow 0$, the pole from $szeta^'(s+1)$ will not cancel out with anything. ($zeta^'(s+1)$ has a double pole)
– Eric Naslund
Mar 18 '12 at 12:29
@EricNaslund, thanks for pointing out typos! I will fix it right now.
– Sangchul Lee
Mar 18 '12 at 12:32
5
@sos440, I just wanted to add, this is one of my favorite proofs I have read so far on Math Stack Exchange.
– Eric Naslund
Apr 14 '12 at 9:33
2
This is nuts, man.
– Pedro Tamaroff♦
Oct 21 '12 at 18:49
Excellent answer. I was trying to figure out how to evaluate $beta^'(1)$, but couldn't get very far.
– Eric Naslund
Mar 18 '12 at 12:18
Excellent answer. I was trying to figure out how to evaluate $beta^'(1)$, but couldn't get very far.
– Eric Naslund
Mar 18 '12 at 12:18
Ahhh, a small typo. Each of the zeta terms should be of the form $zeta(s+1)$, specifically it should be $2^-2s-1zeta(s+1)$ instead of $2^-2s-1zeta(s)$. Otherwise in the line where you take $srightarrow 0$, the pole from $szeta^'(s+1)$ will not cancel out with anything. ($zeta^'(s+1)$ has a double pole)
– Eric Naslund
Mar 18 '12 at 12:29
Ahhh, a small typo. Each of the zeta terms should be of the form $zeta(s+1)$, specifically it should be $2^-2s-1zeta(s+1)$ instead of $2^-2s-1zeta(s)$. Otherwise in the line where you take $srightarrow 0$, the pole from $szeta^'(s+1)$ will not cancel out with anything. ($zeta^'(s+1)$ has a double pole)
– Eric Naslund
Mar 18 '12 at 12:29
@EricNaslund, thanks for pointing out typos! I will fix it right now.
– Sangchul Lee
Mar 18 '12 at 12:32
@EricNaslund, thanks for pointing out typos! I will fix it right now.
– Sangchul Lee
Mar 18 '12 at 12:32
5
5
@sos440, I just wanted to add, this is one of my favorite proofs I have read so far on Math Stack Exchange.
– Eric Naslund
Apr 14 '12 at 9:33
@sos440, I just wanted to add, this is one of my favorite proofs I have read so far on Math Stack Exchange.
– Eric Naslund
Apr 14 '12 at 9:33
2
2
This is nuts, man.
– Pedro Tamaroff♦
Oct 21 '12 at 18:49
This is nuts, man.
– Pedro Tamaroff♦
Oct 21 '12 at 18:49
 |Â
show 2 more comments
up vote
16
down vote
$newcommand+^dagger
newcommandangles[1]leftlangle #1 rightrangle
newcommandbraces[1]leftlbrace #1 rightrbrace
newcommandbracks[1]leftlbrack #1 rightrbrack
newcommandceil[1],leftlceil #1 rightrceil,
newcommandddrm d
newcommanddowndownarrow
newcommandds[1]displaystyle#1
newcommandequalby[1]#1 atop = atop vphantomhuge A
newcommandexpo[1],rm e^#1,
newcommandfermi,rm f
newcommandfloor[1],leftlfloor #1 rightrfloor,
newcommandhalf1 over 2
newcommandicrm i
newcommandiffLongleftrightarrow
newcommandimpLongrightarrow
newcommandisdiv,left.rightvert,
newcommandket[1]leftvert #1rightrangle
newcommandol[1]overline#1
newcommandpars[1]left( #1 right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandppcal P
newcommandroot[2],sqrt[#1],#2,,
newcommandsech,rm sech
newcommandsgn,rm sgn
newcommandtotald[3]fracrm d^#1 #2rm d #3^#1
newcommandul[1]underline#1
newcommandverts[1]leftvert, #1 ,rightvert$
$dsint_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsrootvphantomlarge A2pi,
Gammapars3/4 over Gammapars1/4: Large ?$
With $dsx to 1/x$:
$$
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
int_infty^1lnparslnparsx,pars-,dd x/x^2 over 1 + 1/x^2
=int_1^inftylnparslnparsx over 1 + x^2,dd x
$$
With $x equiv expotquadiffquad t = lnparsx$:
beginalign
&color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
int_0^inftylnparst over 1 + expo2t,expotdd t
=int_0^inftylnparstexpo-t,1 over 1 + expo-2t,dd t
\[3mm]&=int_0^inftylnparstexpo-t,
sum_ell = 0^inftypars-1^ellexpo-2ell t,dd t
=sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
int_0^inftyt^muexpo-pars2ell + 1t,dd t
\[3mm]&=sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
1 over pars2ell + 1^mu + 1
overbraceint_0^inftyt^muexpo-t,dd t^dsGammaparsmu + 1
endalign
where $Gammaparsz$ is the
Gamma Function.
beginalign
&color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
Gammaparsmu + 1 over pars2ell + 1^mu + 1
\[3mm]&=
lim_mu to 0partialdmubracks%
Gammaparsmu + 1sum_ell = 0^infty
pars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=
lim_mu to 0braces%
Gamma'parsmu + 1sum_ell = 0^infty
pars-1^ellover pars2ell + 1^mu + 1
+Gammaparsmu + 1partialdmubracks%
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=-gammasum_ell = 0^inftypars-1^ellover 2ell + 1
+
lim_mu to 0partialdmu
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1tag1
endalign
In this result, we used $Psipars1 = -gamma$ and $Gammapars1 = 1$ where
$Psiparsz equiv ddlnGammaparsz/dd z$ is the
Digamma Function and $gamma$ is the
Euler-Mascheroni constant.
