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

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.

– user243301
Aug 21 '16 at 10:04
• I believe this Laplace transform can be expressed in terms of digamma function. Aug 21 '16 at 10:39
• Very thanks @Nemo, then thanks for your contribution. I know how get easy Laplace transforms and Laplce transform properties, but was surprising an answer for this question from the online calculator. Thus I am asking to obtain the details ans obtain a nice answer.
– user243301
Aug 21 '16 at 10:42
• If now I say that it is neccesary integration by parts, perhaps I am wrong. I don't know how get the Laplace transform. Also I tried use the online calculator combining with the fact that the Laplace transform is linear with identities that satisfies this special function, but seem that with my codes don't work.
– user243301
Aug 21 '16 at 10:49
• Use $gd(x)=\pi/2-2\arctan(e^{-x}),~x>0$, and then expand the arctan and integrate termwise, then convert the resuting series to digamma function. Aug 21 '16 at 11:08

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

– user243301
Aug 21 '16 at 11:02
• Very thanks much for your answer, I try understand your answer in nexts hours. Is the last step, the integration $\int_0^\infty \arctan(e^t)e^{-st}dt$ where was my lack of knowledge, to do a comparison with myself calculation from the online calculator. Your answer is very valuable, many thanks.
– user243301
Aug 21 '16 at 11:20
• @user243301 You're welcome, I'm glad that I could help. Aug 21 '16 at 11:22
• You're generous when you help to all users with your good answers-
– user243301
Aug 21 '16 at 11:23

$$\mathcal{L}[\mathrm{gd}(x)](s) = \int\limits_{0}^{\infty} \mathrm{e}^{-sx}\mathrm{gd}(x)\mathrm{d} x$$ Integration by parts yields \begin{align} \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.
• Very thanks much, I've understand the integration by parts and that it is possible solve $\frac{0}{0}$ easily from L'Hòpital. Also that the Laplace transform of the hyperbolic secant is standard. Very thanks much to you and the other user, now we have these calculations in this site.
– user243301
Aug 21 '16 at 19:45