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Find the infinite sum of the series $sum_n=1^infty frac1n^2 +1$
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This is a homework question whereby I am supposed to evaluate:
$$sum_n=1^infty frac1n^2 +1$$
Wolfram Alpha outputs the answer as
$$frac12(pi coth(pi) - 1)$$
But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked.
calculus sequences-and-series
edited Apr 2 '14 at 17:16
A.D
3,8251235
asked Apr 2 '14 at 16:43
Briandmg
17018
 |Â
show 3 more comments
up vote
21
down vote
favorite
This is a homework question whereby I am supposed to evaluate:
$$sum_n=1^infty frac1n^2 +1$$
Wolfram Alpha outputs the answer as
$$frac12(pi coth(pi) - 1)$$
But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked.
calculus sequences-and-series
edited Apr 2 '14 at 17:16
A.D
3,8251235
asked Apr 2 '14 at 16:43
Briandmg
17018
Have you studied the method of series calculation using Fourier series?
– Frank
Apr 2 '14 at 16:45
(no idea how to use fourier series)You can estimate through $arctan(x)$ though
– Guy
Apr 2 '14 at 16:46
1
@Frank What's the function that works here?
– Git Gud
Apr 2 '14 at 16:47
1
@Git Gud Was wondering the same thing. I have studied it but the function isn't immediately obvious.
– Briandmg
Apr 2 '14 at 16:48
9
This can also be done via residue calculus. Alternatively, with a bit of handwaving, use the Weierstrass prod. for sin: $$x^-1sin x=prod_k Big(1-x^2(pi k)^-2Big)Rightarrow (ipi x)^-1sin ipi x=prod_k Big(1+x^2k^-2Big)$$ Take logs, differentiate, and tada: $sum (k^2+x^2)^-1=frac12x^2(pi xcoth pi x-1)$
– Julien Godawatta
Apr 2 '14 at 16:51
 |Â
show 3 more comments
up vote
21
down vote
favorite
up vote
21
down vote
favorite
This is a homework question whereby I am supposed to evaluate:
$$sum_n=1^infty frac1n^2 +1$$
Wolfram Alpha outputs the answer as
$$frac12(pi coth(pi) - 1)$$
But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked.
calculus sequences-and-series
edited Apr 2 '14 at 17:16
A.D
3,8251235
asked Apr 2 '14 at 16:43
Briandmg
17018
This is a homework question whereby I am supposed to evaluate:
$$sum_n=1^infty frac1n^2 +1$$
Wolfram Alpha outputs the answer as
$$frac12(pi coth(pi) - 1)$$
But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked.
calculus sequences-and-series
edited Apr 2 '14 at 17:16
A.D
3,8251235
asked Apr 2 '14 at 16:43
Briandmg
17018
edited Apr 2 '14 at 17:16
A.D
3,8251235
edited Apr 2 '14 at 17:16
A.D
3,8251235
edited Apr 2 '14 at 17:16
A.D
3,8251235
3,8251235
asked Apr 2 '14 at 16:43
Briandmg
17018
asked Apr 2 '14 at 16:43
Briandmg
17018
asked Apr 2 '14 at 16:43
Briandmg
17018
17018
Have you studied the method of series calculation using Fourier series?
– Frank
Apr 2 '14 at 16:45
(no idea how to use fourier series)You can estimate through $arctan(x)$ though
– Guy
Apr 2 '14 at 16:46
1
@Frank What's the function that works here?
– Git Gud
Apr 2 '14 at 16:47
1
@Git Gud Was wondering the same thing. I have studied it but the function isn't immediately obvious.
– Briandmg
Apr 2 '14 at 16:48
9
This can also be done via residue calculus. Alternatively, with a bit of handwaving, use the Weierstrass prod. for sin: $$x^-1sin x=prod_k Big(1-x^2(pi k)^-2Big)Rightarrow (ipi x)^-1sin ipi x=prod_k Big(1+x^2k^-2Big)$$ Take logs, differentiate, and tada: $sum (k^2+x^2)^-1=frac12x^2(pi xcoth pi x-1)$
– Julien Godawatta
Apr 2 '14 at 16:51
 |Â
show 3 more comments
Have you studied the method of series calculation using Fourier series?
– Frank
Apr 2 '14 at 16:45
(no idea how to use fourier series)You can estimate through $arctan(x)$ though
– Guy
Apr 2 '14 at 16:46
1
@Frank What's the function that works here?
– Git Gud
Apr 2 '14 at 16:47
1
@Git Gud Was wondering the same thing. I have studied it but the function isn't immediately obvious.
– Briandmg
Apr 2 '14 at 16:48
9
This can also be done via residue calculus. Alternatively, with a bit of handwaving, use the Weierstrass prod. for sin: $$x^-1sin x=prod_k Big(1-x^2(pi k)^-2Big)Rightarrow (ipi x)^-1sin ipi x=prod_k Big(1+x^2k^-2Big)$$ Take logs, differentiate, and tada: $sum (k^2+x^2)^-1=frac12x^2(pi xcoth pi x-1)$
– Julien Godawatta
Apr 2 '14 at 16:51
Have you studied the method of series calculation using Fourier series?
– Frank
Apr 2 '14 at 16:45
Have you studied the method of series calculation using Fourier series?
– Frank
Apr 2 '14 at 16:45
(no idea how to use fourier series)You can estimate through $arctan(x)$ though
– Guy
Apr 2 '14 at 16:46
(no idea how to use fourier series)You can estimate through $arctan(x)$ though
– Guy
Apr 2 '14 at 16:46
1
1
@Frank What's the function that works here?
– Git Gud
Apr 2 '14 at 16:47
@Frank What's the function that works here?
– Git Gud
Apr 2 '14 at 16:47
1
1
@Git Gud Was wondering the same thing. I have studied it but the function isn't immediately obvious.
– Briandmg
Apr 2 '14 at 16:48
@Git Gud Was wondering the same thing. I have studied it but the function isn't immediately obvious.
