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Give an asymptotic developement of $I_n=int_0^1 (x^n-x^n-2)ln(1+x^n)dx.$
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Let $$I_n=int_0^1 (x^n+x^n-2)ln(1+x^n)dx.$$
Give an asymptotic developement of $I_n$ at order $O(frac1n^3)$ when $nto infty $.
I wanted to use the fact that $$sum_k=1^infty frac(-1)^k+1 x^knk,$$
and thus $$I_n=int_0^1(x^n+x^n-2)sum_k=1^infty frac(-1)^k+1kx^kndx,$$
but since the convergence of this series is a priori not uniform and the coefficient are not positive on $[0,1]$ I can't permute the sum and the integral.
Any other idea ?
real-analysis sequences-and-series asymptotics
edited Jul 21 at 23:03
Surb
36.3k84274
asked Jul 21 at 20:26
user386627
714214
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Let $$I_n=int_0^1 (x^n+x^n-2)ln(1+x^n)dx.$$
Give an asymptotic developement of $I_n$ at order $O(frac1n^3)$ when $nto infty $.
I wanted to use the fact that $$sum_k=1^infty frac(-1)^k+1 x^knk,$$
and thus $$I_n=int_0^1(x^n+x^n-2)sum_k=1^infty frac(-1)^k+1kx^kndx,$$
but since the convergence of this series is a priori not uniform and the coefficient are not positive on $[0,1]$ I can't permute the sum and the integral.
Any other idea ?
real-analysis sequences-and-series asymptotics
edited Jul 21 at 23:03
Surb
36.3k84274
asked Jul 21 at 20:26
user386627
714214
1
@Michael: If $f_nto f$ uniformly on $[0,r]$ for all $r<1$ but not on $[0,1[$, then$$lim_nto infty lim_rto 1int_0^r f_n=lim_rto 1lim_nto infty int_0^r f_n$$ is false a priori.
– Surb
Jul 21 at 22:38
Maybe you are missing the left hand side for the second equation. Maybe you are missing $dx$ for the third equation. If so, please revise the question.
– norio
Jul 21 at 22:42
@Surb In our case it is true, because the sum of the intergrals convergences uniformly.
– Michael
Jul 22 at 9:07
1
@Surb $sum_k=1^infty int_0^r(x^n+x^n-2)frac(-1)^k+1kx^kndx=sum_k=1^inftyfrac(-1)^k+1k(dfracr^n(k+1)+1n(k+1)+1+dfracr^n(k+1)-1n(k+1)-1)$. The right hand side converges uniformly in $r$, because $rleq 1$ and you essentially have the sum of $1/k(k+1)$.
– Michael
Jul 22 at 9:32
1
The title and question do not match.
– zhw.
Jul 23 at 1:07
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up vote
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Let $$I_n=int_0^1 (x^n+x^n-2)ln(1+x^n)dx.$$
Give an asymptotic developement of $I_n$ at order $O(frac1n^3)$ when $nto infty $.
I wanted to use the fact that $$sum_k=1^infty frac(-1)^k+1 x^knk,$$
and thus $$I_n=int_0^1(x^n+x^n-2)sum_k=1^infty frac(-1)^k+1kx^kndx,$$
but since the convergence of this series is a priori not uniform and the coefficient are not positive on $[0,1]$ I can't permute the sum and the integral.
Any other idea ?
real-analysis sequences-and-series asymptotics
edited Jul 21 at 23:03
Surb
36.3k84274
asked Jul 21 at 20:26
user386627
714214
Let $$I_n=int_0^1 (x^n+x^n-2)ln(1+x^n)dx.$$
Give an asymptotic developement of $I_n$ at order $O(frac1n^3)$ when $nto infty $.
I wanted to use the fact that $$sum_k=1^infty frac(-1)^k+1 x^knk,$$
and thus $$I_n=int_0^1(x^n+x^n-2)sum_k=1^infty frac(-1)^k+1kx^kndx,$$
but since the convergence of this series is a priori not uniform and the coefficient are not positive on $[0,1]$ I can't permute the sum and the integral.
Any other idea ?
real-analysis sequences-and-series asymptotics
edited Jul 21 at 23:03
Surb
36.3k84274
asked Jul 21 at 20:26
user386627
714214
edited Jul 21 at 23:03
Surb
36.3k84274
edited Jul 21 at 23:03
Surb
36.3k84274
edited Jul 21 at 23:03
Surb
36.3k84274
36.3k84274
asked Jul 21 at 20:26
user386627
714214
asked Jul 21 at 20:26
user386627
714214
asked Jul 21 at 20:26
user386627
714214
714214
1
@Michael: If $f_nto f$ uniformly on $[0,r]$ for all $r<1$ but not on $[0,1[$, then$$lim_nto infty lim_rto 1int_0^r f_n=lim_rto 1lim_nto infty int_0^r f_n$$ is false a priori.
– Surb
Jul 21 at 22:38
Maybe you are missing the left hand side for the second equation. Maybe you are missing $dx$ for the third equation. If so, please revise the question.
– norio
Jul 21 at 22:42
@Surb In our case it is true, because the sum of the intergrals convergences uniformly.
– Michael
Jul 22 at 9:07
1
@Surb $sum_k=1^infty int_0^r(x^n+x^n-2)frac(-1)^k+1kx^kndx=sum_k=1^inftyfrac(-1)^k+1k(dfracr^n(k+1)+1n(k+1)+1+dfracr^n(k+1)-1n(k+1)-1)$. The right hand side converges uniformly in $r$, because $rleq 1$ and you essentially have the sum of $1/k(k+1)$.
– Michael
Jul 22 at 9:32
1
The title and question do not match.
– zhw.
Jul 23 at 1:07
add a comment |Â
1
@Michael: If $f_nto f$ uniformly on $[0,r]$ for all $r<1$ but not on $[0,1[$, then$$lim_nto infty lim_rto 1int_0^r f_n=lim_rto 1lim_nto infty int_0^r f_n$$ is false a priori.
– Surb
Jul 21 at 22:38
Maybe you are missing the left hand side for the second equation. Maybe you are missing $dx$ for the third equation. If so, please revise the question.
– norio
Jul 21 at 22:42
@Surb In our case it is true, because the sum of the intergrals convergences uniformly.
– Michael
Jul 22 at 9:07
1
@Surb $sum_k=1^infty int_0^r(x^n+x^n-2)frac(-1)^k+1kx^kndx=sum_k=1^inftyfrac(-1)^k+1k(dfracr^n(k+1)+1n(k+1)+1+dfracr^n(k+1)-1n(k+1)-1)$. The right hand side converges uniformly in $r$, because $rleq 1$ and you essentially have the sum of $1/k(k+1)$.
– Michael
Jul 22 at 9:32
1
The title and question do not match.
– zhw.
Jul 23 at 1:07
1
1
@Michael: If $f_nto f$ uniformly on $[0,r]$ for all $r<1$ but not on $[0,1[$, then$$lim_nto infty lim_rto 1int_0^r f_n=lim_rto 1lim_nto infty int_0^r f_n$$ is false a priori.