The first $ell$-sum in the right member of $pars1$ is given by:
beginalign
&sum_ell = 0pars-^ell over 2ell + 1=
sum_ell = 0pars1 over 4ell + 1 - 1 over 4ell + 3
=1 over 8sum_ell = 01 over parsell + 1/4parsell + 3/4
\[3mm]&=-,1 over 4bracksPsipars1 over 4 - Psipars3 over 4
= pi over 4
endalign
where we used the identities:
beginalign
sum_ell = 0^infty1 over parsell + z_0parsell + z_1
&=Psiparsz_0 - Psiparsz_1 over z_0 - z_1tag1.1
\[3mm]Psiparsz - Psipars1 - z &= -picotparspi ztag1.2
endalign
$$
mboxThen,quad
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
-,1 over 4,gammapi + lim_mu to 0partialdmu
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1tag2
$$
Also,
beginalign
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
&=sum_ell = 0^infty1 over bracks2pars2ell + 1^mu + 1
-sum_ell = 0^infty1 over bracks2pars2ell + 1 + 1^mu + 1
\[3mm]&=2^-2mu - 2bracks%
sum_ell = 0^infty1 over parsell + 1/4^mu + 1
-sum_ell = 0^infty1 over parsell + 3/4^mu + 1
\[3mm]&=2^-2mu - 2bracks%
zetaparsmu + 1,1 over 4 - zetaparsmu + 1,3 over 4
endalign
where $dszetaparss,q equiv sum_n = 0^infty1 over parsq + n^s$
with $Reparss > 1$ and $Reparsq > 0$ i s the
Hurwitz Zeta Function or/and Generalizated Zeta Funcion .
So,
beginalign
&lim_mu to 0partialdmusum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=-,1 over 4,lnpars2
underbraceoverbracesum_ell = 0^infty1 over parsell + 3/4parsell + 1/4
^ds2bracksPsipars3/4 - Psipars1/4 = 2pi
_dsmboxSee pars1.1 mboxand pars1.2
+ 1 over 4
overbracepartialdmubracks%
zetaparsmu,1 over 4 - zetaparsmu,3 over 4_mu = 1
^ds-gamma_1pars1/4 + gamma_1pars3/4
endalign
where $gamma_nparsz$ is a
Generalizated Stieltjes Constant
.
With this result, $pars2$ is reduced to:
beginalign
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2
&=-,1 over 4,braces%
pibracks%
gamma + 2lnpars2 + gamma_1pars1 over 4 - gamma_1pars3 over 4
tag3
endalign
The difference $gamma_1pars1/4 - gamma_1pars3/4$ is evaluated with the 1846 Carl Malmsten identity
:
$$
gamma_1parsm over n - gamma_1pars1 - m over n
=-pibracksgamma + lnpars2pi ncotparsmpi over n
+ 2pisum_ell = 1^n - 1
sinpars2pi m over n,elllnparsGammaparsell over n
$$
With $m = 1$ and $n = 4$:
beginalign
&gamma_1pars1 over 4 - gamma_1pars3 over 4
\[3mm]&=-pibracksgamma + lnpars8picotparspi over 4
+ 2pisum_ell = 1^3sinparspi,ell over 2
lnparsGammaparsell over 4
\[3mm]&=-pibracksgamma + 2lnpars2 + lnpars2pi
\[3mm]&phantom=
+ 2pibrackssinparspi over 2lnparsGammapars1 over 4
+ sinparspilnparsGammapars1 over 2
+ sinpars3pi over 2Gammapars3 over 4
\[3mm]&=-pibracksgamma + 2lnpars2 + lnpars2pi
+2pibrackslnparsGammapars1 over 4 - lnparsGammapars3 over 4
\[3mm]&=-pibracksgamma + 2lnpars2 - 2pibracks%
lnparsroot2pi +lnparsGammapars3 over 4 over Gammapars1 over 4
\[3mm]&=-pibracksgamma + 2lnpars2
-2pilnparsroot2pi,Gammapars3 over 4 over Gammapars1 over 4
endalign
By replacing this result in $pars3$, we find:
$$color#00flarge%
int_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsrootvphantomlarge A2pi,
Gammapars3/4 over Gammapars1/4
$$
As an 'extra-bonus' we can use the identity
$dsGammaparsz = pi over Gammapars1 - zsinparspi z$ to 'kill' one of the $Gamma,$'s functions:
$dsGammapars1 over 4 = root2pi over Gammapars3/4$ which yields:
$$
int_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsGamma^,2pars3/4 over rootpi
$$
answered Mar 2 '14 at 0:28
Felix Marin
65.6k7105135
add a comment |Â
up vote
16
down vote
$newcommand+^dagger
newcommandangles[1]leftlangle #1 rightrangle
newcommandbraces[1]leftlbrace #1 rightrbrace
newcommandbracks[1]leftlbrack #1 rightrbrack