– Briandmg
Apr 2 '14 at 16:48
9
9
This can also be done via residue calculus. Alternatively, with a bit of handwaving, use the Weierstrass prod. for sin: $$x^-1sin x=prod_k Big(1-x^2(pi k)^-2Big)Rightarrow (ipi x)^-1sin ipi x=prod_k Big(1+x^2k^-2Big)$$ Take logs, differentiate, and tada: $sum (k^2+x^2)^-1=frac12x^2(pi xcoth pi x-1)$
– Julien Godawatta
Apr 2 '14 at 16:51
This can also be done via residue calculus. Alternatively, with a bit of handwaving, use the Weierstrass prod. for sin: $$x^-1sin x=prod_k Big(1-x^2(pi k)^-2Big)Rightarrow (ipi x)^-1sin ipi x=prod_k Big(1+x^2k^-2Big)$$ Take logs, differentiate, and tada: $sum (k^2+x^2)^-1=frac12x^2(pi xcoth pi x-1)$
– Julien Godawatta
Apr 2 '14 at 16:51
 |Â
show 3 more comments
5 Answers
5
active
oldest
votes
up vote
14
down vote
accepted
Using David Cardon's method, https://mathoverflow.net/questions/59645/algebraic-proof-of-an-infinite-sum
We can solve a more general sum,
$$sum_-infty^infty frac1n^2+a^2 = fracpia coth(pi a).$$
Note that this sum satisfies the conditions in the above link. The poles lie at $z=ia$ and $z=-ia$, so
$$sum_n=-infty^infty frac1n^2+a^2 = -pileft[operatornameResleft(fraccot(pi z)z^2+a^2,iaright) + operatornameResleft(fraccot(pi z)z^2+a^2,-iaright)right].$$
Computing the residues:
$$operatornameResleft(fraccot(pi z)z^2+a^2,iaright) = lim_zrightarrow iafrac(z-ia)cot(pi z)(z-ia)(z+ia) = fraccot(pi ia)2i a $$
and
$$ operatornameResleft(fraccot(pi z)z^2+a^2,-iaright) = lim_zrightarrow -iafrac(z+ia)cot(pi z)(z+ia)(z-ia) = fraccot(ipi a)2ia.$$
Therefore, summing these we get
$$sum_-infty^infty frac1n^2+a^2 = -fracpicot(ipi a)ia = fracpi coth(pi a)a.$$
You should be able to extend this idea to your sum with some effort.
edited Apr 13 '17 at 12:58
Community♦
1
answered Apr 2 '14 at 17:37
JessicaK
4,29051633
1
The following MSE link describes the same method if the OP is interested in seeing another example.
– Marko Riedel
Apr 2 '14 at 19:44
add a comment |Â
up vote
7
down vote
$newcommand+^dagger
newcommandangles[1]leftlangle, #1 ,rightrangle
newcommandbraces[1]leftlbrace, #1 ,rightrbrace
newcommandbracks[1]leftlbrack, #1 ,rightrbrack
newcommandceil[1],leftlceil, #1 ,rightrceil,
newcommandddrm d
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newcommandexpo[1],rm e^#1,
newcommandfermi,rm f
newcommandfloor[1],leftlfloor #1 rightrfloor,
newcommandhalf1 over 2
newcommandicrm i
newcommandiffLongleftrightarrow
newcommandimpLongrightarrow
newcommandisdiv,left.rightvert,
newcommandket[1]leftvert #1rightrangle
newcommandol[1]overline#1
newcommandpars[1]left(, #1 ,right)
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newcommandppcal P
newcommandroot[2],sqrt[#1]vphantomlarge A,#2,,
newcommandsech,rm sech
newcommandsgn,rm sgn
newcommandtotald[3]fracrm d^#1 #2rm d #3^#1
newcommandul[1]underline#1
newcommandverts[1]leftvert, #1 ,rightvert
newcommandwt[1]widetilde#1$
beginalign
sum_n = 1^infty1 over n^2 + 1&=
sum_n = 0^infty1 over parsn + 1 + icparsn + 1 - ic
=Psipars1 + ic - Psipars1 - ic over pars1 + ic - pars1 - ic
=ImPsipars1 + ic
endalign
where $Psiparsz$ is the Digamma Function.
With the identity $dsImPsipars1 + ic y = -,1 over 2y + half,picothparspi y$ we'll have:
$$color#00flarge%
sum_n = 1^infty1 over n^2 + 1 = halfbrackspicothparspi - 1
$$
answered Apr 2 '14 at 16:57
Felix Marin
65.5k7105135
add a comment |Â
up vote
5
down vote
There's a little bit of calculation you need to do here to make sure Cauchy's Residue Theorem is applicable here (you need to make sure that certain integrals are bounded etc) but this is a sketch:
Let
$$
f(z) = frac pi (1+z^2)tan(pi z)
$$
Then $f$ has simple poles $forall n in mathbbZ$ and also at $pm i$.
You can calculate the residues as
$$
textRes(f(z), npi) = frac 1 1 + n^2
$$
and
$$ textRes(f(z), pm i) = frac -pi 2tanh(pi)
$$
If these are hard to calculate for you, I can give you more detail.
Now, let $Gamma _N$ be the square contour with vertices $(N + frac 1 2) (pm 1 pm i)$
Then Cauchy's Residue Theorem tells us
$$
int_Gamma N f(z) dz = 2pi i sum textRes(f(z), z)
$$
Where the sum is across all the poles inside the contour.
Now, we see that the simple poles inside the contour are all the ones at integers $n$ s.t. $|n| < N$ and the ones at $pm i$.
So
$$
int_Gamma N f(z) dz = 2pi i left [ frac -2pi 2 tanh(pi) + sum_n = -N^N frac1 1 + n^2 right ]
$$
Now we need to show that the integral on the left goes to zero as $N$ tends to infinity. Now, I'm going to leave this as an exercise for you (unlucky) but the basic idea is to find a constant $C_1$ which bounds $frac pi tan(pi z)$ on the top and bottom of the square and another constant $C_2$ on the sides of the square and take $C$ to be the maximum of these.