– Surb
Jul 21 at 22:38
@Michael: If $f_nto f$ uniformly on $[0,r]$ for all $r<1$ but not on $[0,1[$, then$$lim_nto infty lim_rto 1int_0^r f_n=lim_rto 1lim_nto infty int_0^r f_n$$ is false a priori.
– Surb
Jul 21 at 22:38
Maybe you are missing the left hand side for the second equation. Maybe you are missing $dx$ for the third equation. If so, please revise the question.
– norio
Jul 21 at 22:42
Maybe you are missing the left hand side for the second equation. Maybe you are missing $dx$ for the third equation. If so, please revise the question.
– norio
Jul 21 at 22:42
@Surb In our case it is true, because the sum of the intergrals convergences uniformly.
– Michael
Jul 22 at 9:07
@Surb In our case it is true, because the sum of the intergrals convergences uniformly.
– Michael
Jul 22 at 9:07
1
1
@Surb $sum_k=1^infty int_0^r(x^n+x^n-2)frac(-1)^k+1kx^kndx=sum_k=1^inftyfrac(-1)^k+1k(dfracr^n(k+1)+1n(k+1)+1+dfracr^n(k+1)-1n(k+1)-1)$. The right hand side converges uniformly in $r$, because $rleq 1$ and you essentially have the sum of $1/k(k+1)$.
– Michael
Jul 22 at 9:32
@Surb $sum_k=1^infty int_0^r(x^n+x^n-2)frac(-1)^k+1kx^kndx=sum_k=1^inftyfrac(-1)^k+1k(dfracr^n(k+1)+1n(k+1)+1+dfracr^n(k+1)-1n(k+1)-1)$. The right hand side converges uniformly in $r$, because $rleq 1$ and you essentially have the sum of $1/k(k+1)$.
– Michael
Jul 22 at 9:32
1
1
The title and question do not match.
– zhw.
Jul 23 at 1:07
The title and question do not match.
– zhw.
Jul 23 at 1:07
add a comment |Â
5 Answers
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If $, f_n(x) := (x^n - x^n-2) log(1+x^n), ,$ then
$, n f_n(e^-x/n) to -2xe^-xlog(1+e^-x) ,$ as $, nto infty. ,$
Use $, x = e^-t/n ,$ in $, I_n := int_0^1! f_n(x) , dx ,$ with $, dx = -e^-t/nt/n ,$ so
$, n^2I_n !=! int_0^infty! nf_n(e^-t/n) e^-t/nt, dt. ,$
answered Jul 21 at 21:30
Somos
11.5k1933
Sorry, but I don't really see in what it help...
– user386627
Jul 21 at 21:42
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0
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Assuming that
you can reverse the order,
$beginarray\
I_n
&=int_0^1(x^n+x^n-2)sum_k=1^infty frac(-1)^k+1kx^kndx\
&=sum_k=1^infty int_0^1(x^n+x^n-2)frac(-1)^k+1kx^kndx\
&=sum_k=1^inftyfrac(-1)^k+1k int_0^1(x^n+x^n-2)x^kndx\
&=sum_k=1^inftyfrac(-1)^k+1k int_0^1(x^n(k+1)+x^n(k+1)-2)dx\
&=sum_k=1^inftyfrac(-1)^k+1k(frac1n(k+1)+1+frac1n(k+1)-1)\
&=sum_k=1^inftyfrac(-1)^k+1k(frac2n(k+1)n^2(k+1)^2-1)\
&=sum_k=1^inftyfrac(-1)^k+1kn(frac2(k+1)(k+1)^2-1/n^2)\
&=sum_k=1^inftyfrac(-1)^k+1k(k+1)n(frac21-1/(n^2(k+1)^2))\
&=frac2nsum_k=1^inftyfrac(-1)^k+1k(k+1)sum_m=0^inftyfrac1n^2m(k+1)^2m\
&=frac2nsum_m=0^inftyfrac1n^2msum_k=1^inftyfrac(-1)^k+1k(k+1)frac1(k+1)^2m\
&=2sum_m=0^inftyfrac1n^2m+1sum_k=1^inftyfrac(-1)^k+1k(k+1)^2m+1\
&=2(fracln(4/e)n-frac0.110304071913...n^3+O(frac1n^5))
qquadtext(according to Wolfy)\
endarray
$
Note:
The Inverse Symbolic Calculator
did not find anything for
0.110304071913.
answered Jul 21 at 23:05
marty cohen
69.2k446122
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There is a difference in title $(colorred-)$ and body $(colorred+)$; so, I worked both cases.
Let us consider first
$$int x^a log left(1+x^bright),dx=fracx^a+1 log left(1+x^bright)a+1-frac b1+a int fracx^a+b1+x^b,dx$$
$$int fracx^a+b1+x^b,dx=fracx^a+b+1a+b+1 ,,,
_2F_1left(1,fraca+b+1b;fraca+2b+1b;-x^bright)$$ Using the given bounds and simplifying leads to
$$K_a,b=int_0^1 x^a log left(1+x^bright),dx=fraclog (2)-Phi left(-1,1,fraca+b+1bright)a+1qquad textif qquad Re(b)>0$$ where appears the Hurwitz-Lerch transcendent function.