newcommandceil[1],leftlceil #1 rightrceil,
newcommandddrm d
newcommanddowndownarrow
newcommandds[1]displaystyle#1
newcommandequalby[1]#1 atop = atop vphantomhuge A
newcommandexpo[1],rm e^#1,
newcommandfermi,rm f
newcommandfloor[1],leftlfloor #1 rightrfloor,
newcommandhalf1 over 2
newcommandicrm i
newcommandiffLongleftrightarrow
newcommandimpLongrightarrow
newcommandisdiv,left.rightvert,
newcommandket[1]leftvert #1rightrangle
newcommandol[1]overline#1
newcommandpars[1]left( #1 right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandppcal P
newcommandroot[2],sqrt[#1],#2,,
newcommandsech,rm sech
newcommandsgn,rm sgn
newcommandtotald[3]fracrm d^#1 #2rm d #3^#1
newcommandul[1]underline#1
newcommandverts[1]leftvert, #1 ,rightvert$
$dsint_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsrootvphantomlarge A2pi,
Gammapars3/4 over Gammapars1/4: Large ?$
With $dsx to 1/x$:
$$
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
int_infty^1lnparslnparsx,pars-,dd x/x^2 over 1 + 1/x^2
=int_1^inftylnparslnparsx over 1 + x^2,dd x
$$
With $x equiv expotquadiffquad t = lnparsx$:
beginalign
&color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
int_0^inftylnparst over 1 + expo2t,expotdd t
=int_0^inftylnparstexpo-t,1 over 1 + expo-2t,dd t
\[3mm]&=int_0^inftylnparstexpo-t,
sum_ell = 0^inftypars-1^ellexpo-2ell t,dd t
=sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
int_0^inftyt^muexpo-pars2ell + 1t,dd t
\[3mm]&=sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
1 over pars2ell + 1^mu + 1
overbraceint_0^inftyt^muexpo-t,dd t^dsGammaparsmu + 1
endalign
where $Gammaparsz$ is the
Gamma Function.
beginalign
&color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
Gammaparsmu + 1 over pars2ell + 1^mu + 1
\[3mm]&=
lim_mu to 0partialdmubracks%
Gammaparsmu + 1sum_ell = 0^infty
pars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=
lim_mu to 0braces%
Gamma'parsmu + 1sum_ell = 0^infty
pars-1^ellover pars2ell + 1^mu + 1
+Gammaparsmu + 1partialdmubracks%
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=-gammasum_ell = 0^inftypars-1^ellover 2ell + 1
+
lim_mu to 0partialdmu
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1tag1
endalign
In this result, we used $Psipars1 = -gamma$ and $Gammapars1 = 1$ where
$Psiparsz equiv ddlnGammaparsz/dd z$ is the
Digamma Function and $gamma$ is the
Euler-Mascheroni constant.
The first $ell$-sum in the right member of $pars1$ is given by:
beginalign
&sum_ell = 0pars-^ell over 2ell + 1=
sum_ell = 0pars1 over 4ell + 1 - 1 over 4ell + 3
=1 over 8sum_ell = 01 over parsell + 1/4parsell + 3/4
\[3mm]&=-,1 over 4bracksPsipars1 over 4 - Psipars3 over 4
= pi over 4
endalign
where we used the identities:
beginalign
sum_ell = 0^infty1 over parsell + z_0parsell + z_1
&=Psiparsz_0 - Psiparsz_1 over z_0 - z_1tag1.1
\[3mm]Psiparsz - Psipars1 - z &= -picotparspi ztag1.2
endalign
$$
mboxThen,quad
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
-,1 over 4,gammapi + lim_mu to 0partialdmu
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1tag2
$$
Also,
beginalign
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
&=sum_ell = 0^infty1 over bracks2pars2ell + 1^mu + 1
-sum_ell = 0^infty1 over bracks2pars2ell + 1 + 1^mu + 1
\[3mm]&=2^-2mu - 2bracks%
sum_ell = 0^infty1 over parsell + 1/4^mu + 1
-sum_ell = 0^infty1 over parsell + 3/4^mu + 1
\[3mm]&=2^-2mu - 2bracks%
zetaparsmu + 1,1 over 4 - zetaparsmu + 1,3 over 4
endalign
where $dszetaparss,q equiv sum_n = 0^infty1 over parsq + n^s$
with $Reparss > 1$ and $Reparsq > 0$ i s the
Hurwitz Zeta Function or/and Generalizated Zeta Funcion .