Then by the Estimation Theorem we would have that
$$
left | int_Gamma_N f(z) dz right | leq textlength(Gamma_N) textsup_z in Gamma _N |f(z)| leq 4(2N + 1) C textsup_z in Gamma_N left| frac 1 1 + z^2right | leq frac 4C(2N + 1) 1 + N^2 = O(frac 1 N)
$$
So we let $N to infty$ then we get
$$
0 = 2pi i left [ frac -2pi 2 tanh(pi) + sum_n = -infty^infty frac1 1 + n^2 right ]
$$
So then we have
$$
0 = frac -pi tanh(pi) + 2sum_n = 1^infty frac1 1 + n^2 + 1
$$
Where the 1 that has randomly appeared is the $n = 0 $ term
So
$$
sum_n = 1^infty frac1 1 + n^2 = frac 1 2 left [frac pi tanh(pi) - 1 right] = frac 1 2 (pi coth (pi) - 1).
$$
I hope you know some complex analysis otherwise this might have meant nothing to you...
This method works for most sums though (as long as you get the right things tending to zero, which you do in this case) and I don't think I've ever had it not work.
If you want to evaluate
$$ sum_n = 1 ^ infty phi(n)
$$
Where phi can easily be extended to all of $mathbb C$ you just take
$$f(z) = frac pi phi(z) tan(pi z)
$$
And do the same thing, and if you want to evaluate
$$sum_n = 1 ^ infty (-1)^n phi(n)
$$
You just take
$$f(z) = frac pi phi(z) sin(pi z)
$$
In each case integrating over the same contour. Hope this helps, and if you don't know much complex analysis, you should learn more, it is a very interesting and powerful area. Sorry this was the longest answer ever.
edited Jun 26 '14 at 15:20
Meow
2,51732453
answered Apr 2 '14 at 17:25
CameronJWhitehead
1,361517
add a comment |Â
up vote
4
down vote
This is related to this answer, where it is shown that
$$
sum_k=1^inftyfrac1k^2-z^2
=frac12zleft[frac1z-picot(pi z)right]tag1
$$
Equation $(1)$ is valid for all $zinmathbbC$, in particular for $z=i$, which gives
$$
sum_k=1^inftyfrac1k^2+1
=frac12left[picoth(pi)-1right]tag2
$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Apr 7 '14 at 11:45
robjohn♦
258k26297612
add a comment |Â
up vote
4
down vote
We can start from the Weierstrass product for the $sinh$ function:
$$fracsinh zz=prod_n=1^+inftyleft(1+fracz^2pi^2 n^2right)tag1 $$
then consider the logarithmic derivative of both sides. This leads to:
$$coth z-frac1z=sum_n=1^+inftyfrac2zz^2+pi^2 n^2tag2 $$
or to:
$$picoth(pi w)-frac1w=sum_n=1^+inftyfrac2ww^2+ n^2.tag3 $$
Now just set $w=1$ in $(3)$.
Further approach: by the Poisson summation formula, since the Laplace distribution and the Cauchy distribution are related via the Fourier transform, we have that $sum_ngeq 0frac1n^2+1$ is simply related with $sum_ngeq 0e^-pi n$, which is a geometric series.
answered Jan 10 '15 at 22:39
Jack D'Aurizio♦
270k31264629
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
accepted
Using David Cardon's method, https://mathoverflow.net/questions/59645/algebraic-proof-of-an-infinite-sum
We can solve a more general sum,
$$sum_-infty^infty frac1n^2+a^2 = fracpia coth(pi a).$$
Note that this sum satisfies the conditions in the above link. The poles lie at $z=ia$ and $z=-ia$, so
$$sum_n=-infty^infty frac1n^2+a^2 = -pileft[operatornameResleft(fraccot(pi z)z^2+a^2,iaright) + operatornameResleft(fraccot(pi z)z^2+a^2,-iaright)right].$$
Computing the residues:
$$operatornameResleft(fraccot(pi z)z^2+a^2,iaright) = lim_zrightarrow iafrac(z-ia)cot(pi z)(z-ia)(z+ia) = fraccot(pi ia)2i a $$
and
$$ operatornameResleft(fraccot(pi z)z^2+a^2,-iaright) = lim_zrightarrow -iafrac(z+ia)cot(pi z)(z+ia)(z-ia) = fraccot(ipi a)2ia.$$
Therefore, summing these we get
$$sum_-infty^infty frac1n^2+a^2 = -fracpicot(ipi a)ia = fracpi coth(pi a)a.$$
You should be able to extend this idea to your sum with some effort.
edited Apr 13 '17 at 12:58
Community♦
1
answered Apr 2 '14 at 17:37
JessicaK
4,29051633
1
The following MSE link describes the same method if the OP is interested in seeing another example.
– Marko Riedel
Apr 2 '14 at 19:44
add a comment |Â
up vote
14
down vote
accepted
Using David Cardon's method, https://mathoverflow.net/questions/59645/algebraic-proof-of-an-infinite-sum
We can solve a more general sum,
$$sum_-infty^infty frac1n^2+a^2 = fracpia coth(pi a).$$
Note that this sum satisfies the conditions in the above link. The poles lie at $z=ia$ and $z=-ia$, so
$$sum_n=-infty^infty frac1n^2+a^2 = -pileft[operatornameResleft(fraccot(pi z)z^2+a^2,iaright) + operatornameResleft(fraccot(pi z)z^2+a^2,-iaright)right].$$
Computing the residues:
$$operatornameResleft(fraccot(pi z)z^2+a^2,iaright) = lim_zrightarrow iafrac(z-ia)cot(pi z)(z-ia)(z+ia) = fraccot(pi ia)2i a $$
and
$$ operatornameResleft(fraccot(pi z)z^2+a^2,-iaright) = lim_zrightarrow -iafrac(z+ia)cot(pi z)(z+ia)(z-ia) = fraccot(ipi a)2ia.$$
Therefore, summing these we get
$$sum_-infty^infty frac1n^2+a^2 = -fracpicot(ipi a)ia = fracpi coth(pi a)a.$$
You should be able to extend this idea to your sum with some effort.
edited Apr 13 '17 at 12:58
Community♦
1
answered Apr 2 '14 at 17:37
JessicaK
4,29051633
1
The following MSE link describes the same method if the OP is interested in seeing another example.