So,
$$I_n=int_0^1 (x^n-x^n-2)ln(1+x^n),dx=fraclog (2)-Phi left(-1,1,frac2 n+1nright)n+1-fraclog (2)-Phi
left(-1,1,frac2 n-1nright)n-1$$
$$J_n=int_0^1 (x^n+x^n-2)ln(1+x^n),dx=fraclog (2)-Phi
left(-1,1,frac2 n+1nright)n+1+fraclog (2)-Phi left(-1,1,frac2 n-1nright)n-1$$
Now, consider the expansion of $Phi (-1,1,2+epsilon)$; it is
$$(1-log (2))+left(fracpi ^212-1right) epsilon +left(1-frac3 zeta
(3)4right) epsilon ^2+left(frac7 pi ^4720-1right) epsilon
^3+left(1-frac15 zeta (5)16right) epsilon ^4+Oleft(epsilon ^5right)$$ Making $ epsilon=pm frac 1n$ and continuing with Taylor, this gives
$$I_n=frac4-fracpi ^26-4 log (2)n^2+frac-540 zeta (3)+2880-60 pi ^2-7 pi
^4-1440 log (2)360 n^4+Oleft(frac1n^5right)$$
$$J_n=frac4 log (2)-2n+frac9 zeta (3)-36+pi ^2+24 log (2)6
n^3+Oleft(frac1n^5right)$$
For $J_n$, this is the same result as in marty cohen's answer, since
$$frac9 zeta (3)-36+pi ^2+24 log (2)6=2 times 0.110304071913700 $$
This seems to work quite well even for small values of $n$ as shown in the table below
$$left(
beginarrayccccc
n & I_n^exact & I_n^approx & J_n^exact & J_n^approx\
2 & -0.1120487783 & -0.1115478099 & 0.4158382364 & 0.4138703791 \
3 & -0.0478495232 & -0.0478071470 & 0.2659498411 & 0.2657002461 \
4 & -0.0265505658 & -0.0265431189 & 0.1966526582 & 0.1965941828 \
5 & -0.0168863305 & -0.0168843897 & 0.1563016585 & 0.1562826096 \
6 & -0.0116869861 & -0.0116863381 & 0.1297937520 & 0.1297861210 \
7 & -0.0085688906 & -0.0085686341 & 0.1110165137 & 0.1110129899 \
8 & -0.0065519051 & -0.0065517901 & 0.0970062707 & 0.0970044656 \
9 & -0.0051721371 & -0.0051720804 & 0.0861468097 & 0.0861458088 \
10 & -0.0041867254 & -0.0041866953 & 0.0774800710 & 0.0774794804
endarray
right)$$
answered Jul 22 at 4:06
Claude Leibovici
111k1055126
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After using $,t:=x^n,$ we get:
$displaystyle intlimits_0^1 (x^n+x^n-2) ln(1+x^n) dx = frac2nintlimits_0^1 (cosh z)|_z=fracln tn ln(1+t) dx$
$hspace3cmdisplaystyle approx frac2nintlimits_0^1 [z^0](cosh z)|_z=fracln tn ln(1+t) dx = frac2 (ln 4 - 1)n $
$displaystyle intlimits_0^1 (x^n-x^n-2) ln(1+x^n) dx = frac2nintlimits_0^1 (sinh z)|_z=fracln tn ln(1+t) dx $
$hspace3cmdisplaystyle approx frac2nintlimits_0^1 [z^1](sinh z)|_z=fracln tn ln(1+t) dx = frac4(1 - ln 2) - fracpi^26n^2 $
You can use more elements of $,cosh,$ or $,sinh,$ to get closer to your integral.
Example:
$displaystyle intlimits_0^1 (x^n+x^n-2) ln(1+x^n) dx approx frac2nintlimits_0^1 [z^0+z^2](cosh z)|_z=fracln tn ln(1+t) dx $
$hspace3cmdisplaystyle =fracln 16 - 2n + frac1n^3left(frac3zeta(3)2+fracpi^26+ln 16 - 6right)$
answered Jul 23 at 11:50
user90369
7,596925
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If $f(n, x) = O(|g(n, x)|)$ uniformly in $x$ on $a < x < b$ and $f$ and $g$ are measurable, then $int_a^b f(n, x) dx = O(int_a^b |g(n, x)| dx)$. We have
$$J_n = int_0^1 x^n ln(1 + x^n) dx =
frac 1 n int_0^1 xi^1/n ln(1 + xi) dxi, \
|xi^1/n - 1| leq frac 1 n |! ln xi| quad
textwhen 0 < n land 0 < xi < 1, \
left| J_n - frac 1 n int_0^1 ln(1 + xi) dxi right| leq
-frac 1 n^2 int_0^1 ln xi ln(1 + xi) dxi,$$
giving the leading term in the expansion. The complete asymptotic series for $J_n$ in powers of $n$ is
$$J_n sim sum_k=1^infty
int_0^1 ln^k-1 xi ln(1 + xi) dxi frac n^-k (k-1)!.$$
Similarly, for
$$K_n = int_0^1 x^n-2 ln(1 + x^n) dx =
frac 1 n int_0^1 xi^-1/n ln(1 + xi) dxi,$$
we can take
$$|xi^-1/n - 1| leq frac 2 n sqrt xi quad
textwhen 2 < n land 0 < xi < 1.$$
The complete asymptotic series for $K_n$ is
$$K_n sim sum_k=1^infty (-1)^k-1
int_0^1 ln^k-1 xi ln(1 + xi) dxi frac n^-k (k-1)!,$$
which is formally $-J_-n$.
answered Jul 22 at 21:30
Maxim
2,060113
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5 Answers
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5 Answers
5
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active
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active
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up vote
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If $, f_n(x) := (x^n - x^n-2) log(1+x^n), ,$ then
$, n f_n(e^-x/n) to -2xe^-xlog(1+e^-x) ,$ as $, nto infty. ,$
Use $, x = e^-t/n ,$ in $, I_n := int_0^1! f_n(x) , dx ,$ with $, dx = -e^-t/nt/n ,$ so
$, n^2I_n !=! int_0^infty! nf_n(e^-t/n) e^-t/nt, dt. ,$
answered Jul 21 at 21:30
Somos
11.5k1933
Sorry, but I don't really see in what it help...
– user386627
Jul 21 at 21:42
add a comment |Â
up vote
1
down vote
If $, f_n(x) := (x^n - x^n-2) log(1+x^n), ,$ then
$, n f_n(e^-x/n) to -2xe^-xlog(1+e^-x) ,$ as $, nto infty. ,$
Use $, x = e^-t/n ,$ in $, I_n := int_0^1! f_n(x) , dx ,$ with $, dx = -e^-t/nt/n ,$ so
$, n^2I_n !=! int_0^infty! nf_n(e^-t/n) e^-t/nt, dt. ,$
answered Jul 21 at 21:30
Somos
11.5k1933
Sorry, but I don't really see in what it help...
– user386627
Jul 21 at 21:42
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $, f_n(x) := (x^n - x^n-2) log(1+x^n), ,$ then
$, n f_n(e^-x/n) to -2xe^-xlog(1+e^-x) ,$ as $, nto infty. ,$
Use $, x = e^-t/n ,$ in $, I_n := int_0^1! f_n(x) , dx ,$ with $, dx = -e^-t/nt/n ,$ so
$, n^2I_n !=! int_0^infty! nf_n(e^-t/n) e^-t/nt, dt. ,$
answered Jul 21 at 21:30
Somos
11.5k1933
If $, f_n(x) := (x^n - x^n-2) log(1+x^n), ,$ then
$, n f_n(e^-x/n) to -2xe^-xlog(1+e^-x) ,$ as $, nto infty. ,$
Use $, x = e^-t/n ,$ in $, I_n := int_0^1! f_n(x) , dx ,$ with $, dx = -e^-t/nt/n ,$ so
$, n^2I_n !=! int_0^infty! nf_n(e^-t/n) e^-t/nt, dt. ,$
answered Jul 21 at 21:30
Somos
11.5k1933
edited Jul 21 at 22:27
answered Jul 21 at 21:30
Somos
11.5k1933
answered Jul 21 at 21:30
Somos
11.5k1933
answered Jul 21 at 21:30
Somos
11.5k1933
11.5k1933
Sorry, but I don't really see in what it help...
– user386627
Jul 21 at 21:42
add a comment |Â
Sorry, but I don't really see in what it help...
– user386627
Jul 21 at 21:42
Sorry, but I don't really see in what it help...
– user386627
Jul 21 at 21:42
Sorry, but I don't really see in what it help...