So,
beginalign
&lim_mu to 0partialdmusum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=-,1 over 4,lnpars2
underbraceoverbracesum_ell = 0^infty1 over parsell + 3/4parsell + 1/4
^ds2bracksPsipars3/4 - Psipars1/4 = 2pi
_dsmboxSee pars1.1 mboxand pars1.2
+ 1 over 4
overbracepartialdmubracks%
zetaparsmu,1 over 4 - zetaparsmu,3 over 4_mu = 1
^ds-gamma_1pars1/4 + gamma_1pars3/4
endalign
where $gamma_nparsz$ is a
Generalizated Stieltjes Constant
.
With this result, $pars2$ is reduced to:
beginalign
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2
&=-,1 over 4,braces%
pibracks%
gamma + 2lnpars2 + gamma_1pars1 over 4 - gamma_1pars3 over 4
tag3
endalign
The difference $gamma_1pars1/4 - gamma_1pars3/4$ is evaluated with the 1846 Carl Malmsten identity
:
$$
gamma_1parsm over n - gamma_1pars1 - m over n
=-pibracksgamma + lnpars2pi ncotparsmpi over n
+ 2pisum_ell = 1^n - 1
sinpars2pi m over n,elllnparsGammaparsell over n
$$
With $m = 1$ and $n = 4$:
beginalign
&gamma_1pars1 over 4 - gamma_1pars3 over 4
\[3mm]&=-pibracksgamma + lnpars8picotparspi over 4
+ 2pisum_ell = 1^3sinparspi,ell over 2
lnparsGammaparsell over 4
\[3mm]&=-pibracksgamma + 2lnpars2 + lnpars2pi
\[3mm]&phantom=
+ 2pibrackssinparspi over 2lnparsGammapars1 over 4
+ sinparspilnparsGammapars1 over 2
+ sinpars3pi over 2Gammapars3 over 4
\[3mm]&=-pibracksgamma + 2lnpars2 + lnpars2pi
+2pibrackslnparsGammapars1 over 4 - lnparsGammapars3 over 4
\[3mm]&=-pibracksgamma + 2lnpars2 - 2pibracks%
lnparsroot2pi +lnparsGammapars3 over 4 over Gammapars1 over 4
\[3mm]&=-pibracksgamma + 2lnpars2
-2pilnparsroot2pi,Gammapars3 over 4 over Gammapars1 over 4
endalign
By replacing this result in $pars3$, we find:
$$color#00flarge%
int_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsrootvphantomlarge A2pi,
Gammapars3/4 over Gammapars1/4
$$
As an 'extra-bonus' we can use the identity
$dsGammaparsz = pi over Gammapars1 - zsinparspi z$ to 'kill' one of the $Gamma,$'s functions:
$dsGammapars1 over 4 = root2pi over Gammapars3/4$ which yields:
$$
int_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsGamma^,2pars3/4 over rootpi
$$
answered Mar 2 '14 at 0:28
Felix Marin
65.6k7105135
add a comment |Â
up vote
16
down vote
up vote
16
down vote
$newcommand+^dagger
newcommandangles[1]leftlangle #1 rightrangle
newcommandbraces[1]leftlbrace #1 rightrbrace
newcommandbracks[1]leftlbrack #1 rightrbrack
newcommandceil[1],leftlceil #1 rightrceil,
newcommandddrm d
newcommanddowndownarrow
newcommandds[1]displaystyle#1
newcommandequalby[1]#1 atop = atop vphantomhuge A
newcommandexpo[1],rm e^#1,
newcommandfermi,rm f
newcommandfloor[1],leftlfloor #1 rightrfloor,
newcommandhalf1 over 2
newcommandicrm i
newcommandiffLongleftrightarrow
newcommandimpLongrightarrow
newcommandisdiv,left.rightvert,
newcommandket[1]leftvert #1rightrangle
newcommandol[1]overline#1
newcommandpars[1]left( #1 right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandppcal P
newcommandroot[2],sqrt[#1],#2,,
newcommandsech,rm sech
newcommandsgn,rm sgn
newcommandtotald[3]fracrm d^#1 #2rm d #3^#1
newcommandul[1]underline#1
newcommandverts[1]leftvert, #1 ,rightvert$
$dsint_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsrootvphantomlarge A2pi,
Gammapars3/4 over Gammapars1/4: Large ?$
With $dsx to 1/x$:
$$
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
int_infty^1lnparslnparsx,pars-,dd x/x^2 over 1 + 1/x^2
=int_1^inftylnparslnparsx over 1 + x^2,dd x
$$
With $x equiv expotquadiffquad t = lnparsx$:
beginalign
&color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
int_0^inftylnparst over 1 + expo2t,expotdd t
=int_0^inftylnparstexpo-t,1 over 1 + expo-2t,dd t
\[3mm]&=int_0^inftylnparstexpo-t,
sum_ell = 0^inftypars-1^ellexpo-2ell t,dd t
=sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
int_0^inftyt^muexpo-pars2ell + 1t,dd t
\[3mm]&=sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
1 over pars2ell + 1^mu + 1
overbraceint_0^inftyt^muexpo-t,dd t^dsGammaparsmu + 1
endalign
where $Gammaparsz$ is the
Gamma Function.
