# Laplace Transform of the Gudermannian Function

How to find the Laplace transform of the Gudermannian function was a question on Mathematics Stack Exchange. Here is my solution.

\mathcal{L}[\mathrm{gd}(x)](s) = \int\limits_{0}^{\infty} \mathrm{e}^{-sx}\mathrm{gd}(x)\mathrm{d} x

Integration by parts yields
\begin{align}
\int\limits_{0}^{\infty} \mathrm{e}^{-sx}\mathrm{gd}(x)\mathrm{d} x & =
\frac{-1}{s}\mathrm{gd}(x)\mathrm{e}^{-sx}\big|_{0}^{\infty} +
\frac{1}{s}\int\limits_{0}^{\infty} \mathrm{e}^{-sx}\mathrm{sech}(x)\mathrm{d} x \\
& = 0 + \frac{1}{s}\mathcal{L}[\mathrm{sech}(x)](s) \\
& = \frac{1}{2s}\left[\psi\left(\frac{s+3}{4}\right) – \psi\left(\frac{s+1}{4}\right) \right]
\end{align}
for $$Re(s) > 0$$ due to evaluating the limit at $$x = \infty$$, while $$Re(s) > -1$$ for the Laplace transform of the hyperbolic secant.

Notes:

1. $$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{gd}(x) = \mathrm{sech}(x)$$
2. $$\mathrm{gd}(0) = 0\,$$ and $$\,\mathrm{gd}(\infty) = \frac{\pi}{2}$$
3. $$\psi(s)$$ is the digamma function.