– Marko Riedel
Apr 2 '14 at 19:44
add a comment |Â
up vote
14
down vote
accepted
up vote
14
down vote
accepted
Using David Cardon's method, https://mathoverflow.net/questions/59645/algebraic-proof-of-an-infinite-sum
We can solve a more general sum,
$$sum_-infty^infty frac1n^2+a^2 = fracpia coth(pi a).$$
Note that this sum satisfies the conditions in the above link. The poles lie at $z=ia$ and $z=-ia$, so
$$sum_n=-infty^infty frac1n^2+a^2 = -pileft[operatornameResleft(fraccot(pi z)z^2+a^2,iaright) + operatornameResleft(fraccot(pi z)z^2+a^2,-iaright)right].$$
Computing the residues:
$$operatornameResleft(fraccot(pi z)z^2+a^2,iaright) = lim_zrightarrow iafrac(z-ia)cot(pi z)(z-ia)(z+ia) = fraccot(pi ia)2i a $$
and
$$ operatornameResleft(fraccot(pi z)z^2+a^2,-iaright) = lim_zrightarrow -iafrac(z+ia)cot(pi z)(z+ia)(z-ia) = fraccot(ipi a)2ia.$$
Therefore, summing these we get
$$sum_-infty^infty frac1n^2+a^2 = -fracpicot(ipi a)ia = fracpi coth(pi a)a.$$
You should be able to extend this idea to your sum with some effort.
edited Apr 13 '17 at 12:58
Community♦
1
answered Apr 2 '14 at 17:37
JessicaK
4,29051633
Using David Cardon's method, https://mathoverflow.net/questions/59645/algebraic-proof-of-an-infinite-sum
We can solve a more general sum,
$$sum_-infty^infty frac1n^2+a^2 = fracpia coth(pi a).$$
Note that this sum satisfies the conditions in the above link. The poles lie at $z=ia$ and $z=-ia$, so
$$sum_n=-infty^infty frac1n^2+a^2 = -pileft[operatornameResleft(fraccot(pi z)z^2+a^2,iaright) + operatornameResleft(fraccot(pi z)z^2+a^2,-iaright)right].$$
Computing the residues:
$$operatornameResleft(fraccot(pi z)z^2+a^2,iaright) = lim_zrightarrow iafrac(z-ia)cot(pi z)(z-ia)(z+ia) = fraccot(pi ia)2i a $$
and
$$ operatornameResleft(fraccot(pi z)z^2+a^2,-iaright) = lim_zrightarrow -iafrac(z+ia)cot(pi z)(z+ia)(z-ia) = fraccot(ipi a)2ia.$$
Therefore, summing these we get
$$sum_-infty^infty frac1n^2+a^2 = -fracpicot(ipi a)ia = fracpi coth(pi a)a.$$
You should be able to extend this idea to your sum with some effort.
edited Apr 13 '17 at 12:58
Community♦
1
answered Apr 2 '14 at 17:37
JessicaK
4,29051633
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
1
answered Apr 2 '14 at 17:37
JessicaK
4,29051633
answered Apr 2 '14 at 17:37
JessicaK
4,29051633
answered Apr 2 '14 at 17:37
JessicaK
4,29051633
4,29051633
1
The following MSE link describes the same method if the OP is interested in seeing another example.
– Marko Riedel
Apr 2 '14 at 19:44
add a comment |Â
1
The following MSE link describes the same method if the OP is interested in seeing another example.
– Marko Riedel
Apr 2 '14 at 19:44
1
1
The following MSE link describes the same method if the OP is interested in seeing another example.
– Marko Riedel
Apr 2 '14 at 19:44
The following MSE link describes the same method if the OP is interested in seeing another example.
– Marko Riedel
Apr 2 '14 at 19:44
add a comment |Â
up vote
7
down vote
$newcommand+^dagger
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newcommandverts[1]leftvert, #1 ,rightvert
newcommandwt[1]widetilde#1$
beginalign
sum_n = 1^infty1 over n^2 + 1&=
sum_n = 0^infty1 over parsn + 1 + icparsn + 1 - ic
=Psipars1 + ic - Psipars1 - ic over pars1 + ic - pars1 - ic
=ImPsipars1 + ic
endalign
where $Psiparsz$ is the Digamma Function.
With the identity $dsImPsipars1 + ic y = -,1 over 2y + half,picothparspi y$ we'll have:
$$color#00flarge%
sum_n = 1^infty1 over n^2 + 1 = halfbrackspicothparspi - 1
$$
answered Apr 2 '14 at 16:57
Felix Marin
65.5k7105135
add a comment |Â
up vote
7
down vote
$newcommand+^dagger
newcommandangles[1]leftlangle, #1 ,rightrangle
newcommandbraces[1]leftlbrace, #1 ,rightrbrace
newcommandbracks[1]leftlbrack, #1 ,rightrbrack
newcommandceil[1],leftlceil, #1 ,rightrceil,
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newcommanddowndownarrow
newcommandds[1]displaystyle#1
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newcommandsech,rm sech
newcommandsgn,rm sgn
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newcommandul[1]underline#1
newcommandverts[1]leftvert, #1 ,rightvert
newcommandwt[1]widetilde#1$
beginalign
sum_n = 1^infty1 over n^2 + 1&=
sum_n = 0^infty1 over parsn + 1 + icparsn + 1 - ic
=Psipars1 + ic - Psipars1 - ic over pars1 + ic - pars1 - ic
=ImPsipars1 + ic
endalign
where $Psiparsz$ is the Digamma Function.
With the identity $dsImPsipars1 + ic y = -,1 over 2y + half,picothparspi y$ we'll have:
$$color#00flarge%
sum_n = 1^infty1 over n^2 + 1 = halfbrackspicothparspi - 1
$$
answered Apr 2 '14 at 16:57
Felix Marin
65.5k7105135
add a comment |Â
up vote
7
down vote
up vote
7
down vote
$newcommand+^dagger
newcommandangles[1]leftlangle, #1 ,rightrangle
newcommandbraces[1]leftlbrace, #1 ,rightrbrace
newcommandbracks[1]leftlbrack, #1 ,rightrbrack
newcommandceil[1],leftlceil, #1 ,rightrceil,
newcommandddrm d
newcommanddowndownarrow
newcommandds[1]displaystyle#1
newcommandexpo[1],rm e^#1,
newcommandfermi,rm f
newcommandfloor[1],leftlfloor #1 rightrfloor,
newcommandhalf1 over 2
newcommandicrm i
newcommandiffLongleftrightarrow
newcommandimpLongrightarrow
newcommandisdiv,left.rightvert,
newcommandket[1]leftvert #1rightrangle
newcommandol[1]overline#1
newcommandpars[1]left(, #1 ,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandppcal P
newcommandroot[2],sqrt[#1]vphantomlarge A,#2,,
newcommandsech,rm sech
newcommandsgn,rm sgn
newcommandtotald[3]fracrm d^#1 #2rm d #3^#1
newcommandul[1]underline#1
newcommandverts[1]leftvert, #1 ,rightvert
newcommandwt[1]widetilde#1$
beginalign
sum_n = 1^infty1 over n^2 + 1&=
sum_n = 0^infty1 over parsn + 1 + icparsn + 1 - ic
=Psipars1 + ic - Psipars1 - ic over pars1 + ic - pars1 - ic
=ImPsipars1 + ic
endalign
where $Psiparsz$ is the Digamma Function.
