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I want to prove that for $y >0$, $ x \in \mathbb R$, $$ \sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2} = \frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}$$

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$$\Im\left(\frac1{x+i y+n}\right)=-\frac{y}{(x+n)^2+y^2}$$ might be useful. –  J. M. Aug 1 '12 at 1:07
    
@J.M. Thank you J.M. But I couldn't understand exactly what you mean. If we take summation there, then the left side becomes $\scr F \left(\sum \frac{1}{x+iy+n} \right)$ . –  Ann Aug 1 '12 at 1:25
    
@Ann , what is that symbol you use in your comment? –  DonAntonio Aug 1 '12 at 1:59
    
An idea on how to approach it: use the Poisson sum formula $\sum_{n=-\infty}^{\infty}f(n)=\sum_{k=-\infty}^{\infty}\hat{f}(k)$. Making the substitution u=x+n gives an integral that can be evaluated as a contour integral in the upper half plane; I think you will then get a sum that can be broken up and expressed in closed-form via the geometric series and battered into the form you desire. –  Bitrex Aug 1 '12 at 4:19
    
@Ann: I updated my answer : the problem was simply that here (and in the other thread) a $2\pi$ coefficient was missing in the RHS. –  Raymond Manzoni Aug 1 '12 at 23:27

2 Answers 2

up vote 4 down vote accepted

Let's use J.M.'s excellent hint : $$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}=-\ \Im{\sum_{n=-\infty}^\infty \frac 1{x+iy+n}}$$

Setting $z:=x+iy\ $ we will evaluate : $$\sum_{n=-\infty}^\infty \frac 1{z+n}=\frac 1z+\sum_{n=1}^\infty \frac 1{z+n}+\frac 1{z-n}=\frac 1z+\sum_{n=1}^\infty \frac {2z}{z^2-n^2}$$

The series at the right may be simplified (this was proved many times at S.E. for example here or here or here) getting : $$\pi\cot(\pi z)=\frac1{z}+\sum_{n=1}^{\infty}\frac{2z}{z^2-n^2}$$

and the simple result : $$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}= -\pi\;\Im{(\cot\pi (x+iy))}$$ Your remaining problem is that this simple answer doesn't appear related to your more complicated result. Worse your L.H.S and R.H.S. terms don't numerically correspond so that it seems that you have a problem in your question !

Let's add that your R.H.S. term may be rewritten as half : $$\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}=\frac{\sinh(2\pi y)}{\cosh(2 \pi y) - \cos ( 2 \pi x )}$$


UPDATE: In fact your equation should read (simply multiplying your R.H.S. by $2\pi$) : $$\boxed{\displaystyle \sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2} = \pi\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}}$$

We want : $$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}= -\pi\Im{(\cot \pi (x+iy))}$$

Let's observe that $\ \sin a-\sin b=2\sin\frac{a-b}2\cos\frac{a+b}2$ and $\ \cos a-\cos b=-2\sin\frac{a+b}2\sin\frac{a-b}2$
implies (dividing these expressions) : $$\frac {\sin a-\sin b}{\cos a-\cos b}=-\cot \frac{a+b}2$$

Set $a:=2\pi x,\ b:=2\pi i y\ $ to get : $$-\cot\pi (x+iy)=\frac {\sin 2\pi x-\sin 2\pi i y}{\cos 2\pi x-\cos 2\pi i y}=\frac {\sin 2\pi x-i\sinh 2\pi y}{\cos 2\pi x-\cosh 2\pi y}$$ so that : $$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}=-\pi\frac {\sinh 2\pi y}{\cos 2\pi x-\cosh 2\pi y}$$

$$= \pi\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}$$

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Good follow-through. :) I didn't have time to write it, but this indeed was the route I had in mind. –  J. M. Aug 1 '12 at 10:23
    
Thanks @J.M. always glad to follow-through ! :-) –  Raymond Manzoni Aug 1 '12 at 11:20

For appropriate functions $ƒ$, the Poisson summation formula may be stated as:

$$ \sum_{k=-\infty}^{\infty} f(k+x) {\rm e}^{-ikx} = \sum_{n=-\infty}^\infty F(n+x)\,, $$

where $F$ is the fourier transform of $f$. Let $ f(\omega) = \frac{1}{2}{\rm e}^{-y|\omega|}\, $ then

$$ F(t) = \frac{y}{y^2+t^2}\,. $$

Using the above Poisson summation formula, we have,

$$ \sum_{n=-\infty}^{\infty} \frac{y}{y^2+(x+n)^2} = \sum_{k-\infty}^{\infty} {\rm e}^{-y|k+x|} {\rm e}^{-ikx} \,.$$.

Now, all you need to do is to manipulate the sum on the right hand side in the above equation.

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