– user386627
Jul 21 at 21:42
add a comment |Â
up vote
0
down vote
Assuming that
you can reverse the order,
$beginarray\
I_n
&=int_0^1(x^n+x^n-2)sum_k=1^infty frac(-1)^k+1kx^kndx\
&=sum_k=1^infty int_0^1(x^n+x^n-2)frac(-1)^k+1kx^kndx\
&=sum_k=1^inftyfrac(-1)^k+1k int_0^1(x^n+x^n-2)x^kndx\
&=sum_k=1^inftyfrac(-1)^k+1k int_0^1(x^n(k+1)+x^n(k+1)-2)dx\
&=sum_k=1^inftyfrac(-1)^k+1k(frac1n(k+1)+1+frac1n(k+1)-1)\
&=sum_k=1^inftyfrac(-1)^k+1k(frac2n(k+1)n^2(k+1)^2-1)\
&=sum_k=1^inftyfrac(-1)^k+1kn(frac2(k+1)(k+1)^2-1/n^2)\
&=sum_k=1^inftyfrac(-1)^k+1k(k+1)n(frac21-1/(n^2(k+1)^2))\
&=frac2nsum_k=1^inftyfrac(-1)^k+1k(k+1)sum_m=0^inftyfrac1n^2m(k+1)^2m\
&=frac2nsum_m=0^inftyfrac1n^2msum_k=1^inftyfrac(-1)^k+1k(k+1)frac1(k+1)^2m\
&=2sum_m=0^inftyfrac1n^2m+1sum_k=1^inftyfrac(-1)^k+1k(k+1)^2m+1\
&=2(fracln(4/e)n-frac0.110304071913...n^3+O(frac1n^5))
qquadtext(according to Wolfy)\
endarray
$
Note:
The Inverse Symbolic Calculator
did not find anything for
0.110304071913.
answered Jul 21 at 23:05
marty cohen
69.2k446122
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up vote
0
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Assuming that
you can reverse the order,
$beginarray\
I_n
&=int_0^1(x^n+x^n-2)sum_k=1^infty frac(-1)^k+1kx^kndx\
&=sum_k=1^infty int_0^1(x^n+x^n-2)frac(-1)^k+1kx^kndx\
&=sum_k=1^inftyfrac(-1)^k+1k int_0^1(x^n+x^n-2)x^kndx\
&=sum_k=1^inftyfrac(-1)^k+1k int_0^1(x^n(k+1)+x^n(k+1)-2)dx\
&=sum_k=1^inftyfrac(-1)^k+1k(frac1n(k+1)+1+frac1n(k+1)-1)\
&=sum_k=1^inftyfrac(-1)^k+1k(frac2n(k+1)n^2(k+1)^2-1)\
&=sum_k=1^inftyfrac(-1)^k+1kn(frac2(k+1)(k+1)^2-1/n^2)\
&=sum_k=1^inftyfrac(-1)^k+1k(k+1)n(frac21-1/(n^2(k+1)^2))\
&=frac2nsum_k=1^inftyfrac(-1)^k+1k(k+1)sum_m=0^inftyfrac1n^2m(k+1)^2m\
&=frac2nsum_m=0^inftyfrac1n^2msum_k=1^inftyfrac(-1)^k+1k(k+1)frac1(k+1)^2m\
&=2sum_m=0^inftyfrac1n^2m+1sum_k=1^inftyfrac(-1)^k+1k(k+1)^2m+1\
&=2(fracln(4/e)n-frac0.110304071913...n^3+O(frac1n^5))
qquadtext(according to Wolfy)\
endarray
$
Note:
The Inverse Symbolic Calculator
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0.110304071913.
answered Jul 21 at 23:05
marty cohen
69.2k446122
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Assuming that
you can reverse the order,
$beginarray\
I_n
&=int_0^1(x^n+x^n-2)sum_k=1^infty frac(-1)^k+1kx^kndx\
&=sum_k=1^infty int_0^1(x^n+x^n-2)frac(-1)^k+1kx^kndx\
&=sum_k=1^inftyfrac(-1)^k+1k int_0^1(x^n+x^n-2)x^kndx\
&=sum_k=1^inftyfrac(-1)^k+1k int_0^1(x^n(k+1)+x^n(k+1)-2)dx\
&=sum_k=1^inftyfrac(-1)^k+1k(frac1n(k+1)+1+frac1n(k+1)-1)\
&=sum_k=1^inftyfrac(-1)^k+1k(frac2n(k+1)n^2(k+1)^2-1)\
&=sum_k=1^inftyfrac(-1)^k+1kn(frac2(k+1)(k+1)^2-1/n^2)\
&=sum_k=1^inftyfrac(-1)^k+1k(k+1)n(frac21-1/(n^2(k+1)^2))\
&=frac2nsum_k=1^inftyfrac(-1)^k+1k(k+1)sum_m=0^inftyfrac1n^2m(k+1)^2m\
&=frac2nsum_m=0^inftyfrac1n^2msum_k=1^inftyfrac(-1)^k+1k(k+1)frac1(k+1)^2m\
&=2sum_m=0^inftyfrac1n^2m+1sum_k=1^inftyfrac(-1)^k+1k(k+1)^2m+1\
&=2(fracln(4/e)n-frac0.110304071913...n^3+O(frac1n^5))
qquadtext(according to Wolfy)\
endarray
$
Note:
The Inverse Symbolic Calculator
did not find anything for
0.110304071913.
answered Jul 21 at 23:05
marty cohen
69.2k446122
Assuming that
you can reverse the order,
$beginarray\
I_n
&=int_0^1(x^n+x^n-2)sum_k=1^infty frac(-1)^k+1kx^kndx\
&=sum_k=1^infty int_0^1(x^n+x^n-2)frac(-1)^k+1kx^kndx\
&=sum_k=1^inftyfrac(-1)^k+1k int_0^1(x^n+x^n-2)x^kndx\
&=sum_k=1^inftyfrac(-1)^k+1k int_0^1(x^n(k+1)+x^n(k+1)-2)dx\
&=sum_k=1^inftyfrac(-1)^k+1k(frac1n(k+1)+1+frac1n(k+1)-1)\
&=sum_k=1^inftyfrac(-1)^k+1k(frac2n(k+1)n^2(k+1)^2-1)\
&=sum_k=1^inftyfrac(-1)^k+1kn(frac2(k+1)(k+1)^2-1/n^2)\
&=sum_k=1^inftyfrac(-1)^k+1k(k+1)n(frac21-1/(n^2(k+1)^2))\
&=frac2nsum_k=1^inftyfrac(-1)^k+1k(k+1)sum_m=0^inftyfrac1n^2m(k+1)^2m\
&=frac2nsum_m=0^inftyfrac1n^2msum_k=1^inftyfrac(-1)^k+1k(k+1)frac1(k+1)^2m\
&=2sum_m=0^inftyfrac1n^2m+1sum_k=1^inftyfrac(-1)^k+1k(k+1)^2m+1\
&=2(fracln(4/e)n-frac0.110304071913...n^3+O(frac1n^5))
qquadtext(according to Wolfy)\
endarray
$
Note:
The Inverse Symbolic Calculator
did not find anything for
0.110304071913.
answered Jul 21 at 23:05
marty cohen
69.2k446122
answered Jul 21 at 23:05
marty cohen
69.2k446122
answered Jul 21 at 23:05
marty cohen
69.2k446122
answered Jul 21 at 23:05
marty cohen
69.2k446122
69.2k446122
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There is a difference in title $(colorred-)$ and body $(colorred+)$; so, I worked both cases.