With the identity $dsImPsipars1 + ic y = -,1 over 2y + half,picothparspi y$ we'll have:
$$color#00flarge%
sum_n = 1^infty1 over n^2 + 1 = halfbrackspicothparspi - 1
$$
answered Apr 2 '14 at 16:57
Felix Marin
65.5k7105135
$newcommand+^dagger
newcommandangles[1]leftlangle, #1 ,rightrangle
newcommandbraces[1]leftlbrace, #1 ,rightrbrace
newcommandbracks[1]leftlbrack, #1 ,rightrbrack
newcommandceil[1],leftlceil, #1 ,rightrceil,
newcommandddrm d
newcommanddowndownarrow
newcommandds[1]displaystyle#1
newcommandexpo[1],rm e^#1,
newcommandfermi,rm f
newcommandfloor[1],leftlfloor #1 rightrfloor,
newcommandhalf1 over 2
newcommandicrm i
newcommandiffLongleftrightarrow
newcommandimpLongrightarrow
newcommandisdiv,left.rightvert,
newcommandket[1]leftvert #1rightrangle
newcommandol[1]overline#1
newcommandpars[1]left(, #1 ,right)
newcommandpartiald[3]fracpartial^#1 #2partial #3^#1
newcommandppcal P
newcommandroot[2],sqrt[#1]vphantomlarge A,#2,,
newcommandsech,rm sech
newcommandsgn,rm sgn
newcommandtotald[3]fracrm d^#1 #2rm d #3^#1
newcommandul[1]underline#1
newcommandverts[1]leftvert, #1 ,rightvert
newcommandwt[1]widetilde#1$
beginalign
sum_n = 1^infty1 over n^2 + 1&=
sum_n = 0^infty1 over parsn + 1 + icparsn + 1 - ic
=Psipars1 + ic - Psipars1 - ic over pars1 + ic - pars1 - ic
=ImPsipars1 + ic
endalign
where $Psiparsz$ is the Digamma Function.
With the identity $dsImPsipars1 + ic y = -,1 over 2y + half,picothparspi y$ we'll have:
$$color#00flarge%
sum_n = 1^infty1 over n^2 + 1 = halfbrackspicothparspi - 1
$$
answered Apr 2 '14 at 16:57
Felix Marin
65.5k7105135
edited Apr 2 '14 at 17:10
answered Apr 2 '14 at 16:57
Felix Marin
65.5k7105135
answered Apr 2 '14 at 16:57
Felix Marin
65.5k7105135
answered Apr 2 '14 at 16:57
Felix Marin
65.5k7105135
65.5k7105135
add a comment |Â
add a comment |Â
up vote
5
down vote
There's a little bit of calculation you need to do here to make sure Cauchy's Residue Theorem is applicable here (you need to make sure that certain integrals are bounded etc) but this is a sketch:
Let
$$
f(z) = frac pi (1+z^2)tan(pi z)
$$
Then $f$ has simple poles $forall n in mathbbZ$ and also at $pm i$.
You can calculate the residues as
$$
textRes(f(z), npi) = frac 1 1 + n^2
$$
and
$$ textRes(f(z), pm i) = frac -pi 2tanh(pi)
$$
If these are hard to calculate for you, I can give you more detail.
Now, let $Gamma _N$ be the square contour with vertices $(N + frac 1 2) (pm 1 pm i)$
Then Cauchy's Residue Theorem tells us
$$
int_Gamma N f(z) dz = 2pi i sum textRes(f(z), z)
$$
Where the sum is across all the poles inside the contour.
Now, we see that the simple poles inside the contour are all the ones at integers $n$ s.t. $|n| < N$ and the ones at $pm i$.
So
$$
int_Gamma N f(z) dz = 2pi i left [ frac -2pi 2 tanh(pi) + sum_n = -N^N frac1 1 + n^2 right ]
$$
Now we need to show that the integral on the left goes to zero as $N$ tends to infinity. Now, I'm going to leave this as an exercise for you (unlucky) but the basic idea is to find a constant $C_1$ which bounds $frac pi tan(pi z)$ on the top and bottom of the square and another constant $C_2$ on the sides of the square and take $C$ to be the maximum of these.
Then by the Estimation Theorem we would have that
$$
left | int_Gamma_N f(z) dz right | leq textlength(Gamma_N) textsup_z in Gamma _N |f(z)| leq 4(2N + 1) C textsup_z in Gamma_N left| frac 1 1 + z^2right | leq frac 4C(2N + 1) 1 + N^2 = O(frac 1 N)
$$
So we let $N to infty$ then we get
$$
0 = 2pi i left [ frac -2pi 2 tanh(pi) + sum_n = -infty^infty frac1 1 + n^2 right ]
$$
So then we have
$$
0 = frac -pi tanh(pi) + 2sum_n = 1^infty frac1 1 + n^2 + 1
$$
Where the 1 that has randomly appeared is the $n = 0 $ term
So
$$
sum_n = 1^infty frac1 1 + n^2 = frac 1 2 left [frac pi tanh(pi) - 1 right] = frac 1 2 (pi coth (pi) - 1).
$$
I hope you know some complex analysis otherwise this might have meant nothing to you...
This method works for most sums though (as long as you get the right things tending to zero, which you do in this case) and I don't think I've ever had it not work.
