# Express $\int_0^{\pi/2}\{ \operatorname{gd}^{-1}(x)\}dx$ as series of special functions, with $\operatorname{gd}^{-1}(z)$ the inverse Gudermannian

I know that in the literature there are a lot of integrals involving the fractional part (and other floor and ceiling functions). For some of these integrals is provide its evaluation as a series or sum of series involving particular values of special functions when isn't possible to get the sum in closed-form.

I've considered the following proposal that I believe that isn't in the literature, I hope that the resulting series are interesting for this site.

Problem. We denote the inverse of the Gudermannian function, see this Wikipedia dedicated to this function, as $$\operatorname{gd}^{-1}(x)$$. Calculate an expression for the evaluation of $$I=\int_0^{\pi/2}\{ \operatorname{gd}^{-1}(x)\}dx,$$ where $$\{u\}$$ denotes the fractional part function.

Question. Is it possible to get an expression for $$I$$ as the sum of certain series involving particular values of special functions and presented, if possible, in a simplified expression? Many thanks.

I am asking about an expression of series because I think that isn't feasible to get the closed-form of $$I$$.

Then if there aren't mistakes in my change of variable $$x=\operatorname{gd}(y)$$ one has $$I=\sum_{k=0}^\infty\int_k^{k+1}\frac{y-k}{\cosh y}dy.$$ I know the indefinite integral using Wolfram Alpha online calculator

int (y-k)/cosh(y)dy

My main problem is to glimpse if will be possible to simplify some of those series, and how to group them in the result.

## References:

Also the encyclopedia MathWorld has an article dedicated to the function, that is the article with title Inverse Gudermannian.

The integral is twice Catalan's constant $$G = L(1,\chi_4) = 1 - \frac1{3^2} + \frac1{5^2} - \frac1{7^2} + \frac1{9^2} - + \cdots.$$ This constant can be computed efficiently to high precision, even though no further "closed form" is known or expected.

I guessed this as follows. Since $${\rm gd}^{-1}(x) = \int_0^x \sec t \, dt$$, your integral is $$\int_0^{\pi/2} (\frac\pi2 - t) \sec t \, dt$$. Ask gp to compute this numerically with

intnum(x=0,Pi/2,1/cos(x)*(Pi/2-x))


and get

1.8319311883544380301092070298647682217


; then ask the inverse symbolic calculator for a "simple lookup" of the first 1+24 digits, and the ISC recognizes the number as $$2G$$. The constant $$G$$ is known to gp as Catalan(), and indeed

2 * Catalan()


exactly matches the numerically computed integral.

This answer is simple enough that one expects it to be "well-known", and indeed this definite integral is given by Gradshteyn and Ryzhik as 3.747 #2, citing the Nouvelles tables d'intégrales définies of Bierens de Haan (1867).

• I didn't know it, many thanks you provide all details for the solutions of the integral, and in this case also the reference from the literature, required to accept it as an answer. My thoughts were that it was unknown from the outputs of Wolfram Alpha online calculator int frac(gd^(-1)(x))dx, from x=0 to pi/2 and int (y-k)/cosh(y)dy, from which one hopes tedious calculations for many series, and not a nice constant. – user142929 Sep 10 at 6:36

$$I=\int_0^{\pi/2}\{ \operatorname{gd}^{-1}(x)\}dx=\int_0^{\frac{\pi}{2}} \frac{\frac{\pi} {2} - x}{\cos x} dx\overset{\frac{\pi} {2} - x=t} =\int_{0}^\frac{\pi}{2}\frac{t}{\sin t}dt$$ $$\overset{IBP}=-\int_0^\frac{\pi}{2}\ln\left(\tan\frac{t}{2}\right)dt\overset{\tan \frac{t}{2}=x}=-2\int_0^1 \frac{\ln x}{1+x^2}dx$$ $$=-2\sum_{n=0}^\infty \int_0^1 x^{2n}\ln x dx=2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}=2G$$

• Many thanks for your answer. – user142929 Sep 12 at 8:21