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
\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]
for \(Re(s) > 0\) due to evaluating the limit at \(x = \infty\), while \(Re(s) > -1\) for the Laplace transform of the hyperbolic secant.


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.

A Laplace Transform Proof of \(\mathrm{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\)

This proof appeared in Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals by George Boros and Victor Moll. Here, I fill in the steps of the proof.

We begin with 3 fundamental results of the Laplace transform: the basic definition, convolution, and the Laplace transform of a convolution.

\mathcal{L}[f(t)](s) = \int\limits_{0}^{\infty} \mathrm{e}^{-st} f(t) \mathrm{d} t

(f*g)(t) = \int\limits_{0}^{t} f(\tau) g(t-\tau) \mathrm{d} \tau

\mathcal{L}[f*g] = \mathcal{L}[f]\mathcal{L}[g]

We begin with the following two functions
f(t) = t^{x-1} \quad \mathrm{and} \quad g(t) = t^{y-1}

These functions were chosen for two reasons. First, they appear in the basic integral definition of the beta function
\mathrm{B}(x,y) = \int\limits_{0}^{1} t^{x-1} (1-t)^{y-1} \mathrm{d} t
and, as we now show, their Laplace transforms result in gamma functions
\mathcal{L}[f(t)] & = \mathcal{L}[t^{x-1}] = \int\limits_{0}^{\infty} \mathrm{e}^{-st} t^{x-1} \mathrm{d} t \\
& = \frac{1}{s^{x}} \int\limits_{0}^{\infty} \mathrm{e}^{-v} v^{x-1} \mathrm{d} v \\
& = \frac{\Gamma(x)}{s^{x}}
where we have used the substitution \(v = st\). Likewise, taking the Laplace transform of \(g(t)\) yields,
\mathcal{L}[g(t)] = \mathcal{L}[t^{y-1}] = \frac{\Gamma(y)}{s^{y}}

Now we substitute equations \eqref{eq:1608067}, \eqref{eq:1608066}, and \eqref{eq:1608062} into \eqref{eq:1608063}
\frac{\Gamma(x)}{s^{x}} \frac{\Gamma(y)}{s^{y}} = \mathcal{L}\left[\int\limits_{0}^{t} \tau^{x-1} (t – \tau)^{y-1} \mathrm{d} \tau \right]

To evaluate the integral, we let \(\tau = tu\)
\int\limits_{0}^{t} \tau^{x-1} (t – \tau)^{y-1} \mathrm{d} \tau & = t^{x+y-1} \int\limits_{0}^{1} u^{x-1} (1 – u)^{y-1} \mathrm{d} u \\
& = t^{x+y-1} \mathrm{B}(x,y)

Thus, we have
\mathcal{L}[t^{x+y-1} \mathrm{B}(x,y)] & = \frac{\Gamma(x)}{s^{x}} \frac{\Gamma(y)}{s^{y}} \\
& = \mathrm{B}(x,y) \int\limits_{0}^{\infty} \mathrm{e}^{-st} t^{x+y-1} \mathrm{d} t \\
& = \mathrm{B}(x,y) \frac{1}{s^{x+y}} \int\limits_{0}^{\infty} \mathrm{e}^{-v} v^{x+y-1} \mathrm{d} v \\
& = \mathrm{B}(x,y) \frac{\Gamma(x+y)}{s^{x+y}}

Combining the first and last results from the right hand side of the above equation yields our final result
\mathrm{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}