If you want to evaluate
$$ sum_n = 1 ^ infty phi(n)
$$
Where phi can easily be extended to all of $mathbb C$ you just take
$$f(z) = frac pi phi(z) tan(pi z)
$$
And do the same thing, and if you want to evaluate
$$sum_n = 1 ^ infty (-1)^n phi(n)
$$
You just take
$$f(z) = frac pi phi(z) sin(pi z)
$$
In each case integrating over the same contour. Hope this helps, and if you don't know much complex analysis, you should learn more, it is a very interesting and powerful area. Sorry this was the longest answer ever.
edited Jun 26 '14 at 15:20
Meow
2,51732453
answered Apr 2 '14 at 17:25
CameronJWhitehead
1,361517
add a comment |Â
up vote
5
down vote
There's a little bit of calculation you need to do here to make sure Cauchy's Residue Theorem is applicable here (you need to make sure that certain integrals are bounded etc) but this is a sketch:
Let
$$
f(z) = frac pi (1+z^2)tan(pi z)
$$
Then $f$ has simple poles $forall n in mathbbZ$ and also at $pm i$.
You can calculate the residues as
$$
textRes(f(z), npi) = frac 1 1 + n^2
$$
and
$$ textRes(f(z), pm i) = frac -pi 2tanh(pi)
$$
If these are hard to calculate for you, I can give you more detail.
Now, let $Gamma _N$ be the square contour with vertices $(N + frac 1 2) (pm 1 pm i)$
Then Cauchy's Residue Theorem tells us
$$
int_Gamma N f(z) dz = 2pi i sum textRes(f(z), z)
$$
Where the sum is across all the poles inside the contour.
Now, we see that the simple poles inside the contour are all the ones at integers $n$ s.t. $|n| < N$ and the ones at $pm i$.
So
$$
int_Gamma N f(z) dz = 2pi i left [ frac -2pi 2 tanh(pi) + sum_n = -N^N frac1 1 + n^2 right ]
$$
Now we need to show that the integral on the left goes to zero as $N$ tends to infinity. Now, I'm going to leave this as an exercise for you (unlucky) but the basic idea is to find a constant $C_1$ which bounds $frac pi tan(pi z)$ on the top and bottom of the square and another constant $C_2$ on the sides of the square and take $C$ to be the maximum of these.
Then by the Estimation Theorem we would have that
$$
left | int_Gamma_N f(z) dz right | leq textlength(Gamma_N) textsup_z in Gamma _N |f(z)| leq 4(2N + 1) C textsup_z in Gamma_N left| frac 1 1 + z^2right | leq frac 4C(2N + 1) 1 + N^2 = O(frac 1 N)
$$
So we let $N to infty$ then we get
$$
0 = 2pi i left [ frac -2pi 2 tanh(pi) + sum_n = -infty^infty frac1 1 + n^2 right ]
$$
So then we have
$$
0 = frac -pi tanh(pi) + 2sum_n = 1^infty frac1 1 + n^2 + 1
$$
Where the 1 that has randomly appeared is the $n = 0 $ term
So
$$
sum_n = 1^infty frac1 1 + n^2 = frac 1 2 left [frac pi tanh(pi) - 1 right] = frac 1 2 (pi coth (pi) - 1).
$$
I hope you know some complex analysis otherwise this might have meant nothing to you...
This method works for most sums though (as long as you get the right things tending to zero, which you do in this case) and I don't think I've ever had it not work.
If you want to evaluate
$$ sum_n = 1 ^ infty phi(n)
$$
Where phi can easily be extended to all of $mathbb C$ you just take
$$f(z) = frac pi phi(z) tan(pi z)
$$
And do the same thing, and if you want to evaluate
$$sum_n = 1 ^ infty (-1)^n phi(n)
$$
You just take
$$f(z) = frac pi phi(z) sin(pi z)
$$
In each case integrating over the same contour. Hope this helps, and if you don't know much complex analysis, you should learn more, it is a very interesting and powerful area. Sorry this was the longest answer ever.
edited Jun 26 '14 at 15:20
Meow
2,51732453
answered Apr 2 '14 at 17:25
CameronJWhitehead
1,361517
add a comment |Â
up vote
5
down vote
up vote
5
down vote
There's a little bit of calculation you need to do here to make sure Cauchy's Residue Theorem is applicable here (you need to make sure that certain integrals are bounded etc) but this is a sketch:
Let
$$
f(z) = frac pi (1+z^2)tan(pi z)
$$
Then $f$ has simple poles $forall n in mathbbZ$ and also at $pm i$.
You can calculate the residues as
$$
textRes(f(z), npi) = frac 1 1 + n^2
$$
and
$$ textRes(f(z), pm i) = frac -pi 2tanh(pi)
$$
If these are hard to calculate for you, I can give you more detail.
Now, let $Gamma _N$ be the square contour with vertices $(N + frac 1 2) (pm 1 pm i)$
Then Cauchy's Residue Theorem tells us
$$
int_Gamma N f(z) dz = 2pi i sum textRes(f(z), z)
$$
Where the sum is across all the poles inside the contour.
Now, we see that the simple poles inside the contour are all the ones at integers $n$ s.t. $|n| < N$ and the ones at $pm i$.
So
$$
int_Gamma N f(z) dz = 2pi i left [ frac -2pi 2 tanh(pi) + sum_n = -N^N frac1 1 + n^2 right ]
$$
Now we need to show that the integral on the left goes to zero as $N$ tends to infinity. Now, I'm going to leave this as an exercise for you (unlucky) but the basic idea is to find a constant $C_1$ which bounds $frac pi tan(pi z)$ on the top and bottom of the square and another constant $C_2$ on the sides of the square and take $C$ to be the maximum of these.
Then by the Estimation Theorem we would have that
$$
left | int_Gamma_N f(z) dz right | leq textlength(Gamma_N) textsup_z in Gamma _N |f(z)| leq 4(2N + 1) C textsup_z in Gamma_N left| frac 1 1 + z^2right | leq frac 4C(2N + 1) 1 + N^2 = O(frac 1 N)
$$
So we let $N to infty$ then we get
$$
0 = 2pi i left [ frac -2pi 2 tanh(pi) + sum_n = -infty^infty frac1 1 + n^2 right ]
$$
So then we have
$$
0 = frac -pi tanh(pi) + 2sum_n = 1^infty frac1 1 + n^2 + 1
$$
Where the 1 that has randomly appeared is the $n = 0 $ term
So
$$
sum_n = 1^infty frac1 1 + n^2 = frac 1 2 left [frac pi tanh(pi) - 1 right] = frac 1 2 (pi coth (pi) - 1).
$$
I hope you know some complex analysis otherwise this might have meant nothing to you...
