Evaluating infinite series $\sum_{n=0}^{\infty} \frac{1}{a^{2}+(2n+1)^2}$

Question

I have no idea to approach this problem. Mathematica gave the sum to be $$ \sum_{n=0}^{\infty} \frac{1}{a^{2}+(2n+1)^2} = \frac{\pi}{4a} \tanh(\frac{a \pi}{2}) $$

How can I analyze this?

Answer 1

If you can get $s(a) =\sum \dfrac1{a^2+n^2} $, then $s(a/2) =\sum \dfrac1{a^2/4+n^2} =4\sum \dfrac1{a^2+4n^2} =4\sum \dfrac1{a^2+(2n)^2} $, so $s(a)-\frac14 s(a/2) =\sum \dfrac1{a^2+n^2}-\sum \dfrac1{a^2+(2n)^2} =\sum \dfrac1{a^2+(2n+1)^2} $.

Answer 2

I don't know if this has been considered resolved yet, but combining hints from @user1952009: Let $$f(t)=e^{-at}, -\pi<t<\pi$$ and have period $2\pi$. Then we want to find the Fourier series of $f(t)$ $$f(t)=\sum_{n=-\infty}^{\infty}a_ne^{int}$$ Then $$\int_{-\pi}^{\pi}e^{-at}e^{-int}dt=\left.\frac{e^{-(a+in)t}}{-(a+in)}\right|_{-\pi}^{\pi}=\frac{(-1)^n\left(e^{-a\pi}-e^{a\pi}\right)}{-(a+in)}=2\pi a_n$$ So we conclude that for $-\pi<t<\pi$, $$e^{-at}=\sum_{n=-\infty}^{\infty}\frac{(-1)^n\sinh a\pi}{\pi(a+in)}e^{int}$$ Then we are ready to evalute $$\int_{-\pi}^{\pi}\left|e^{-at}\right|^2dt=\left.\frac{e^{-2at}}{(-2a)}\right|_{\pi}^{\pi}=\frac{e^{-2a\pi}-e^{2a\pi}}{(-2a)}=\frac{\sinh2a\pi}a=\sum_{n=-\infty}^{\infty}2\pi\frac{\sinh^2a\pi}{\pi^2(a^2+n^2)}$$ So we have established $$\sum_{n=-\infty}^{\infty}\frac1{a^2+n^2}=\frac{\pi}{2a}\frac{\sinh2a\pi}{\sinh^2a\pi}=\frac{\pi}a\coth a\pi$$ Now we are ready to use @marty cohen's hint: $$\frac14\sum_{n=-\infty}^{\infty}\frac1{\left(\frac a2\right)^2+n^2}=\sum_{n=-\infty}^{\infty}\frac1{a^2+(2n)^2}=\frac14\frac{\pi}{\frac a2}\coth\left(\frac{a\pi}2\right)=\frac{\pi}{2a}\coth\left(\frac{a\pi}2\right)$$ So now $$\begin{align}\sum_{n=0}^{\infty}\frac1{a^2+(2n+1)^2}&=\frac12\left\{\sum_{n=-\infty}^{\infty}\frac1{a^2+n^2}-\sum_{n=-\infty}^{\infty}\frac1{a^2+(2n)^2}\right\}\\& =\frac{\pi}{2a}\coth a\pi-\frac{\pi}{4a}\coth\left(\frac{a\pi}2\right)\\ &=\frac{\pi}{2a}\left(\frac{1+\tanh^2\left(\frac{\pi a}2\right)}{2\tanh\left(\frac{\pi a}2\right)}\right)-\frac{\pi}{4a}\frac1{\tanh\left(\frac{\pi a}2\right)}\\ &=\frac{\pi}{4a}\tanh\left(\frac{\pi a}2\right)\end{align}$$ I don't know if this was being Captain Obvious or useful sewing up of the patient by Nurse @user5713492.

EDIT: Looking at the companion thread it seems more direct to try $$\oint_C\frac{\tan\pi z}{a^2+4z^2}dz$$ where the contour is around a box bounded by $\pm N$ on the left and right and by $\pm iN$ top and bottom. Then it can be seen that as $N\rightarrow\infty$ the contour integral goes to zero, so $$2\pi i\left\{\frac{\tan\left(-i\pi\frac a2\right)}{4(-ia)}+\frac{\tan\left(i\pi\frac a2\right)}{4(ia)}+2\sum_{n=0}^{\infty} \frac{-\frac1{\pi}}{a^2+(2n+1)^2}\right\}=0$$ And so the result follows directly.