$sum_{n=-infty}^{infty}frac{1}{(n+alpha)^2}=frac{pi^2}{(sin pi alpha)^2}$?












3












$begingroup$


Now we want to prove:
$$sum_{n=-infty}^{infty}frac{1}{(n+alpha)^2}=frac{pi^2}{(sin pi alpha)^2}$$
$alpha$ >0 and not an integer.
According to poisson summation formula
$$sum_{n=-infty}^{infty}f(x+n)=sum_{n=-infty}^{infty}hat{f}(n)e^{2pi inx}.$$
So, if we let $f(x)=frac{1}{x^2}$, then
$$sum_{n=-infty}^{infty}f(x+n)=sum_{n=-infty}^{infty}frac{1}{(n+x)^2}$$
Therefore, if we can prove that
$$sum_{n=-infty}^{infty}left(int_{-infty}^{infty}frac{1}{x^2} e^{-2pi inx}dx e^{2pi inx}right)=frac{pi^2}{(sin pi x)^2}$$
Then it will be done.
But I don't know how to prove
$$sum_{n=-infty}^{infty}left(int_{-infty}^{infty}frac{1}{x^2} e^{-2pi inx}dx e^{2pi inx}right)=frac{pi^2}{(sin pi x)^2}$$
Who could give me some hints? Thanks!






calculus fourier-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Interesing... we can square termwise.
    $endgroup$
    – Pedro Tamaroff♦
    Mar 11 '13 at 22:55












  • $begingroup$
    Here is a related problem 1, related problem 2.
    $endgroup$
    – Mhenni Benghorbal
    Mar 11 '13 at 23:07












  • $begingroup$
    @user39843: I found your question under my post. See here for the answer.
    $endgroup$
    – Mhenni Benghorbal
    Mar 13 '13 at 23:45










  • $begingroup$
    @MhenniBenghorbal thx! It would be helpful.
    $endgroup$
    – user39843
    Mar 18 '13 at 4:56










  • $begingroup$
    Does this series converge uniformly ?
    $endgroup$
    – Fardad Pouran
    Dec 29 '15 at 19:05
















3












$begingroup$


Now we want to prove:
$$sum_{n=-infty}^{infty}frac{1}{(n+alpha)^2}=frac{pi^2}{(sin pi alpha)^2}$$
$alpha$ >0 and not an integer.
According to poisson summation formula
$$sum_{n=-infty}^{infty}f(x+n)=sum_{n=-infty}^{infty}hat{f}(n)e^{2pi inx}.$$
So, if we let $f(x)=frac{1}{x^2}$, then
$$sum_{n=-infty}^{infty}f(x+n)=sum_{n=-infty}^{infty}frac{1}{(n+x)^2}$$
Therefore, if we can prove that
$$sum_{n=-infty}^{infty}left(int_{-infty}^{infty}frac{1}{x^2} e^{-2pi inx}dx e^{2pi inx}right)=frac{pi^2}{(sin pi x)^2}$$
Then it will be done.
But I don't know how to prove
$$sum_{n=-infty}^{infty}left(int_{-infty}^{infty}frac{1}{x^2} e^{-2pi inx}dx e^{2pi inx}right)=frac{pi^2}{(sin pi x)^2}$$
Who could give me some hints? Thanks!






calculus fourier-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Interesing... we can square termwise.
    $endgroup$
    – Pedro Tamaroff♦
    Mar 11 '13 at 22:55












  • $begingroup$
    Here is a related problem 1, related problem 2.
    $endgroup$
    – Mhenni Benghorbal
    Mar 11 '13 at 23:07












  • $begingroup$
    @user39843: I found your question under my post. See here for the answer.
    $endgroup$
    – Mhenni Benghorbal
    Mar 13 '13 at 23:45










  • $begingroup$
    @MhenniBenghorbal thx! It would be helpful.
    $endgroup$
    – user39843
    Mar 18 '13 at 4:56










  • $begingroup$
    Does this series converge uniformly ?
    $endgroup$
    – Fardad Pouran
    Dec 29 '15 at 19:05














3












3








3


4



$begingroup$


Now we want to prove:
$$sum_{n=-infty}^{infty}frac{1}{(n+alpha)^2}=frac{pi^2}{(sin pi alpha)^2}$$
$alpha$ >0 and not an integer.
According to poisson summation formula
$$sum_{n=-infty}^{infty}f(x+n)=sum_{n=-infty}^{infty}hat{f}(n)e^{2pi inx}.$$
So, if we let $f(x)=frac{1}{x^2}$, then
$$sum_{n=-infty}^{infty}f(x+n)=sum_{n=-infty}^{infty}frac{1}{(n+x)^2}$$
Therefore, if we can prove that
$$sum_{n=-infty}^{infty}left(int_{-infty}^{infty}frac{1}{x^2} e^{-2pi inx}dx e^{2pi inx}right)=frac{pi^2}{(sin pi x)^2}$$
Then it will be done.
But I don't know how to prove
$$sum_{n=-infty}^{infty}left(int_{-infty}^{infty}frac{1}{x^2} e^{-2pi inx}dx e^{2pi inx}right)=frac{pi^2}{(sin pi x)^2}$$
Who could give me some hints? Thanks!






