# Infinite sum and the conditions with tanh

Is there a way to prove this relation?:

$$\sum_{\text{n}=0}^\infty\frac{1}{\text{s}^2+\left(1+2\text{n}\right)^2\omega^2}=\frac{\pi\tanh\left(\frac{\pi\text{s}}{2\omega}\right)}{4\text{s}\omega}$$

And find the conditions for which this equality hold?

• hint1: mittag leffler theorem. hint2: product representation of $\sin$. hint3: contour integration – tired Nov 20 '16 at 16:30
• then consult the internet – tired Nov 20 '16 at 16:36
• i give you a starting point: math.stackexchange.com/questions/141470/… – tired Nov 20 '16 at 16:39
• Besides the very great hint given by @tired I would suggest you to read and study: " Complex Variables, theory and applications " by Murray S. Spiegel, in which you will learn about infinite series, residues, complex variables, contour integration et cetera. After that, I bet you'll be able to solve that. – Von Neumann Nov 20 '16 at 16:41
• @SimpleArt The problem is that it sound strange to me that he has to evaluate that with Real Analysis. Sure it has to be possible, but.. quite sadistic, isn't it? :D – Von Neumann Nov 20 '16 at 17:06

You may remove a useless parameter by setting $s=\omega t$ then look for a closed form for $$S(t)=\sum_{n\geq 0}\frac{1}{t^2+(2n+1)^2}\tag{1}$$ Integration by parts gives a useful lemma: $$\forall a,b>0,\qquad\int_{0}^{+\infty}\frac{\sin(a x)}{a} e^{-bx}\,dx = \frac{1}{a^2+b^2}\tag{2}$$ and $$\sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1}=W(x)\tag{3}$$ is a $2\pi$-periodic rectangle wave that equals $\frac{\pi}{4}$ over $(0,\pi)$ and $-\frac{\pi}{4}$ over $(\pi,2\pi)$.
By exploiting $(2)$ and $(3)$, $$S(t) = \int_{0}^{+\infty}W(x) e^{-tx}\,dx =\frac{\pi}{4}\sum_{n\geq 0}(-1)^n\int_{n\pi}^{(n+1)\pi}e^{-tx}\,dx\tag{4}$$ and by computing the last integrals and the resulting geometric series, $$\sum_{n\geq 0}\frac{1}{t^2+(2n+1)^2}=\color{red}{\frac{\pi}{4t}\,\tanh\left(\frac{\pi t}{2}\right)}\tag{5}$$ follows. The same can be achieved through the Poisson summation formula.
Using the identity $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{Residues of }\pi\cot\left(\pi z\right)f\left(z\right)\textrm{ at }f'\textrm{s poles}\right\}$$ which follows from the residue theorem we have $$\sum_{n\geq0}\frac{1}{s^{2}+\left(2n+1\right)^{2}\omega^{2}}=\frac{1}{2}\sum_{n\in\mathbb{Z}}\frac{1}{s^{2}+\left(2n+1\right)^{2}\omega^{2}}$$ and since we have poles at $z=-\frac{1}{2}\pm\frac{i\sqrt{s^{2}}}{2\sqrt{\omega^{2}}}$ we can conclude that $$\sum_{n\geq0}\frac{1}{s^{2}+\left(2n+1\right)^{2}\omega^{2}}=\color{red}{\frac{\pi\tanh\left(\frac{\pi s}{2\omega}\right)}{4s\omega}}$$ as wanted.