beginalign
&color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
Gammaparsmu + 1 over pars2ell + 1^mu + 1
\[3mm]&=
lim_mu to 0partialdmubracks%
Gammaparsmu + 1sum_ell = 0^infty
pars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=
lim_mu to 0braces%
Gamma'parsmu + 1sum_ell = 0^infty
pars-1^ellover pars2ell + 1^mu + 1
+Gammaparsmu + 1partialdmubracks%
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=-gammasum_ell = 0^inftypars-1^ellover 2ell + 1
+
lim_mu to 0partialdmu
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1tag1
endalign
In this result, we used $Psipars1 = -gamma$ and $Gammapars1 = 1$ where
$Psiparsz equiv ddlnGammaparsz/dd z$ is the
Digamma Function and $gamma$ is the
Euler-Mascheroni constant.
The first $ell$-sum in the right member of $pars1$ is given by:
beginalign
&sum_ell = 0pars-^ell over 2ell + 1=
sum_ell = 0pars1 over 4ell + 1 - 1 over 4ell + 3
=1 over 8sum_ell = 01 over parsell + 1/4parsell + 3/4
\[3mm]&=-,1 over 4bracksPsipars1 over 4 - Psipars3 over 4
= pi over 4
endalign
where we used the identities:
beginalign
sum_ell = 0^infty1 over parsell + z_0parsell + z_1
&=Psiparsz_0 - Psiparsz_1 over z_0 - z_1tag1.1
\[3mm]Psiparsz - Psipars1 - z &= -picotparspi ztag1.2
endalign
$$
mboxThen,quad
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
-,1 over 4,gammapi + lim_mu to 0partialdmu
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1tag2
$$
Also,
beginalign
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
&=sum_ell = 0^infty1 over bracks2pars2ell + 1^mu + 1
-sum_ell = 0^infty1 over bracks2pars2ell + 1 + 1^mu + 1
\[3mm]&=2^-2mu - 2bracks%
sum_ell = 0^infty1 over parsell + 1/4^mu + 1
-sum_ell = 0^infty1 over parsell + 3/4^mu + 1
\[3mm]&=2^-2mu - 2bracks%
zetaparsmu + 1,1 over 4 - zetaparsmu + 1,3 over 4
endalign
where $dszetaparss,q equiv sum_n = 0^infty1 over parsq + n^s$
with $Reparss > 1$ and $Reparsq > 0$ i s the
Hurwitz Zeta Function or/and Generalizated Zeta Funcion .
So,
beginalign
&lim_mu to 0partialdmusum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=-,1 over 4,lnpars2
underbraceoverbracesum_ell = 0^infty1 over parsell + 3/4parsell + 1/4
^ds2bracksPsipars3/4 - Psipars1/4 = 2pi
_dsmboxSee pars1.1 mboxand pars1.2
+ 1 over 4
overbracepartialdmubracks%
zetaparsmu,1 over 4 - zetaparsmu,3 over 4_mu = 1
^ds-gamma_1pars1/4 + gamma_1pars3/4
endalign
where $gamma_nparsz$ is a
Generalizated Stieltjes Constant
.
With this result, $pars2$ is reduced to:
beginalign
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2
&=-,1 over 4,braces%
pibracks%
gamma + 2lnpars2 + gamma_1pars1 over 4 - gamma_1pars3 over 4
tag3
endalign
The difference $gamma_1pars1/4 - gamma_1pars3/4$ is evaluated with the 1846 Carl Malmsten identity
:
$$
gamma_1parsm over n - gamma_1pars1 - m over n
=-pibracksgamma + lnpars2pi ncotparsmpi over n
+ 2pisum_ell = 1^n - 1
sinpars2pi m over n,elllnparsGammaparsell over n
$$
With $m = 1$ and $n = 4$:
beginalign
&gamma_1pars1 over 4 - gamma_1pars3 over 4
\[3mm]&=-pibracksgamma + lnpars8picotparspi over 4
+ 2pisum_ell = 1^3sinparspi,ell over 2
lnparsGammaparsell over 4
\[3mm]&=-pibracksgamma + 2lnpars2 + lnpars2pi
\[3mm]&phantom=
+ 2pibrackssinparspi over 2lnparsGammapars1 over 4
+ sinparspilnparsGammapars1 over 2
+ sinpars3pi over 2Gammapars3 over 4
\[3mm]&=-pibracksgamma + 2lnpars2 + lnpars2pi
+2pibrackslnparsGammapars1 over 4 - lnparsGammapars3 over 4
\[3mm]&=-pibracksgamma + 2lnpars2 - 2pibracks%
lnparsroot2pi +lnparsGammapars3 over 4 over Gammapars1 over 4
\[3mm]&=-pibracksgamma + 2lnpars2
-2pilnparsroot2pi,Gammapars3 over 4 over Gammapars1 over 4
endalign
By replacing this result in $pars3$, we find:
$$color#00flarge%
int_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsrootvphantomlarge A2pi,
Gammapars3/4 over Gammapars1/4
$$
As an 'extra-bonus' we can use the identity
$dsGammaparsz = pi over Gammapars1 - zsinparspi z$ to 'kill' one of the $Gamma,$'s functions:
$dsGammapars1 over 4 = root2pi over Gammapars3/4$ which yields:
