The Fourier transform of the hyperbolic secant

A classic property of the hyperbolic secant \(\operatorname{sech}(\pi x) = \frac{1}{\cosh(\pi x)}\) is that, like the Gaussian, it is an eigenfunction of the Fourier transform. We will prove that for \(a > 0\),

\[ \int_{-\infty}^\infty \frac{e^{ia x}}{\cosh(\pi x)}\,dx = \frac{1}{\cosh\left(\frac{a}{2}\right)}. \]

Since \(\cosh(\pi x)\) is an even function, the imaginary part \(\int_{-\infty}^\infty \frac{\sin(ax)}{\cosh(\pi x)}\,dx\) vanishes by symmetry, so the formula is symmetric in \(a\) and holds for all real \(a \in \mathbb{R}\).

Method 1: An expanding rectangular contour

Let \(f(z) = \frac{e^{iaz}}{\cosh(\pi z)}\). The poles of \(f(z)\) occur where \(\cosh(\pi z) = 0\), namely at

\[ z_n = \frac{(2n+1)i}{2}, \qquad n \in \mathbb{Z}, \]

all of which lie on the imaginary axis with spacing \(i\). For each positive integer \(N \in \{1, 2, \dots\}\), define \(B_N\) to be the positively oriented rectangle with vertices \(-N\), \(N\), \(N+iN\), and \(-N+iN\):

Expanding rectangular contour B_N in the upper half-plane enclosing N poles on the imaginary axis

Choosing the top edge at height \(\operatorname{Im} z = N\) ensures that the contour passes halfway between the poles \(z_{N-1} = \left(N-\frac{1}{2}\right)i\) and \(z_N = \left(N+\frac{1}{2}\right)i\), enclosing exactly the \(N\) simple poles \(z_0, z_1, \dots, z_{N-1}\).

At \(z_n = \frac{(2n+1)i}{2}\), we have \(\frac{d}{dz}\cosh(\pi z)\big|_{z=z_n} = \pi\sinh\left(i\pi\left(n+\frac{1}{2}\right)\right) = \pi i (-1)^n\), so the residue is

\[ \mathop{\operatorname{Res}}_{z=\frac{(2n+1)i}{2}} \frac{e^{iaz}}{\cosh(\pi z)} = \frac{e^{-a\left(n+\frac{1}{2}\right)}}{\pi i (-1)^n} = \frac{(-1)^n e^{-\frac{a}{2}} e^{-an}}{\pi i}. \]

Applying the residue theorem to \(B_N\) and taking \(N \to \infty\) gives a convergent geometric series (since \(a > 0\) implies \(e^{-a} < 1\)):

\[ \begin{aligned} \lim_{N\to \infty}\int_{B_N}\frac{e^{ia z}}{\cosh(\pi z)}\,dz &= 2\pi i \sum_{n=0}^\infty \mathop{\operatorname{Res}}_{z=\frac{(2n+1)i}{2}} \frac{e^{ia z}}{\cosh(\pi z)} \\ &= 2\pi i \left(\frac{e^{-\frac{a}{2}}}{\pi i}\sum_{n=0}^\infty (-1)^n e^{-an}\right) \\ &= \frac{2e^{-\frac{a}{2}}}{1 + e^{-a}} \\ &= \frac{1}{\cosh\left(\frac{a}{2}\right)}. \end{aligned} \]

To verify that only the integral along the bottom edge \([-N, N]\) survives as \(N \to \infty\), we estimate the other three sides:

Top edge (\(z = x + iN\), \(-N \leq x \leq N\)): Since \(\cosh(\pi(x+iN)) = (-1)^N\cosh(\pi x)\) and \(|e^{ia(x+iN)}| = e^{-aN}\),

\[ \left|\int_{-N}^{N} \frac{e^{ia(x+iN)}}{\cosh(\pi(x+iN))}\,dx\right| \leq e^{-aN}\int_{-\infty}^\infty \frac{dx}{\cosh(\pi x)} = e^{-aN} \to 0 \quad (N \to \infty). \]

Vertical edges (\(z = \pm N + iy\), \(0 \leq y \leq N\)): By the reverse triangle inequality, \(|\cosh(\pi(\pm N + iy))| \geq \sinh(\pi N)\), while \(|e^{ia(\pm N + iy)}| = e^{-ay} \leq 1\), so

\[ \left|\int_0^N \frac{e^{ia(\pm N + iy)}}{\cosh(\pi(\pm N + iy))}\,i\,dy\right| \leq \frac{N}{\sinh(\pi N)} \to 0 \quad (N \to \infty). \]

Consequently,

\[ \int_{-\infty}^\infty \frac{e^{ia x}}{\cosh(\pi x)}\,dx = \lim_{N\to \infty}\int_{B_N}\frac{e^{ia z}}{\cosh(\pi z)}\,dz = \frac{1}{\cosh\left(\frac{a}{2}\right)}. \]

Method 2: A fixed-height strip enclosing a single pole

We can also avoid summing an infinite series of residues by using a rectangle \(\Gamma_R\) of fixed height \(1\) with vertices \(-R\), \(R\), \(R+i\), and \(-R+i\). Since \(\cosh(\pi(x+i)) = -\cosh(\pi x)\), the top edge is a constant multiple of the bottom edge:

\[ \int_{R+i}^{-R+i} \frac{e^{iaz}}{\cosh(\pi z)}\,dz = e^{-a}\int_{-R}^R \frac{e^{iax}}{\cosh(\pi x)}\,dx. \]

The vertical sides are bounded by \(\frac{1}{\sinh(\pi R)} \to 0\) as \(R \to \infty\), and \(\Gamma_R\) encloses only the single pole at \(z_0 = \frac{i}{2}\). The residue theorem then gives

\[ (1 + e^{-a})\int_{-\infty}^\infty \frac{e^{iax}}{\cosh(\pi x)}\,dx = 2\pi i \mathop{\operatorname{Res}}_{z=\frac{i}{2}} \frac{e^{iaz}}{\cosh(\pi z)} = 2\pi i \left(\frac{e^{-\frac{a}{2}}}{\pi i}\right) = 2e^{-\frac{a}{2}}, \]

which immediately divides through to \(\frac{1}{\cosh(a/2)}\) for every \(a \in \mathbb{R}\) (and, by analytic continuation, for every complex \(a\) in the strip \(|\operatorname{Im} a| < \pi\)).