Let us consider first
$$int x^a log left(1+x^bright),dx=fracx^a+1 log left(1+x^bright)a+1-frac b1+a int fracx^a+b1+x^b,dx$$
$$int fracx^a+b1+x^b,dx=fracx^a+b+1a+b+1 ,,,
_2F_1left(1,fraca+b+1b;fraca+2b+1b;-x^bright)$$ Using the given bounds and simplifying leads to
$$K_a,b=int_0^1 x^a log left(1+x^bright),dx=fraclog (2)-Phi left(-1,1,fraca+b+1bright)a+1qquad textif qquad Re(b)>0$$ where appears the Hurwitz-Lerch transcendent function.
So,
$$I_n=int_0^1 (x^n-x^n-2)ln(1+x^n),dx=fraclog (2)-Phi left(-1,1,frac2 n+1nright)n+1-fraclog (2)-Phi
left(-1,1,frac2 n-1nright)n-1$$
$$J_n=int_0^1 (x^n+x^n-2)ln(1+x^n),dx=fraclog (2)-Phi
left(-1,1,frac2 n+1nright)n+1+fraclog (2)-Phi left(-1,1,frac2 n-1nright)n-1$$
Now, consider the expansion of $Phi (-1,1,2+epsilon)$; it is
$$(1-log (2))+left(fracpi ^212-1right) epsilon +left(1-frac3 zeta
(3)4right) epsilon ^2+left(frac7 pi ^4720-1right) epsilon
^3+left(1-frac15 zeta (5)16right) epsilon ^4+Oleft(epsilon ^5right)$$ Making $ epsilon=pm frac 1n$ and continuing with Taylor, this gives
$$I_n=frac4-fracpi ^26-4 log (2)n^2+frac-540 zeta (3)+2880-60 pi ^2-7 pi
^4-1440 log (2)360 n^4+Oleft(frac1n^5right)$$
$$J_n=frac4 log (2)-2n+frac9 zeta (3)-36+pi ^2+24 log (2)6
n^3+Oleft(frac1n^5right)$$
For $J_n$, this is the same result as in marty cohen's answer, since
$$frac9 zeta (3)-36+pi ^2+24 log (2)6=2 times 0.110304071913700 $$
This seems to work quite well even for small values of $n$ as shown in the table below
$$left(
beginarrayccccc
n & I_n^exact & I_n^approx & J_n^exact & J_n^approx\
2 & -0.1120487783 & -0.1115478099 & 0.4158382364 & 0.4138703791 \
3 & -0.0478495232 & -0.0478071470 & 0.2659498411 & 0.2657002461 \
4 & -0.0265505658 & -0.0265431189 & 0.1966526582 & 0.1965941828 \
5 & -0.0168863305 & -0.0168843897 & 0.1563016585 & 0.1562826096 \
6 & -0.0116869861 & -0.0116863381 & 0.1297937520 & 0.1297861210 \
7 & -0.0085688906 & -0.0085686341 & 0.1110165137 & 0.1110129899 \
8 & -0.0065519051 & -0.0065517901 & 0.0970062707 & 0.0970044656 \
9 & -0.0051721371 & -0.0051720804 & 0.0861468097 & 0.0861458088 \
10 & -0.0041867254 & -0.0041866953 & 0.0774800710 & 0.0774794804
endarray
right)$$
answered Jul 22 at 4:06
Claude Leibovici
111k1055126
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There is a difference in title $(colorred-)$ and body $(colorred+)$; so, I worked both cases.
Let us consider first
$$int x^a log left(1+x^bright),dx=fracx^a+1 log left(1+x^bright)a+1-frac b1+a int fracx^a+b1+x^b,dx$$
$$int fracx^a+b1+x^b,dx=fracx^a+b+1a+b+1 ,,,
_2F_1left(1,fraca+b+1b;fraca+2b+1b;-x^bright)$$ Using the given bounds and simplifying leads to
$$K_a,b=int_0^1 x^a log left(1+x^bright),dx=fraclog (2)-Phi left(-1,1,fraca+b+1bright)a+1qquad textif qquad Re(b)>0$$ where appears the Hurwitz-Lerch transcendent function.
So,
$$I_n=int_0^1 (x^n-x^n-2)ln(1+x^n),dx=fraclog (2)-Phi left(-1,1,frac2 n+1nright)n+1-fraclog (2)-Phi
left(-1,1,frac2 n-1nright)n-1$$
$$J_n=int_0^1 (x^n+x^n-2)ln(1+x^n),dx=fraclog (2)-Phi
left(-1,1,frac2 n+1nright)n+1+fraclog (2)-Phi left(-1,1,frac2 n-1nright)n-1$$
Now, consider the expansion of $Phi (-1,1,2+epsilon)$; it is
$$(1-log (2))+left(fracpi ^212-1right) epsilon +left(1-frac3 zeta
(3)4right) epsilon ^2+left(frac7 pi ^4720-1right) epsilon
^3+left(1-frac15 zeta (5)16right) epsilon ^4+Oleft(epsilon ^5right)$$ Making $ epsilon=pm frac 1n$ and continuing with Taylor, this gives
$$I_n=frac4-fracpi ^26-4 log (2)n^2+frac-540 zeta (3)+2880-60 pi ^2-7 pi
^4-1440 log (2)360 n^4+Oleft(frac1n^5right)$$
$$J_n=frac4 log (2)-2n+frac9 zeta (3)-36+pi ^2+24 log (2)6
n^3+Oleft(frac1n^5right)$$
For $J_n$, this is the same result as in marty cohen's answer, since
$$frac9 zeta (3)-36+pi ^2+24 log (2)6=2 times 0.110304071913700 $$
This seems to work quite well even for small values of $n$ as shown in the table below
$$left(
beginarrayccccc
n & I_n^exact & I_n^approx & J_n^exact & J_n^approx\
2 & -0.1120487783 & -0.1115478099 & 0.4158382364 & 0.4138703791 \
3 & -0.0478495232 & -0.0478071470 & 0.2659498411 & 0.2657002461 \
4 & -0.0265505658 & -0.0265431189 & 0.1966526582 & 0.1965941828 \
5 & -0.0168863305 & -0.0168843897 & 0.1563016585 & 0.1562826096 \
6 & -0.0116869861 & -0.0116863381 & 0.1297937520 & 0.1297861210 \
7 & -0.0085688906 & -0.0085686341 & 0.1110165137 & 0.1110129899 \
8 & -0.0065519051 & -0.0065517901 & 0.0970062707 & 0.0970044656 \
9 & -0.0051721371 & -0.0051720804 & 0.0861468097 & 0.0861458088 \
10 & -0.0041867254 & -0.0041866953 & 0.0774800710 & 0.0774794804
endarray
right)$$
answered Jul 22 at 4:06
Claude Leibovici
111k1055126
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There is a difference in title $(colorred-)$ and body $(colorred+)$; so, I worked both cases.