$$
int_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsGamma^,2pars3/4 over rootpi
$$
answered Mar 2 '14 at 0:28
Felix Marin
65.6k7105135
$newcommand+^dagger
newcommandangles[1]leftlangle #1 rightrangle
newcommandbraces[1]leftlbrace #1 rightrbrace
newcommandbracks[1]leftlbrack #1 rightrbrack
newcommandceil[1],leftlceil #1 rightrceil,
newcommandddrm d
newcommanddowndownarrow
newcommandds[1]displaystyle#1
newcommandequalby[1]#1 atop = atop vphantomhuge A
newcommandexpo[1],rm e^#1,
newcommandfermi,rm f
newcommandfloor[1],leftlfloor #1 rightrfloor,
newcommandhalf1 over 2
newcommandicrm i
newcommandiffLongleftrightarrow
newcommandimpLongrightarrow
newcommandisdiv,left.rightvert,
newcommandket[1]leftvert #1rightrangle
newcommandol[1]overline#1
newcommandpars[1]left( #1 right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandppcal P
newcommandroot[2],sqrt[#1],#2,,
newcommandsech,rm sech
newcommandsgn,rm sgn
newcommandtotald[3]fracrm d^#1 #2rm d #3^#1
newcommandul[1]underline#1
newcommandverts[1]leftvert, #1 ,rightvert$
$dsint_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsrootvphantomlarge A2pi,
Gammapars3/4 over Gammapars1/4: Large ?$
With $dsx to 1/x$:
$$
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
int_infty^1lnparslnparsx,pars-,dd x/x^2 over 1 + 1/x^2
=int_1^inftylnparslnparsx over 1 + x^2,dd x
$$
With $x equiv expotquadiffquad t = lnparsx$:
beginalign
&color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
int_0^inftylnparst over 1 + expo2t,expotdd t
=int_0^inftylnparstexpo-t,1 over 1 + expo-2t,dd t
\[3mm]&=int_0^inftylnparstexpo-t,
sum_ell = 0^inftypars-1^ellexpo-2ell t,dd t
=sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
int_0^inftyt^muexpo-pars2ell + 1t,dd t
\[3mm]&=sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
1 over pars2ell + 1^mu + 1
overbraceint_0^inftyt^muexpo-t,dd t^dsGammaparsmu + 1
endalign
where $Gammaparsz$ is the
Gamma Function.
beginalign
&color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
sum_ell = 0^inftypars-1^elllim_mu to 0partialdmubracks%
Gammaparsmu + 1 over pars2ell + 1^mu + 1
\[3mm]&=
lim_mu to 0partialdmubracks%
Gammaparsmu + 1sum_ell = 0^infty
pars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=
lim_mu to 0braces%
Gamma'parsmu + 1sum_ell = 0^infty
pars-1^ellover pars2ell + 1^mu + 1
+Gammaparsmu + 1partialdmubracks%
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=-gammasum_ell = 0^inftypars-1^ellover 2ell + 1
+
lim_mu to 0partialdmu
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1tag1
endalign
In this result, we used $Psipars1 = -gamma$ and $Gammapars1 = 1$ where
$Psiparsz equiv ddlnGammaparsz/dd z$ is the
Digamma Function and $gamma$ is the
Euler-Mascheroni constant.
The first $ell$-sum in the right member of $pars1$ is given by:
beginalign
&sum_ell = 0pars-^ell over 2ell + 1=
sum_ell = 0pars1 over 4ell + 1 - 1 over 4ell + 3
=1 over 8sum_ell = 01 over parsell + 1/4parsell + 3/4
\[3mm]&=-,1 over 4bracksPsipars1 over 4 - Psipars3 over 4
= pi over 4
endalign
where we used the identities:
beginalign
sum_ell = 0^infty1 over parsell + z_0parsell + z_1
&=Psiparsz_0 - Psiparsz_1 over z_0 - z_1tag1.1
\[3mm]Psiparsz - Psipars1 - z &= -picotparspi ztag1.2
endalign
$$
mboxThen,quad
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2=
-,1 over 4,gammapi + lim_mu to 0partialdmu
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1tag2
$$
Also,
beginalign
sum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
&=sum_ell = 0^infty1 over bracks2pars2ell + 1^mu + 1
-sum_ell = 0^infty1 over bracks2pars2ell + 1 + 1^mu + 1
\[3mm]&=2^-2mu - 2bracks%
sum_ell = 0^infty1 over parsell + 1/4^mu + 1
-sum_ell = 0^infty1 over parsell + 3/4^mu + 1
\[3mm]&=2^-2mu - 2bracks%
zetaparsmu + 1,1 over 4 - zetaparsmu + 1,3 over 4
endalign
where $dszetaparss,q equiv sum_n = 0^infty1 over parsq + n^s$
with $Reparss > 1$ and $Reparsq > 0$ i s the
Hurwitz Zeta Function or/and Generalizated Zeta Funcion .
