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