Let us consider first
$$int x^a log left(1+x^bright),dx=fracx^a+1 log left(1+x^bright)a+1-frac b1+a int fracx^a+b1+x^b,dx$$
$$int fracx^a+b1+x^b,dx=fracx^a+b+1a+b+1 ,,,
_2F_1left(1,fraca+b+1b;fraca+2b+1b;-x^bright)$$ Using the given bounds and simplifying leads to
$$K_a,b=int_0^1 x^a log left(1+x^bright),dx=fraclog (2)-Phi left(-1,1,fraca+b+1bright)a+1qquad textif qquad Re(b)>0$$ where appears the Hurwitz-Lerch transcendent function.
So,
$$I_n=int_0^1 (x^n-x^n-2)ln(1+x^n),dx=fraclog (2)-Phi left(-1,1,frac2 n+1nright)n+1-fraclog (2)-Phi
left(-1,1,frac2 n-1nright)n-1$$
$$J_n=int_0^1 (x^n+x^n-2)ln(1+x^n),dx=fraclog (2)-Phi
left(-1,1,frac2 n+1nright)n+1+fraclog (2)-Phi left(-1,1,frac2 n-1nright)n-1$$
Now, consider the expansion of $Phi (-1,1,2+epsilon)$; it is
$$(1-log (2))+left(fracpi ^212-1right) epsilon +left(1-frac3 zeta
(3)4right) epsilon ^2+left(frac7 pi ^4720-1right) epsilon
^3+left(1-frac15 zeta (5)16right) epsilon ^4+Oleft(epsilon ^5right)$$ Making $ epsilon=pm frac 1n$ and continuing with Taylor, this gives
$$I_n=frac4-fracpi ^26-4 log (2)n^2+frac-540 zeta (3)+2880-60 pi ^2-7 pi
^4-1440 log (2)360 n^4+Oleft(frac1n^5right)$$
$$J_n=frac4 log (2)-2n+frac9 zeta (3)-36+pi ^2+24 log (2)6
n^3+Oleft(frac1n^5right)$$
For $J_n$, this is the same result as in marty cohen's answer, since
$$frac9 zeta (3)-36+pi ^2+24 log (2)6=2 times 0.110304071913700 $$
This seems to work quite well even for small values of $n$ as shown in the table below
$$left(
beginarrayccccc
n & I_n^exact & I_n^approx & J_n^exact & J_n^approx\
2 & -0.1120487783 & -0.1115478099 & 0.4158382364 & 0.4138703791 \
3 & -0.0478495232 & -0.0478071470 & 0.2659498411 & 0.2657002461 \
4 & -0.0265505658 & -0.0265431189 & 0.1966526582 & 0.1965941828 \
5 & -0.0168863305 & -0.0168843897 & 0.1563016585 & 0.1562826096 \
6 & -0.0116869861 & -0.0116863381 & 0.1297937520 & 0.1297861210 \
7 & -0.0085688906 & -0.0085686341 & 0.1110165137 & 0.1110129899 \
8 & -0.0065519051 & -0.0065517901 & 0.0970062707 & 0.0970044656 \
9 & -0.0051721371 & -0.0051720804 & 0.0861468097 & 0.0861458088 \
10 & -0.0041867254 & -0.0041866953 & 0.0774800710 & 0.0774794804
endarray
right)$$
answered Jul 22 at 4:06
Claude Leibovici
111k1055126
There is a difference in title $(colorred-)$ and body $(colorred+)$; so, I worked both cases.
Let us consider first
$$int x^a log left(1+x^bright),dx=fracx^a+1 log left(1+x^bright)a+1-frac b1+a int fracx^a+b1+x^b,dx$$
$$int fracx^a+b1+x^b,dx=fracx^a+b+1a+b+1 ,,,
_2F_1left(1,fraca+b+1b;fraca+2b+1b;-x^bright)$$ Using the given bounds and simplifying leads to
$$K_a,b=int_0^1 x^a log left(1+x^bright),dx=fraclog (2)-Phi left(-1,1,fraca+b+1bright)a+1qquad textif qquad Re(b)>0$$ where appears the Hurwitz-Lerch transcendent function.
So,
$$I_n=int_0^1 (x^n-x^n-2)ln(1+x^n),dx=fraclog (2)-Phi left(-1,1,frac2 n+1nright)n+1-fraclog (2)-Phi
left(-1,1,frac2 n-1nright)n-1$$
$$J_n=int_0^1 (x^n+x^n-2)ln(1+x^n),dx=fraclog (2)-Phi
left(-1,1,frac2 n+1nright)n+1+fraclog (2)-Phi left(-1,1,frac2 n-1nright)n-1$$
Now, consider the expansion of $Phi (-1,1,2+epsilon)$; it is
$$(1-log (2))+left(fracpi ^212-1right) epsilon +left(1-frac3 zeta
(3)4right) epsilon ^2+left(frac7 pi ^4720-1right) epsilon
^3+left(1-frac15 zeta (5)16right) epsilon ^4+Oleft(epsilon ^5right)$$ Making $ epsilon=pm frac 1n$ and continuing with Taylor, this gives
$$I_n=frac4-fracpi ^26-4 log (2)n^2+frac-540 zeta (3)+2880-60 pi ^2-7 pi
^4-1440 log (2)360 n^4+Oleft(frac1n^5right)$$
$$J_n=frac4 log (2)-2n+frac9 zeta (3)-36+pi ^2+24 log (2)6
n^3+Oleft(frac1n^5right)$$
For $J_n$, this is the same result as in marty cohen's answer, since
$$frac9 zeta (3)-36+pi ^2+24 log (2)6=2 times 0.110304071913700 $$
This seems to work quite well even for small values of $n$ as shown in the table below
$$left(
beginarrayccccc
n & I_n^exact & I_n^approx & J_n^exact & J_n^approx\
2 & -0.1120487783 & -0.1115478099 & 0.4158382364 & 0.4138703791 \
3 & -0.0478495232 & -0.0478071470 & 0.2659498411 & 0.2657002461 \
4 & -0.0265505658 & -0.0265431189 & 0.1966526582 & 0.1965941828 \
5 & -0.0168863305 & -0.0168843897 & 0.1563016585 & 0.1562826096 \
6 & -0.0116869861 & -0.0116863381 & 0.1297937520 & 0.1297861210 \
7 & -0.0085688906 & -0.0085686341 & 0.1110165137 & 0.1110129899 \
8 & -0.0065519051 & -0.0065517901 & 0.0970062707 & 0.0970044656 \
9 & -0.0051721371 & -0.0051720804 & 0.0861468097 & 0.0861458088 \
10 & -0.0041867254 & -0.0041866953 & 0.0774800710 & 0.0774794804
endarray
right)$$
answered Jul 22 at 4:06
Claude Leibovici
111k1055126
edited Jul 23 at 3:27
answered Jul 22 at 4:06
Claude Leibovici
111k1055126
answered Jul 22 at 4:06
Claude Leibovici
111k1055126
answered Jul 22 at 4:06
Claude Leibovici
111k1055126
111k1055126
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After using $,t:=x^n,$ we get:
$displaystyle intlimits_0^1 (x^n+x^n-2) ln(1+x^n) dx = frac2nintlimits_0^1 (cosh z)|_z=fracln tn ln(1+t) dx$
$hspace3cmdisplaystyle approx frac2nintlimits_0^1 [z^0](cosh z)|_z=fracln tn ln(1+t) dx = frac2 (ln 4 - 1)n $
$displaystyle intlimits_0^1 (x^n-x^n-2) ln(1+x^n) dx = frac2nintlimits_0^1 (sinh z)|_z=fracln tn ln(1+t) dx $
$hspace3cmdisplaystyle approx frac2nintlimits_0^1 [z^1](sinh z)|_z=fracln tn ln(1+t) dx = frac4(1 - ln 2) - fracpi^26n^2 $
You can use more elements of $,cosh,$ or $,sinh,$ to get closer to your integral.