So,
beginalign
&lim_mu to 0partialdmusum_ell = 0^inftypars-1^ellover pars2ell + 1^mu + 1
\[3mm]&=-,1 over 4,lnpars2
underbraceoverbracesum_ell = 0^infty1 over parsell + 3/4parsell + 1/4
^ds2bracksPsipars3/4 - Psipars1/4 = 2pi
_dsmboxSee pars1.1 mboxand pars1.2
+ 1 over 4
overbracepartialdmubracks%
zetaparsmu,1 over 4 - zetaparsmu,3 over 4_mu = 1
^ds-gamma_1pars1/4 + gamma_1pars3/4
endalign
where $gamma_nparsz$ is a
Generalizated Stieltjes Constant
.
With this result, $pars2$ is reduced to:
beginalign
color#f00int_0^1lnparslnpars1 over x,dd x over 1 + x^2
&=-,1 over 4,braces%
pibracks%
gamma + 2lnpars2 + gamma_1pars1 over 4 - gamma_1pars3 over 4
tag3
endalign
The difference $gamma_1pars1/4 - gamma_1pars3/4$ is evaluated with the 1846 Carl Malmsten identity
:
$$
gamma_1parsm over n - gamma_1pars1 - m over n
=-pibracksgamma + lnpars2pi ncotparsmpi over n
+ 2pisum_ell = 1^n - 1
sinpars2pi m over n,elllnparsGammaparsell over n
$$
With $m = 1$ and $n = 4$:
beginalign
&gamma_1pars1 over 4 - gamma_1pars3 over 4
\[3mm]&=-pibracksgamma + lnpars8picotparspi over 4
+ 2pisum_ell = 1^3sinparspi,ell over 2
lnparsGammaparsell over 4
\[3mm]&=-pibracksgamma + 2lnpars2 + lnpars2pi
\[3mm]&phantom=
+ 2pibrackssinparspi over 2lnparsGammapars1 over 4
+ sinparspilnparsGammapars1 over 2
+ sinpars3pi over 2Gammapars3 over 4
\[3mm]&=-pibracksgamma + 2lnpars2 + lnpars2pi
+2pibrackslnparsGammapars1 over 4 - lnparsGammapars3 over 4
\[3mm]&=-pibracksgamma + 2lnpars2 - 2pibracks%
lnparsroot2pi +lnparsGammapars3 over 4 over Gammapars1 over 4
\[3mm]&=-pibracksgamma + 2lnpars2
-2pilnparsroot2pi,Gammapars3 over 4 over Gammapars1 over 4
endalign
By replacing this result in $pars3$, we find:
$$color#00flarge%
int_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsrootvphantomlarge A2pi,
Gammapars3/4 over Gammapars1/4
$$
As an 'extra-bonus' we can use the identity
$dsGammaparsz = pi over Gammapars1 - zsinparspi z$ to 'kill' one of the $Gamma,$'s functions:
$dsGammapars1 over 4 = root2pi over Gammapars3/4$ which yields:
$$
int_0^1lnparslnpars1 over x,dd x over 1 + x^2
=pi over 2,lnparsGamma^,2pars3/4 over rootpi
$$
answered Mar 2 '14 at 0:28
Felix Marin
65.6k7105135
edited Jul 5 '14 at 19:44
answered Mar 2 '14 at 0:28
Felix Marin
65.6k7105135
answered Mar 2 '14 at 0:28
Felix Marin
65.6k7105135
answered Mar 2 '14 at 0:28
Felix Marin
65.6k7105135
65.6k7105135
add a comment |Â
add a comment |Â
up vote
5
down vote
See also V. Adamchik's formula $$int_0^1 fracx^p-11+x^nlog log frac1xdx = fracgamma+log(2n)2n(psi(fracp2n)-psi(fracn+p2n))+frac12n(zeta'(1,fracp2n)-zeta'(1,fracn+p2n))$$ in http://dx.doi.org/10.1145/258726.258736 .
answered Mar 1 '14 at 18:40
R. J. Mathar
6912
add a comment |Â
up vote
5
down vote
See also V. Adamchik's formula $$int_0^1 fracx^p-11+x^nlog log frac1xdx = fracgamma+log(2n)2n(psi(fracp2n)-psi(fracn+p2n))+frac12n(zeta'(1,fracp2n)-zeta'(1,fracn+p2n))$$ in http://dx.doi.org/10.1145/258726.258736 .