This method works for most sums though (as long as you get the right things tending to zero, which you do in this case) and I don't think I've ever had it not work.
If you want to evaluate
$$ sum_n = 1 ^ infty phi(n)
$$
Where phi can easily be extended to all of $mathbb C$ you just take
$$f(z) = frac pi phi(z) tan(pi z)
$$
And do the same thing, and if you want to evaluate
$$sum_n = 1 ^ infty (-1)^n phi(n)
$$
You just take
$$f(z) = frac pi phi(z) sin(pi z)
$$
In each case integrating over the same contour. Hope this helps, and if you don't know much complex analysis, you should learn more, it is a very interesting and powerful area. Sorry this was the longest answer ever.
edited Jun 26 '14 at 15:20
Meow
2,51732453
answered Apr 2 '14 at 17:25
CameronJWhitehead
1,361517
There's a little bit of calculation you need to do here to make sure Cauchy's Residue Theorem is applicable here (you need to make sure that certain integrals are bounded etc) but this is a sketch:
Let
$$
f(z) = frac pi (1+z^2)tan(pi z)
$$
Then $f$ has simple poles $forall n in mathbbZ$ and also at $pm i$.
You can calculate the residues as
$$
textRes(f(z), npi) = frac 1 1 + n^2
$$
and
$$ textRes(f(z), pm i) = frac -pi 2tanh(pi)
$$
If these are hard to calculate for you, I can give you more detail.
Now, let $Gamma _N$ be the square contour with vertices $(N + frac 1 2) (pm 1 pm i)$
Then Cauchy's Residue Theorem tells us
$$
int_Gamma N f(z) dz = 2pi i sum textRes(f(z), z)
$$
Where the sum is across all the poles inside the contour.
Now, we see that the simple poles inside the contour are all the ones at integers $n$ s.t. $|n| < N$ and the ones at $pm i$.
So
$$
int_Gamma N f(z) dz = 2pi i left [ frac -2pi 2 tanh(pi) + sum_n = -N^N frac1 1 + n^2 right ]
$$
Now we need to show that the integral on the left goes to zero as $N$ tends to infinity. Now, I'm going to leave this as an exercise for you (unlucky) but the basic idea is to find a constant $C_1$ which bounds $frac pi tan(pi z)$ on the top and bottom of the square and another constant $C_2$ on the sides of the square and take $C$ to be the maximum of these.
Then by the Estimation Theorem we would have that
$$
left | int_Gamma_N f(z) dz right | leq textlength(Gamma_N) textsup_z in Gamma _N |f(z)| leq 4(2N + 1) C textsup_z in Gamma_N left| frac 1 1 + z^2right | leq frac 4C(2N + 1) 1 + N^2 = O(frac 1 N)
$$
So we let $N to infty$ then we get
$$
0 = 2pi i left [ frac -2pi 2 tanh(pi) + sum_n = -infty^infty frac1 1 + n^2 right ]
$$
So then we have
$$
0 = frac -pi tanh(pi) + 2sum_n = 1^infty frac1 1 + n^2 + 1
$$
Where the 1 that has randomly appeared is the $n = 0 $ term
So
$$
sum_n = 1^infty frac1 1 + n^2 = frac 1 2 left [frac pi tanh(pi) - 1 right] = frac 1 2 (pi coth (pi) - 1).
$$
I hope you know some complex analysis otherwise this might have meant nothing to you...
This method works for most sums though (as long as you get the right things tending to zero, which you do in this case) and I don't think I've ever had it not work.
If you want to evaluate
$$ sum_n = 1 ^ infty phi(n)
$$
Where phi can easily be extended to all of $mathbb C$ you just take
$$f(z) = frac pi phi(z) tan(pi z)
$$
And do the same thing, and if you want to evaluate
$$sum_n = 1 ^ infty (-1)^n phi(n)
$$
You just take
$$f(z) = frac pi phi(z) sin(pi z)
$$
In each case integrating over the same contour. Hope this helps, and if you don't know much complex analysis, you should learn more, it is a very interesting and powerful area. Sorry this was the longest answer ever.