Example:
$displaystyle intlimits_0^1 (x^n+x^n-2) ln(1+x^n) dx approx frac2nintlimits_0^1 [z^0+z^2](cosh z)|_z=fracln tn ln(1+t) dx $
$hspace3cmdisplaystyle =fracln 16 - 2n + frac1n^3left(frac3zeta(3)2+fracpi^26+ln 16 - 6right)$
answered Jul 23 at 11:50
user90369
7,596925
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After using $,t:=x^n,$ we get:
$displaystyle intlimits_0^1 (x^n+x^n-2) ln(1+x^n) dx = frac2nintlimits_0^1 (cosh z)|_z=fracln tn ln(1+t) dx$
$hspace3cmdisplaystyle approx frac2nintlimits_0^1 [z^0](cosh z)|_z=fracln tn ln(1+t) dx = frac2 (ln 4 - 1)n $
$displaystyle intlimits_0^1 (x^n-x^n-2) ln(1+x^n) dx = frac2nintlimits_0^1 (sinh z)|_z=fracln tn ln(1+t) dx $
$hspace3cmdisplaystyle approx frac2nintlimits_0^1 [z^1](sinh z)|_z=fracln tn ln(1+t) dx = frac4(1 - ln 2) - fracpi^26n^2 $
You can use more elements of $,cosh,$ or $,sinh,$ to get closer to your integral.
Example:
$displaystyle intlimits_0^1 (x^n+x^n-2) ln(1+x^n) dx approx frac2nintlimits_0^1 [z^0+z^2](cosh z)|_z=fracln tn ln(1+t) dx $
$hspace3cmdisplaystyle =fracln 16 - 2n + frac1n^3left(frac3zeta(3)2+fracpi^26+ln 16 - 6right)$
answered Jul 23 at 11:50
user90369
7,596925
add a comment |Â
up vote
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up vote
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After using $,t:=x^n,$ we get:
$displaystyle intlimits_0^1 (x^n+x^n-2) ln(1+x^n) dx = frac2nintlimits_0^1 (cosh z)|_z=fracln tn ln(1+t) dx$
$hspace3cmdisplaystyle approx frac2nintlimits_0^1 [z^0](cosh z)|_z=fracln tn ln(1+t) dx = frac2 (ln 4 - 1)n $
$displaystyle intlimits_0^1 (x^n-x^n-2) ln(1+x^n) dx = frac2nintlimits_0^1 (sinh z)|_z=fracln tn ln(1+t) dx $
$hspace3cmdisplaystyle approx frac2nintlimits_0^1 [z^1](sinh z)|_z=fracln tn ln(1+t) dx = frac4(1 - ln 2) - fracpi^26n^2 $
You can use more elements of $,cosh,$ or $,sinh,$ to get closer to your integral.
Example:
$displaystyle intlimits_0^1 (x^n+x^n-2) ln(1+x^n) dx approx frac2nintlimits_0^1 [z^0+z^2](cosh z)|_z=fracln tn ln(1+t) dx $
$hspace3cmdisplaystyle =fracln 16 - 2n + frac1n^3left(frac3zeta(3)2+fracpi^26+ln 16 - 6right)$
answered Jul 23 at 11:50
user90369
7,596925
After using $,t:=x^n,$ we get:
$displaystyle intlimits_0^1 (x^n+x^n-2) ln(1+x^n) dx = frac2nintlimits_0^1 (cosh z)|_z=fracln tn ln(1+t) dx$
$hspace3cmdisplaystyle approx frac2nintlimits_0^1 [z^0](cosh z)|_z=fracln tn ln(1+t) dx = frac2 (ln 4 - 1)n $
$displaystyle intlimits_0^1 (x^n-x^n-2) ln(1+x^n) dx = frac2nintlimits_0^1 (sinh z)|_z=fracln tn ln(1+t) dx $
$hspace3cmdisplaystyle approx frac2nintlimits_0^1 [z^1](sinh z)|_z=fracln tn ln(1+t) dx = frac4(1 - ln 2) - fracpi^26n^2 $
You can use more elements of $,cosh,$ or $,sinh,$ to get closer to your integral.
Example:
$displaystyle intlimits_0^1 (x^n+x^n-2) ln(1+x^n) dx approx frac2nintlimits_0^1 [z^0+z^2](cosh z)|_z=fracln tn ln(1+t) dx $
$hspace3cmdisplaystyle =fracln 16 - 2n + frac1n^3left(frac3zeta(3)2+fracpi^26+ln 16 - 6right)$
answered Jul 23 at 11:50
user90369
7,596925
edited Jul 23 at 13:04
answered Jul 23 at 11:50
user90369
7,596925
answered Jul 23 at 11:50
user90369
7,596925
answered Jul 23 at 11:50
user90369
7,596925
7,596925
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If $f(n, x) = O(|g(n, x)|)$ uniformly in $x$ on $a < x < b$ and $f$ and $g$ are measurable, then $int_a^b f(n, x) dx = O(int_a^b |g(n, x)| dx)$. We have
$$J_n = int_0^1 x^n ln(1 + x^n) dx =
frac 1 n int_0^1 xi^1/n ln(1 + xi) dxi, \
|xi^1/n - 1| leq frac 1 n |! ln xi| quad
textwhen 0 < n land 0 < xi < 1, \
left| J_n - frac 1 n int_0^1 ln(1 + xi) dxi right| leq
-frac 1 n^2 int_0^1 ln xi ln(1 + xi) dxi,$$
giving the leading term in the expansion. The complete asymptotic series for $J_n$ in powers of $n$ is
$$J_n sim sum_k=1^infty
int_0^1 ln^k-1 xi ln(1 + xi) dxi frac n^-k (k-1)!.$$
Similarly, for
$$K_n = int_0^1 x^n-2 ln(1 + x^n) dx =
frac 1 n int_0^1 xi^-1/n ln(1 + xi) dxi,$$
we can take
$$|xi^-1/n - 1| leq frac 2 n sqrt xi quad
textwhen 2 < n land 0 < xi < 1.$$
The complete asymptotic series for $K_n$ is
$$K_n sim sum_k=1^infty (-1)^k-1
int_0^1 ln^k-1 xi ln(1 + xi) dxi frac n^-k (k-1)!,$$
which is formally $-J_-n$.
