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](https://en.wikipedia.org/wiki/Gudermannian_function#Inverse) 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*.



Express $\int_0^{\pi/2}\{ \operatorname{gd}^{-1}(x)\}dx$ as series of special functions, with $\operatorname{gd}^{-1}(z)$ the inverse Gudermannian
The integral is twice [Catalan's constant](https://en.wikipedia.org/wiki/Catalan%27s_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](http://wayback.cecm.sfu.ca/projects/ISC/ISCmain.html) 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).