calculus fourier-analysis







share|cite|improve this question









$endgroup$




Now we want to prove:
$$sum_{n=-infty}^{infty}frac{1}{(n+alpha)^2}=frac{pi^2}{(sin pi alpha)^2}$$
$alpha$ >0 and not an integer.
According to poisson summation formula
$$sum_{n=-infty}^{infty}f(x+n)=sum_{n=-infty}^{infty}hat{f}(n)e^{2pi inx}.$$
So, if we let $f(x)=frac{1}{x^2}$, then
$$sum_{n=-infty}^{infty}f(x+n)=sum_{n=-infty}^{infty}frac{1}{(n+x)^2}$$
Therefore, if we can prove that
$$sum_{n=-infty}^{infty}left(int_{-infty}^{infty}frac{1}{x^2} e^{-2pi inx}dx e^{2pi inx}right)=frac{pi^2}{(sin pi x)^2}$$
Then it will be done.
But I don't know how to prove
$$sum_{n=-infty}^{infty}left(int_{-infty}^{infty}frac{1}{x^2} e^{-2pi inx}dx e^{2pi inx}right)=frac{pi^2}{(sin pi x)^2}$$
Who could give me some hints? Thanks!






calculus fourier-analysis




calculus fourier-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 11 '13 at 12:15









user39843user39843

529212




529212












  • $begingroup$
    Interesing... we can square termwise.
    $endgroup$
    – Pedro Tamaroff♦
    Mar 11 '13 at 22:55












  • $begingroup$
    Here is a related problem 1, related problem 2.
    $endgroup$
    – Mhenni Benghorbal
    Mar 11 '13 at 23:07












  • $begingroup$
    @user39843: I found your question under my post. See here for the answer.
    $endgroup$
    – Mhenni Benghorbal
    Mar 13 '13 at 23:45










  • $begingroup$
    @MhenniBenghorbal thx! It would be helpful.
    $endgroup$
    – user39843
    Mar 18 '13 at 4:56










  • $begingroup$
    Does this series converge uniformly ?
    $endgroup$
    – Fardad Pouran
    Dec 29 '15 at 19:05


















  • $begingroup$
    Interesing... we can square termwise.
    $endgroup$
    – Pedro Tamaroff♦
    Mar 11 '13 at 22:55












  • $begingroup$
    Here is a related problem 1, related problem 2.
    $endgroup$
    – Mhenni Benghorbal
    Mar 11 '13 at 23:07












  • $begingroup$
    @user39843: I found your question under my post. See here for the answer.
    $endgroup$
    – Mhenni Benghorbal
    Mar 13 '13 at 23:45










  • $begingroup$
    @MhenniBenghorbal thx! It would be helpful.
    $endgroup$
    – user39843
    Mar 18 '13 at 4:56










  • $begingroup$
    Does this series converge uniformly ?
    $endgroup$
    – Fardad Pouran
    Dec 29 '15 at 19:05
















$begingroup$
Interesing... we can square termwise.
$endgroup$
– Pedro Tamaroff♦
Mar 11 '13 at 22:55






$begingroup$
Interesing... we can square termwise.
$endgroup$
– Pedro Tamaroff♦
Mar 11 '13 at 22:55














$begingroup$
Here is a related problem 1, related problem 2.
$endgroup$
– Mhenni Benghorbal
Mar 11 '13 at 23:07






$begingroup$
Here is a related problem 1, related problem 2.
$endgroup$
– Mhenni Benghorbal
Mar 11 '13 at 23:07














$begingroup$
@user39843: I found your question under my post. See here for the answer.
$endgroup$
– Mhenni Benghorbal
Mar 13 '13 at 23:45




$begingroup$
@user39843: I found your question under my post. See here for the answer.
$endgroup$
– Mhenni Benghorbal
Mar 13 '13 at 23:45












$begingroup$
@MhenniBenghorbal thx! It would be helpful.
$endgroup$
– user39843
Mar 18 '13 at 4:56




$begingroup$
@MhenniBenghorbal thx! It would be helpful.
$endgroup$
– user39843
Mar 18 '13 at 4:56












$begingroup$
Does this series converge uniformly ?
$endgroup$
– Fardad Pouran
Dec 29 '15 at 19:05




$begingroup$
Does this series converge uniformly ?
$endgroup$
– Fardad Pouran
Dec 29 '15 at 19:05










1 Answer
1






active

oldest

votes


















4












$begingroup$

The proper technique is documented at MSE 112161 and at MSE 3056578.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This one is better done with a rectangular contour aligned with half-integers (real and imaginary).
    $endgroup$
    – Marko Riedel
    Dec 30 '13 at 21:15












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The proper technique is documented at MSE 112161 and at MSE 3056578.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This one is better done with a rectangular contour aligned with half-integers (real and imaginary).
    $endgroup$
    – Marko Riedel
    Dec 30 '13 at 21:15
















4












$begingroup$

The proper technique is documented at MSE 112161 and at MSE 3056578.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This one is better done with a rectangular contour aligned with half-integers (real and imaginary).
    $endgroup$
    – Marko Riedel
    Dec 30 '13 at 21:15














4












4








4





$begingroup$

The proper technique is documented at MSE 112161 and at MSE 3056578.






share|cite|improve this answer











$endgroup$



The proper technique is documented at MSE 112161 and at MSE 3056578.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 17:23

























answered Mar 11 '13 at 22:52









Marko RiedelMarko Riedel

41.8k341112




41.8k341112












  • $begingroup$
    This one is better done with a rectangular contour aligned with half-integers (real and imaginary).
    $endgroup$
    – Marko Riedel
    Dec 30 '13 at 21:15


















  • $begingroup$
    This one is better done with a rectangular contour aligned with half-integers (real and imaginary).
    $endgroup$
    – Marko Riedel
    Dec 30 '13 at 21:15
















$begingroup$
This one is better done with a rectangular contour aligned with half-integers (real and imaginary).
$endgroup$
– Marko Riedel
Dec 30 '13 at 21:15




$begingroup$
This one is better done with a rectangular contour aligned with half-integers (real and imaginary).
$endgroup$
– Marko Riedel
Dec 30 '13 at 21:15


















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