answered Jul 22 at 21:30
Maxim
2,060113
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If $f(n, x) = O(|g(n, x)|)$ uniformly in $x$ on $a < x < b$ and $f$ and $g$ are measurable, then $int_a^b f(n, x) dx = O(int_a^b |g(n, x)| dx)$. We have
$$J_n = int_0^1 x^n ln(1 + x^n) dx =
frac 1 n int_0^1 xi^1/n ln(1 + xi) dxi, \
|xi^1/n - 1| leq frac 1 n |! ln xi| quad
textwhen 0 < n land 0 < xi < 1, \
left| J_n - frac 1 n int_0^1 ln(1 + xi) dxi right| leq
-frac 1 n^2 int_0^1 ln xi ln(1 + xi) dxi,$$
giving the leading term in the expansion. The complete asymptotic series for $J_n$ in powers of $n$ is
$$J_n sim sum_k=1^infty
int_0^1 ln^k-1 xi ln(1 + xi) dxi frac n^-k (k-1)!.$$
Similarly, for
$$K_n = int_0^1 x^n-2 ln(1 + x^n) dx =
frac 1 n int_0^1 xi^-1/n ln(1 + xi) dxi,$$
we can take
$$|xi^-1/n - 1| leq frac 2 n sqrt xi quad
textwhen 2 < n land 0 < xi < 1.$$
The complete asymptotic series for $K_n$ is
$$K_n sim sum_k=1^infty (-1)^k-1
int_0^1 ln^k-1 xi ln(1 + xi) dxi frac n^-k (k-1)!,$$
which is formally $-J_-n$.
answered Jul 22 at 21:30
Maxim
2,060113
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $f(n, x) = O(|g(n, x)|)$ uniformly in $x$ on $a < x < b$ and $f$ and $g$ are measurable, then $int_a^b f(n, x) dx = O(int_a^b |g(n, x)| dx)$. We have
$$J_n = int_0^1 x^n ln(1 + x^n) dx =
frac 1 n int_0^1 xi^1/n ln(1 + xi) dxi, \
|xi^1/n - 1| leq frac 1 n |! ln xi| quad
textwhen 0 < n land 0 < xi < 1, \
left| J_n - frac 1 n int_0^1 ln(1 + xi) dxi right| leq
-frac 1 n^2 int_0^1 ln xi ln(1 + xi) dxi,$$
giving the leading term in the expansion. The complete asymptotic series for $J_n$ in powers of $n$ is
$$J_n sim sum_k=1^infty
int_0^1 ln^k-1 xi ln(1 + xi) dxi frac n^-k (k-1)!.$$
Similarly, for
$$K_n = int_0^1 x^n-2 ln(1 + x^n) dx =
frac 1 n int_0^1 xi^-1/n ln(1 + xi) dxi,$$
we can take
$$|xi^-1/n - 1| leq frac 2 n sqrt xi quad
textwhen 2 < n land 0 < xi < 1.$$
The complete asymptotic series for $K_n$ is
$$K_n sim sum_k=1^infty (-1)^k-1
int_0^1 ln^k-1 xi ln(1 + xi) dxi frac n^-k (k-1)!,$$
which is formally $-J_-n$.
answered Jul 22 at 21:30
Maxim
2,060113
If $f(n, x) = O(|g(n, x)|)$ uniformly in $x$ on $a < x < b$ and $f$ and $g$ are measurable, then $int_a^b f(n, x) dx = O(int_a^b |g(n, x)| dx)$. We have
$$J_n = int_0^1 x^n ln(1 + x^n) dx =
frac 1 n int_0^1 xi^1/n ln(1 + xi) dxi, \
|xi^1/n - 1| leq frac 1 n |! ln xi| quad
textwhen 0 < n land 0 < xi < 1, \
left| J_n - frac 1 n int_0^1 ln(1 + xi) dxi right| leq
-frac 1 n^2 int_0^1 ln xi ln(1 + xi) dxi,$$
giving the leading term in the expansion. The complete asymptotic series for $J_n$ in powers of $n$ is
$$J_n sim sum_k=1^infty
int_0^1 ln^k-1 xi ln(1 + xi) dxi frac n^-k (k-1)!.$$
Similarly, for
$$K_n = int_0^1 x^n-2 ln(1 + x^n) dx =
frac 1 n int_0^1 xi^-1/n ln(1 + xi) dxi,$$
we can take
$$|xi^-1/n - 1| leq frac 2 n sqrt xi quad
textwhen 2 < n land 0 < xi < 1.$$
The complete asymptotic series for $K_n$ is
$$K_n sim sum_k=1^infty (-1)^k-1
int_0^1 ln^k-1 xi ln(1 + xi) dxi frac n^-k (k-1)!,$$
which is formally $-J_-n$.
answered Jul 22 at 21:30
Maxim
2,060113
edited Jul 23 at 17:14
answered Jul 22 at 21:30
Maxim
2,060113
answered Jul 22 at 21:30
Maxim
2,060113
answered Jul 22 at 21:30
Maxim
2,060113
2,060113
add a comment |Â
add a comment |Â
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1
@Michael: If $f_nto f$ uniformly on $[0,r]$ for all $r<1$ but not on $[0,1[$, then$$lim_nto infty lim_rto 1int_0^r f_n=lim_rto 1lim_nto infty int_0^r f_n$$ is false a priori.
– Surb
Jul 21 at 22:38
Maybe you are missing the left hand side for the second equation. Maybe you are missing $dx$ for the third equation. If so, please revise the question.
– norio
Jul 21 at 22:42
@Surb In our case it is true, because the sum of the intergrals convergences uniformly.
– Michael
Jul 22 at 9:07
1
@Surb $sum_k=1^infty int_0^r(x^n+x^n-2)frac(-1)^k+1kx^kndx=sum_k=1^inftyfrac(-1)^k+1k(dfracr^n(k+1)+1n(k+1)+1+dfracr^n(k+1)-1n(k+1)-1)$. The right hand side converges uniformly in $r$, because $rleq 1$ and you essentially have the sum of $1/k(k+1)$.
– Michael
Jul 22 at 9:32
1
The title and question do not match.
– zhw.
Jul 23 at 1:07