Show $\frac{\pi}{\alpha \sinh (\pi \alpha)} = \sum_{n= -\infty}^\infty \frac{(-1)^n}{\alpha^2 + n^2} $

Question

The question has 2 parts.

First it asks to show that, given $\;f(t) = e^{\alpha t}\;$ on $\;(-\pi, \pi):$

$$ e^{\alpha t} = \frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}(\alpha + i n) e^{i n t} $$

which I did using complex version of Fourier expansion:

$$ f(t) = \sum_{n= -\infty}^\infty c_n e^{i n t} $$

with

$$ c_n = \frac{1}{2\pi} \int_{-\pi}^\pi f(t) e^{-i n t} dt, \quad \forall\; n \in \mathbb{Z}. $$

Next it asks using the result from first part, to show that

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

and that it where I am stuck.

any hints?

Answer 1

You're almost there. Note that from odd symmetry that

$$\color{blue}{\sum_{n=-\infty}^\infty \frac{(-1)^n\,n}{n^2+\alpha^2}=\sum_{n=-\infty}^{-1} \frac{(-1)^n\,n}{n^2+\alpha^2}+\sum_{n=1}^\infty \frac{(-1)^n\,n}{n^2+\alpha^2}=0}$$

Now, setting $t=0$ in the equation

$$e^{\alpha t} = \frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}(\alpha + i n) e^{i n t}$$

reveals

$$\begin{align}1 &= \frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}(\alpha + i n) \\\\ &=\frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}\,\alpha}{\alpha^2 + n^2}+i\frac{\sinh (\pi \alpha)}{\pi}\color{blue}{\overbrace{ \sum_{n= -\infty}^\infty \frac{(-1)^{n}\,n}{\alpha^2 + n^2}}^{=0}} \\\\ &=\frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}\,\alpha}{\alpha^2 + n^2} \end{align}$$

whereupon solving for the series yields

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

as was to be shown!

Answer 2

Hint: Since $f$ is differentiable at $0$, its Fourier series at $0$ converges to $f(0)$.

Complete answer: You get

\begin{align} 1 &= \frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}(\alpha + i n) \\ &= \frac{\alpha\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}\tag{$\star$} \end{align}

hence $$ \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2} = \frac{\pi}{\alpha \sinh(\pi \alpha)} $$

Note. From the three following facts:

• if the series $\displaystyle\sum_{-\infty}^{\infty}$ converges then its value is equal to $\displaystyle\lim_{m\to\infty}\sum_{n=-m}^m$
• the term $a_n:=\frac{(-1)^n}{\alpha^2+n^2}in$ equals $0$ for $n=0$
• $a_n=-a_{-n}$ for $n>0$

we see that $$ \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}(\alpha + i n) = \alpha \lim_{m\to\infty}\sum_{n= -m}^m\frac{(-1)^{n}}{\alpha^2 + n^2} $$ But this last series converges absolutely hence the order of summation doesn't matter and indeed we have $$ \alpha \lim_{m\to\infty}\sum_{n= -m}^m\frac{(-1)^{n}}{\alpha^2 + n^2} = \alpha \sum_{n= -\infty}^{\infty}\frac{(-1)^{n}}{\alpha^2 + n^2} $$ This justifies $(\star)$.