What's about the Laplace transform of the Gudermannian function?

Question

Let the Laplace transform, of a function $f(t)$, defined by (I say this unilateral integral transform)

$$F(s)=\int_0^\infty e^{-st}f(t)dt.$$ I am interested in special functions, one of the more recent that I've found in Internet is the Gudermannian function, here in Wikipedia you can see its definition. Also I'm interested in integral transforms, and was surprised because it seems possible calculate its Laplace transform. I don't know if it is possible to calculate or makes sense different integral transforms for this special function. But Wolfram Alpha online calculator knows how do calculations for this tak involving the digamma function and the secant function

LaplaceTransform[gd(t),t,s]

My question is

Question. Can you provide us calculations for $$\int_0^\infty e^{-st}\text{gd}(t)dt$$ where $\text{gd}(t)$ is the Gudermannian function? Thus I am asking about previous identity from this online calculator involving the digamma function and the secant, or other calculations that you can do with the purpose to justify and understand the Laplace transform of the Gudermannian function. Thanks in advance.

I tried with this online calculator different codes, from identities that satisfies the Gudermannian function, to get the Laplace transform of this function.

Answer 1

\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} \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.

Answer 2

Use:

$$\text{gd}(t)=2\arctan\left(e^t\right)-\frac{\pi}{2}$$

So, for the Laplace transform:

$$\int_0^\infty\text{gd}(t)e^{-\text{s}t}\space\text{d}t=\int_0^\infty\left(2\arctan\left(e^t\right)-\frac{\pi}{2}\right)e^{-\text{s}t}\space\text{d}t=$$ $$2\int_0^\infty\arctan\left(e^t\right)e^{-\text{s}t}\space\text{d}t-\frac{\pi}{2}\int_0^\infty e^{-\text{s}t}\space\text{d}t$$

Now, use ($\Re[s]>0$):

• $$\int_0^\infty e^{-\text{s}t}\space\text{d}t=\int_0^\infty1\cdot e^{-\text{s}t}\space\text{d}t=\mathcal{L}_t\left[1\right]_{(s)}=\frac{1}{s}$$

For the left integral ($\Re[s]>0$), you've to use the incomplete beta function:

$$\int_0^\infty\arctan\left(e^t\right)e^{-\text{s}t}\space\text{d}t$$

Using WolframAlpha:

The code:

integral_ (0)^(infinity) (arctan(e^t)e^(-st)) dt