Compute the Fourier transform integral: \[I(\xi) = \int_{-\infty}^{\infty} \frac{e^{-2\pi i x \xi}}{\cosh(\pi x)} \, dx\] where \(\xi \in \mathbb{R}\) and \(\cosh z = \frac{e^z + e^{-z}}{2}\).
Solution
Let \[ F(z)=\frac{e^{-2\pi i\xi z}}{\cosh(\pi z)}. \] Integrate over the rectangle with vertices \(-R,R,R+i,-R+i\). The only pole in \(0<\Im z<1\) is \(z=i/2\), and \[ \operatorname{Res}_{z=i/2}F = \frac{e^{\pi\xi}}{\pi\sinh(i\pi/2)} = \frac{e^{\pi\xi}}{i\pi}. \] Hence the contour integral equals \[ 2\pi i\operatorname{Res}_{i/2}F=2e^{\pi\xi}. \]
On the top edge, using \[ \cosh(\pi(x+i))=-\cosh(\pi x), \] and its reversed orientation, \[ \int_{R+i}^{-R+i}F(z)\,dz =e^{2\pi\xi} \int_{-R}^{R}\frac{e^{-2\pi i\xi x}}{\cosh(\pi x)}\,dx. \] The vertical-edge integrals tend to \(0\) as \(R\to\infty\), since for \(z=\pm R+iy\), \(0\le y\le1\), \[ |e^{-2\pi i\xi z}|\le e^{2\pi|\xi|}, \qquad |\cosh(\pi z)|\ge \sinh(\pi R). \] Therefore \[ (1+e^{2\pi\xi})I(\xi)=2e^{\pi\xi}. \] Thus \[ I(\xi)=\frac{2e^{\pi\xi}}{1+e^{2\pi\xi}} =\frac1{\cosh(\pi\xi)}. \]