Computation of integral

Calculus (Integrals, Series)
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Tsakanikas Nickos
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Computation of integral

#1

Post by Tsakanikas Nickos »

Show that

$$\int_{- \infty}^{+ \infty} {e}^{-2 \pi i x \xi} \frac{ \sin \pi \alpha }{ \cosh \pi x + \cos \pi \alpha } \mathrm{d} x = 2 \frac{ \sinh 2 \pi \alpha \xi }{ \sinh 2 \pi \xi }$$

where $\xi \in \mathbb{R}, \; \alpha \in (0, 1)$.
r9m
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Re: Computation of integral

#2

Post by r9m »

We have: $$\begin{align*}\frac{\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha} &= \frac{1}{1+e^{-\pi(x+i\alpha)}}-\frac{1}{1+e^{-\pi(x - i\alpha)}}\\&= 2\sum\limits_{n=1}^{\infty} (-1)^{n}e^{-n\pi x}\sin n\pi\alpha\end{align*}$$

Thus, we have:

\begin{align*}\int_{-\infty}^{\infty} \frac{e^{-2\pi i \xi x}\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha}\,dx &= \int_{0}^{\infty} \frac{2\sin \pi \alpha \cos 2\pi\xi x}{\cosh \pi x + \cos \pi \alpha}\,dx\\&= 4\int_0^{\infty}\sum\limits_{n=1}^{\infty}(-1)^n e^{-n\pi x}\cos 2\pi\xi x \sin n\pi \alpha\,dx\\&= \frac{4}{\pi}\sum\limits_{n=1}^{\infty}(-1)^n \frac{n\sin n\pi\alpha}{n^2+4\xi^2}\\&= \frac{2}{\pi}\sum\limits_{n=1}^{\infty} (-1)^n\left(\frac{1}{n+2i\xi}+\frac{1}{n-2i\xi}\right)\sin n\pi \alpha\\&= \frac{2}{\pi}\sum\limits_{n = -\infty}^{\infty} \frac{(-1)^n\sin n\pi\alpha}{n+2i\xi}\\&= 2\frac{\sin 2\pi i \alpha\xi}{\sin 2\pi i\xi} \qquad \textrm{ (by residue theorem) }\\&= 2\frac{\sinh 2\pi \alpha\xi}{\sinh 2\pi \xi}\end{align*}

Justification for application of residue theorem$^\star$:

Choose $\displaystyle f(z) = \frac{\sin \pi\alpha z}{w+z}$, then by residue theorem:

$\displaystyle \frac{1}{2\pi i}\int_{\gamma_N} \pi\csc (\pi z) f(z)\,dz = \sum\limits_{n=-N}^{N} (-1)^nf(n) + \operatorname{Res}\limits_{z = -w} \left(\pi \csc (\pi z)f(z)\right)$

where, $\gamma_N$ is a rectangular contour with vertices: $\displaystyle A = \left(N+\frac{1}{2}\right)(1+i), B=\left(N+\frac{1}{2}\right)(1-i),C=\left(N+\frac{1}{2}\right)(-1-i)$ and $\displaystyle D=\left(N+\frac{1}{2}\right)(-1+i)$.

Since, $\displaystyle \lim\limits_{N \to \infty} \int_{\gamma_N} \pi\csc (\pi z) f(z)\,dz = 0$, we have: $\displaystyle \sum\limits_{n = -\infty}^{\infty} (-1)^nf(n) = -\operatorname{Res}\limits_{z = -w} \left(\pi \csc (\pi z)f(z)\right) = \frac{\pi\sin \pi \alpha w}{\sin \pi w}$
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