Infinite sums and integrals using residues

Question

I have no idea how to solve these two, any help?

$\mathtt{i)}$ $$\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\frac{e^{tz}}{\sqrt{z+1}}dz$$ $$ a,t\gt0$$ $\mathtt{ii)}$ $$ \sum_{n=1}^\infty \frac{\coth (n\pi)}{n^3}=\frac{7\pi^3}{180}$$

Answer 1

I'll do (i). Consider the contour integral

$$\oint_C dz \frac{e^{t z}}{\sqrt{z+1}}$$

where $C$ is the contour consisting of the line $\Re{z}=a$, $\Im{z} \in [-R,R]$; a circular arc of radius $R$ from $a+i R$ to $R e^{i \pi}$; a line from $R e^{i \pi}$ to $((1+\epsilon) e^{i \pi}$; a circle arc about $z=-1$ of radius $\epsilon$ from $((1+\epsilon) e^{i \pi}$ to $((1+\epsilon) e^{-i \pi}$; a line from $((1+\epsilon) e^{-i \pi}$ to $R e^{-i \pi}$; and a circular arc of radius $R$ from $R e^{-i \pi}$ to $a-i R$. Note that the integrals over all the circular arcs vanish as $R \to \infty$ and $\epsilon \to 0$. Thus we have, by Cauchy's theorem:

$$ \int_{a-i \infty}^{a+i \infty} dz \frac{e^{z t}}{\sqrt{z+1}} + e^{i \pi} e^{-t} \int_{\infty}^0 dx \, \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x}}+ e^{-i \pi} e^{-t} \int_{\infty}^0 dx \, \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x}} =0$$

Thus,

$$\frac1{i 2 \pi} \int_{a-i \infty}^{a+i \infty} dz \frac{e^{z t}}{\sqrt{z+1}} = \frac{1}{\pi} e^{-t} \int_0^{\infty} dx \, x^{-1/2} \, e^{-x t} $$

Using a substitution $x=u^2$ and a rescaling in the integral, we find that

$$\frac1{i 2 \pi} \int_{a-i \infty}^{a+i \infty} dz \frac{e^{z t}}{\sqrt{z+1}} = \frac{e^{-t}}{\sqrt{\pi t}}$$

Answer 2

For the second problem, let $ \displaystyle f(z) = \frac{\pi\cot (\pi z) \coth(\pi z)}{z^{3}}$ and integrate around a square ($C_{N}$) with vertices at $\pm (N+\frac{1}{2}) \pm (N+\frac{1}{2})$ where $N$ is a positive integer.

Both $\cot (\pi z)$ and $\coth(\pi z)$ are uniformly bounded on the contour.

So $ \displaystyle \int_{C_{N}} f(z) \ dz$ vanishes as $N$ goes to infinity through the integers..

And we have

$$ 2 \sum_{n=1}^{\infty} \frac{\coth (\pi n)}{n^{3}} + \text{Res}[f(z),0] + \sum_{n=1}^{\infty} \text{Res} [f(z),in] + \sum_{n=1}^{\infty} \text{Res} [f(z),-in] =0 $$

where

$$ \text{Res} [f(z),in] = \lim_{z \to in} \frac{\pi \cot (\pi z)}{\frac{d}{dz} (z^{3} \tanh (\pi z) )} = \lim_{z \to in} \frac{\pi \cot (\pi z)}{3z^{2} \tanh (\pi z) + \pi z^{3} \text{sech}^{2} (\pi z) } = \frac{\coth(\pi n)}{n^{3}}$$

and similarly

$$ \text{Res} [f(z),-in] = \frac{\coth (\pi n)}{n^{3}} \ .$$

And expanding $\cot (\pi z)$ and $\coth(\pi z)$ in Laurent expansions about the origin,

$$ \begin{align} \frac{\pi\cot (\pi z) \coth(\pi z)}{z^{3}} &= \frac{\pi}{z^{3}} \left( \frac{1}{\pi z} - \frac{\pi z}{3} - \frac{\pi^{3}z^{3}}{45} + \ldots \right) \left( \frac{1}{\pi z} + \frac{\pi z}{3} - \frac{\pi^{3}z^{3}}{45} + \ldots \right) \\ &= \frac{\pi}{z^{3}} \left(\frac{1}{\pi^{2}z^{2}} + \frac{1}{3} - \frac{\pi^{2}z^{2}}{45} - \frac{1}{3} - \frac{\pi^{2} z^{2}}{9} - \frac{\pi^{2}z^{2}}{45} + \ldots \right) \\ &= \frac{1}{\pi z^{5}} - \frac{7 \pi^{3}}{45 z} + \ldots \ \ .\end{align}$$

Therefore, $$\text{Res}[f(z),0] = - \frac{7 \pi^{3}}{45}$$

and $$\sum_{n=1}^{\infty} \frac{\coth (\pi n)}{n^{3}} = \frac{1}{4} \left( \frac{7 \pi^{3}}{45} \right) = \frac{7 \pi^{3}}{180} .$$