With $\operatorname{gd}(x)$ we denote the so-called Gudermannian function, see for example this MathWorld.

I know the closed-form of the following integral for the more simple cases of integers $n,m\geq 1$ $$\int_0^\infty\frac{(\operatorname{gd}(x))^n}{(\cosh(x))^m}dx.\tag{1}$$

I don't know if such family was in the literature.

Question. Can you provide me a summary of the strategy to get the closed-form of an integral in $(1)$, for example with $n=3$ and $m=4$? Are not required all tedious calculations, nor the exact closed-form: what it is required is the strategy to get the result. Thanks in advance.

  • $\begingroup$ As I've said only is required some idea of what should a strategy to simplify or systematize the calculations. $\endgroup$ – user243301 Dec 2 '17 at 14:24

$$I(m,n)=\int_{0}^{+\infty}\frac{\left(\arctan\sinh x\right)^n}{\left(\cosh x\right)^m }\,dx=\int_{0}^{+\infty}\frac{\left(\arctan u\right)^n}{\left(1+u^2\right)^{m+1/2}}\,du$$ can be simply computed by repeated integration by parts or by Fourier series, since $$ I(m,n) = \int_{0}^{\pi/2}\theta^n \left(\cos\theta\right)^{2m-1}\,d\theta. $$ In particular $$ I(4,3)=\frac{44658302}{13505625}-\frac{206656}{128625}\,\pi +\frac{2}{35}\,\pi^3.$$

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    $\begingroup$ Many thanks tomorrow I am going to follow your calculations and try see how works your method. My problem was that the CAS that I've used calculated a closed-form for the indefinite integral $(1)$ (and also computes the closed-form for the definite integral) but such calculations seem to me very complicated to my eyes. $\endgroup$ – user243301 Dec 2 '17 at 19:22

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