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I am supposed to evaluate this infinite series using contour integration. However, since a and b can't be integers, I'm assuming the denominator can never vanish, so I'm not sure if the idea of residues would apply here. Just need a push in the right direction to get started. Thanks.

$\sum_{- \infty}^{\infty} \frac{1}{(n+a)(n+b)} $ where $ a \neq b$ and $a,b \notin Z $

gorgy
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2 Answers2

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Going from Phillipe's hint, the basic result for sums of this type (rational function of $n$) is

$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_{k=1}^N \operatorname*{Res}_{z=z_k} \pi \, \cot{(\pi z)} \, f(z)$$

In your case, $f(z)=1/[(z-a)(z-b)]$. This is fairly simple because the poles of $f$ are non-integral and simple. We then have

$$\sum_{n=-\infty}^{\infty} \frac{1}{(n-a)(n-b)} = -\pi \left [\frac{\cot{(\pi a)}}{a-b} + \frac{\cot{(\pi b)}}{b-a} \right ] =-\pi \frac{\cot{(\pi b)}-\cot{(\pi a)}}{b-a} $$

Note that when $a=b$, the sum approaches $\pi^2 \csc^2{(\pi a)}$.

Ron Gordon
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Hint : Use the residue theorem with the meromorphic function $f:z\mapsto \dfrac{\pi\cot(\pi z)}{(z+a)(z+b)}$ and the contour $C_n=\partial D\left(0,n+\dfrac 12\right)$, and let $n$ tends to $+\infty$

Also check my message here if needed : Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$

Community
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Philippe Malot
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  • will try that. thanks – gorgy Oct 23 '13 at 20:37
  • I'm confused about what you wrote on that thread: when taking the Residue of f at n, how do you get that tan'(pi n) (which I'm assuming is sec^2) term to be equal to 1/pi? – gorgy Oct 23 '13 at 23:21
  • Well, I forgot a $\pi$ in this message and just added it. The derivative of $z\mapsto \tan(\pi z)$ is $z\mapsto \pi(1+\tan^2(\pi z))$. Then take $z=n$. You can also take $z\mapsto \pi\cot(\pi z)/((z+a)(z+b))$ for $f$ to avoid the $1/\pi$ factor. – Philippe Malot Oct 24 '13 at 06:52
  • I've edited it again, and my answer to your message, for easier reading. – Philippe Malot Oct 25 '13 at 16:34