## 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.

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

## 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.

\begin{equation}
\mathcal{L}[f(t)](s) = \int\limits_{0}^{\infty} \mathrm{e}^{-st} f(t) \mathrm{d} t
\label{eq:1608061}
\tag{1}
\end{equation}

\begin{equation}
(f*g)(t) = \int\limits_{0}^{t} f(\tau) g(t-\tau) \mathrm{d} \tau
\label{eq:1608062}
\tag{2}
\end{equation}

\begin{equation}
\mathcal{L}[f*g] = \mathcal{L}[f]\mathcal{L}[g]
\label{eq:1608063}
\tag{3}
\end{equation}

We begin with the following two functions
\begin{equation}
f(t) = t^{x-1} \quad \mathrm{and} \quad g(t) = t^{y-1}
\label{eq:1608064}
\tag{4}
\end{equation}

These functions were chosen for two reasons. First, they appear in the basic integral definition of the beta function
\begin{equation}
\mathrm{B}(x,y) = \int\limits_{0}^{1} t^{x-1} (1-t)^{y-1} \mathrm{d} t
\label{eq:1608065}
\tag{5}
\end{equation}
and, as we now show, their Laplace transforms result in gamma functions
\begin{align}
\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}}
\label{eq:1608066}
\tag{6}
\end{align}
where we have used the substitution $$v = st$$. Likewise, taking the Laplace transform of $$g(t)$$ yields,
\begin{equation}
\mathcal{L}[g(t)] = \mathcal{L}[t^{y-1}] = \frac{\Gamma(y)}{s^{y}}
\label{eq:1608067}
\tag{7}
\end{equation}

Now we substitute equations \eqref{eq:1608067}, \eqref{eq:1608066}, and \eqref{eq:1608062} into \eqref{eq:1608063}
\begin{equation}
\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]
\label{eq:1608068}
\tag{8}
\end{equation}

To evaluate the integral, we let $$\tau = tu$$
\begin{align}
\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)
\label{eq:1608069}
\tag{9}
\end{align}

Thus, we have
\begin{align}
\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}}
\label{eq:16080610}
\tag{10}
\end{align}

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