edited Jun 26 '14 at 15:20
Meow
2,51732453
answered Apr 2 '14 at 17:25
CameronJWhitehead
1,361517
edited Jun 26 '14 at 15:20
Meow
2,51732453
edited Jun 26 '14 at 15:20
Meow
2,51732453
edited Jun 26 '14 at 15:20
Meow
2,51732453
2,51732453
answered Apr 2 '14 at 17:25
CameronJWhitehead
1,361517
answered Apr 2 '14 at 17:25
CameronJWhitehead
1,361517
answered Apr 2 '14 at 17:25
CameronJWhitehead
1,361517
1,361517
add a comment |Â
add a comment |Â
up vote
4
down vote
This is related to this answer, where it is shown that
$$
sum_k=1^inftyfrac1k^2-z^2
=frac12zleft[frac1z-picot(pi z)right]tag1
$$
Equation $(1)$ is valid for all $zinmathbbC$, in particular for $z=i$, which gives
$$
sum_k=1^inftyfrac1k^2+1
=frac12left[picoth(pi)-1right]tag2
$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Apr 7 '14 at 11:45
robjohn♦
258k26297612
add a comment |Â
up vote
4
down vote
This is related to this answer, where it is shown that
$$
sum_k=1^inftyfrac1k^2-z^2
=frac12zleft[frac1z-picot(pi z)right]tag1
$$
Equation $(1)$ is valid for all $zinmathbbC$, in particular for $z=i$, which gives
$$
sum_k=1^inftyfrac1k^2+1
=frac12left[picoth(pi)-1right]tag2
$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Apr 7 '14 at 11:45
robjohn♦
258k26297612
add a comment |Â
up vote
4
down vote
up vote
4
down vote
This is related to this answer, where it is shown that
$$
sum_k=1^inftyfrac1k^2-z^2
=frac12zleft[frac1z-picot(pi z)right]tag1
$$
Equation $(1)$ is valid for all $zinmathbbC$, in particular for $z=i$, which gives
$$
sum_k=1^inftyfrac1k^2+1
=frac12left[picoth(pi)-1right]tag2
$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Apr 7 '14 at 11:45
robjohn♦
258k26297612
This is related to this answer, where it is shown that
$$
sum_k=1^inftyfrac1k^2-z^2
=frac12zleft[frac1z-picot(pi z)right]tag1
$$
Equation $(1)$ is valid for all $zinmathbbC$, in particular for $z=i$, which gives
$$
sum_k=1^inftyfrac1k^2+1
=frac12left[picoth(pi)-1right]tag2
$$
edited Apr 13 '17 at 12:20
Community♦
1
answered Apr 7 '14 at 11:45
robjohn♦
258k26297612
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Apr 7 '14 at 11:45
robjohn♦
258k26297612
answered Apr 7 '14 at 11:45
robjohn♦
258k26297612
answered Apr 7 '14 at 11:45
robjohn♦
258k26297612
258k26297612
add a comment |Â
add a comment |Â
up vote
4
down vote
We can start from the Weierstrass product for the $sinh$ function:
$$fracsinh zz=prod_n=1^+inftyleft(1+fracz^2pi^2 n^2right)tag1 $$
then consider the logarithmic derivative of both sides. This leads to:
$$coth z-frac1z=sum_n=1^+inftyfrac2zz^2+pi^2 n^2tag2 $$
or to:
$$picoth(pi w)-frac1w=sum_n=1^+inftyfrac2ww^2+ n^2.tag3 $$
Now just set $w=1$ in $(3)$.
Further approach: by the Poisson summation formula, since the Laplace distribution and the Cauchy distribution are related via the Fourier transform, we have that $sum_ngeq 0frac1n^2+1$ is simply related with $sum_ngeq 0e^-pi n$, which is a geometric series.
answered Jan 10 '15 at 22:39
Jack D'Aurizio♦
270k31264629
add a comment |Â
up vote
4
down vote
We can start from the Weierstrass product for the $sinh$ function:
$$fracsinh zz=prod_n=1^+inftyleft(1+fracz^2pi^2 n^2right)tag1 $$
then consider the logarithmic derivative of both sides. This leads to:
$$coth z-frac1z=sum_n=1^+inftyfrac2zz^2+pi^2 n^2tag2 $$
or to:
$$picoth(pi w)-frac1w=sum_n=1^+inftyfrac2ww^2+ n^2.tag3 $$
Now just set $w=1$ in $(3)$.
Further approach: by the Poisson summation formula, since the Laplace distribution and the Cauchy distribution are related via the Fourier transform, we have that $sum_ngeq 0frac1n^2+1$ is simply related with $sum_ngeq 0e^-pi n$, which is a geometric series.
answered Jan 10 '15 at 22:39
Jack D'Aurizio♦
270k31264629
add a comment |Â
up vote
4
down vote
up vote
4
down vote
We can start from the Weierstrass product for the $sinh$ function:
$$fracsinh zz=prod_n=1^+inftyleft(1+fracz^2pi^2 n^2right)tag1 $$
then consider the logarithmic derivative of both sides. This leads to:
$$coth z-frac1z=sum_n=1^+inftyfrac2zz^2+pi^2 n^2tag2 $$
or to:
$$picoth(pi w)-frac1w=sum_n=1^+inftyfrac2ww^2+ n^2.tag3 $$
Now just set $w=1$ in $(3)$.
Further approach: by the Poisson summation formula, since the Laplace distribution and the Cauchy distribution are related via the Fourier transform, we have that $sum_ngeq 0frac1n^2+1$ is simply related with $sum_ngeq 0e^-pi n$, which is a geometric series.
answered Jan 10 '15 at 22:39
Jack D'Aurizio♦
270k31264629
We can start from the Weierstrass product for the $sinh$ function:
$$fracsinh zz=prod_n=1^+inftyleft(1+fracz^2pi^2 n^2right)tag1 $$
then consider the logarithmic derivative of both sides. This leads to:
$$coth z-frac1z=sum_n=1^+inftyfrac2zz^2+pi^2 n^2tag2 $$
or to:
$$picoth(pi w)-frac1w=sum_n=1^+inftyfrac2ww^2+ n^2.tag3 $$
Now just set $w=1$ in $(3)$.
Further approach: by the Poisson summation formula, since the Laplace distribution and the Cauchy distribution are related via the Fourier transform, we have that $sum_ngeq 0frac1n^2+1$ is simply related with $sum_ngeq 0e^-pi n$, which is a geometric series.
answered Jan 10 '15 at 22:39
Jack D'Aurizio♦
270k31264629
edited Oct 27 '17 at 19:34
answered Jan 10 '15 at 22:39
Jack D'Aurizio♦
270k31264629
answered Jan 10 '15 at 22:39
Jack D'Aurizio♦
270k31264629
answered Jan 10 '15 at 22:39
Jack D'Aurizio♦
270k31264629
270k31264629
add a comment |Â
add a comment |Â
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Have you studied the method of series calculation using Fourier series?
– Frank
Apr 2 '14 at 16:45
(no idea how to use fourier series)You can estimate through $arctan(x)$ though
– Guy
Apr 2 '14 at 16:46
1
@Frank What's the function that works here?
– Git Gud
Apr 2 '14 at 16:47
1
@Git Gud Was wondering the same thing. I have studied it but the function isn't immediately obvious.
– Briandmg
Apr 2 '14 at 16:48
9
This can also be done via residue calculus. Alternatively, with a bit of handwaving, use the Weierstrass prod. for sin: $$x^-1sin x=prod_k Big(1-x^2(pi k)^-2Big)Rightarrow (ipi x)^-1sin ipi x=prod_k Big(1+x^2k^-2Big)$$ Take logs, differentiate, and tada: $sum (k^2+x^2)^-1=frac12x^2(pi xcoth pi x-1)$
– Julien Godawatta
Apr 2 '14 at 16:51