answered Mar 1 '14 at 18:40
R. J. Mathar
6912
add a comment |Â
up vote
5
down vote
up vote
5
down vote
See also V. Adamchik's formula $$int_0^1 fracx^p-11+x^nlog log frac1xdx = fracgamma+log(2n)2n(psi(fracp2n)-psi(fracn+p2n))+frac12n(zeta'(1,fracp2n)-zeta'(1,fracn+p2n))$$ in http://dx.doi.org/10.1145/258726.258736 .
answered Mar 1 '14 at 18:40
R. J. Mathar
6912
See also V. Adamchik's formula $$int_0^1 fracx^p-11+x^nlog log frac1xdx = fracgamma+log(2n)2n(psi(fracp2n)-psi(fracn+p2n))+frac12n(zeta'(1,fracp2n)-zeta'(1,fracn+p2n))$$ in http://dx.doi.org/10.1145/258726.258736 .
answered Mar 1 '14 at 18:40
R. J. Mathar
6912
answered Mar 1 '14 at 18:40
R. J. Mathar
6912
answered Mar 1 '14 at 18:40
R. J. Mathar
6912
answered Mar 1 '14 at 18:40
R. J. Mathar
6912
6912
add a comment |Â
add a comment |Â
up vote
4
down vote
The above integral was long-time known as the Vardi's integral. Quite recently, an interesting research work devoted to this integral was published by Iaroslav Blagouchine. It appears that this integral was first evaluated by Carl Malmsten in 1842 (and not by Ilan Vardi in 1988). Blagouchine describes two different methods for its evaluation: the original Malmsten's method and the contour integration method. In the same paper, numerous integrals similar to the above one are also treated, and it is shown that many of them may be evaluated by the contour integration methods.
More recent paper by Blagouchine (mentioned above in the context of Malmsten's formula derived in 1846) is also integersting, but it is less directly related to the integral in question than the first one.
edited Nov 29 '14 at 2:42
Iaroslav Blagouchine
32727
answered Jun 24 '14 at 21:50
Mark
812
add a comment |Â
up vote
4
down vote
The above integral was long-time known as the Vardi's integral. Quite recently, an interesting research work devoted to this integral was published by Iaroslav Blagouchine. It appears that this integral was first evaluated by Carl Malmsten in 1842 (and not by Ilan Vardi in 1988). Blagouchine describes two different methods for its evaluation: the original Malmsten's method and the contour integration method. In the same paper, numerous integrals similar to the above one are also treated, and it is shown that many of them may be evaluated by the contour integration methods.
More recent paper by Blagouchine (mentioned above in the context of Malmsten's formula derived in 1846) is also integersting, but it is less directly related to the integral in question than the first one.
edited Nov 29 '14 at 2:42
Iaroslav Blagouchine
32727
answered Jun 24 '14 at 21:50
Mark
812
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The above integral was long-time known as the Vardi's integral. Quite recently, an interesting research work devoted to this integral was published by Iaroslav Blagouchine. It appears that this integral was first evaluated by Carl Malmsten in 1842 (and not by Ilan Vardi in 1988). Blagouchine describes two different methods for its evaluation: the original Malmsten's method and the contour integration method. In the same paper, numerous integrals similar to the above one are also treated, and it is shown that many of them may be evaluated by the contour integration methods.
More recent paper by Blagouchine (mentioned above in the context of Malmsten's formula derived in 1846) is also integersting, but it is less directly related to the integral in question than the first one.
edited Nov 29 '14 at 2:42
Iaroslav Blagouchine
32727
answered Jun 24 '14 at 21:50
Mark
812
The above integral was long-time known as the Vardi's integral. Quite recently, an interesting research work devoted to this integral was published by Iaroslav Blagouchine. It appears that this integral was first evaluated by Carl Malmsten in 1842 (and not by Ilan Vardi in 1988). Blagouchine describes two different methods for its evaluation: the original Malmsten's method and the contour integration method. In the same paper, numerous integrals similar to the above one are also treated, and it is shown that many of them may be evaluated by the contour integration methods.
More recent paper by Blagouchine (mentioned above in the context of Malmsten's formula derived in 1846) is also integersting, but it is less directly related to the integral in question than the first one.
edited Nov 29 '14 at 2:42
Iaroslav Blagouchine
32727
answered Jun 24 '14 at 21:50
Mark
812
edited Nov 29 '14 at 2:42
Iaroslav Blagouchine
32727
edited Nov 29 '14 at 2:42
Iaroslav Blagouchine
32727
edited Nov 29 '14 at 2:42
Iaroslav Blagouchine
32727
32727
answered Jun 24 '14 at 21:50
Mark
812
answered Jun 24 '14 at 21:50
Mark
812
answered Jun 24 '14 at 21:50
Mark
812
812
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f121545%2fevaluating-int-01-log-log-left-frac1x-right-fracdx1x2%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
@PeterT.off: That isn't true at all.
– Eric Naslund
Mar 18 '12 at 12:18
For some variants of this integral see C.J. Malmsten.
– Raymond Manzoni
Mar 9 '14 at 19:30