In this post I invoke certain function from a post of this site MathOverflow it is [1] (please see further references from the post, authors from the Springer link of the cited literature and answers from [1]: the integral mentioned in the post [1] seems thus that is due Lobachevskii, Application of imaginary geometry to certain integrals (1836)).

The idea was take certain multiple of the RHS of [1], multiply by the inverse function of the Gudermannian raised to a positive integer and after integrate the result, it is consider the integrals $$\mathcal{G}_n=\int_0^{\pi/2}\frac{\left(\operatorname{gd}^{-1}(z)\right)^n}{\sin z}\log\left(\frac{1+\sin z}{1-\sin z}\right)dz,$$ where as was said $\operatorname{gd}^{-1}(z)$ is the inverse function of the Gudermannian (see if you want the corresponding section from the Wikipedia's article Gudermannian function or the Mathworld's article with title Inverse Gudermannian) that is a function with a good mathematical content.

I tried to get a conjecture about the closed-form of those first few cases with the help of Wolfram Alpha online calculator (with standard time of computation for my computer). My belief is that the following conjecture holds (thus it is example of those that I evoke)

Conjecture. One gets $\mathcal{G}_5=\frac{5715}{2}\zeta(7).$

It is thus a case that I think enough difficult, but for which I think that due $\log (a/b)=\log a-\log b$ and the equivalent forms for the inverse of the Gudermannian function, maybe some user can to explain how get the closed-form from our integral $\mathcal{G}_5$.

Question. Prove or refute $$\int_0^{\pi/2}\frac{\left(\operatorname{gd}^{-1}(z)\right)^5}{\sin z}\log\left(\frac{1+\sin z}{1-\sin z}\right)dz=\frac{5715}{2}\zeta(7),$$ where $\zeta(s)$ denotes the Riemann zeta function. Please if the example $\mathcal{G}_5$, or the generalization $\mathcal{G}_n$ reduces easily to some integral that is in the literature refer/comment it, and I try to search and read the equivalent form of the integral from the literature. Many thanks.

I've used codes written in Wolfram Language similar than

int gd^(-1)(z)^5/sin(z) log((1+sin z)/(1-sin z))dz, from z=0 to pi/2


20 digits int gd^(-1)(z)^5/sin(z) log((1+sin z)/(1-sin z))dz, from z=0 to pi/2


[1] Interesting integral, from this MathOverflow (February, 2015).

  • $\begingroup$ In particular I don't know if the equivalent form of my integrand is explicitly in the literature, but I suspect that the corresponding indefinite integral for $\mathcal{G}_5$ should be very difficult. $\endgroup$
    – user142929
    Jul 23, 2019 at 20:26
  • 2
    $\begingroup$ Note that substitution $t=\mathrm{tg}(z/2)$ makes the picture much more clear, i'll post more a little bit later. $\endgroup$ Jul 24, 2019 at 14:25
  • $\begingroup$ Truly Lobachevsky was a great mathematician @RybinDmitry $\endgroup$
    – user142929
    Jul 24, 2019 at 16:47

1 Answer 1


First, $\mathrm{gd}^{-1}(z) = \ln \frac{1 + \tan(\frac{z}{2})}{1 + \tan(\frac{z}{2})}$. So we substitute $t = \tan(\frac{z}{2})$, obtaining:

$$\mathcal{G}_n = \int\limits_{0}^{1}2\cdot\ln^{n + 1}\left(\frac{1 + t}{1 - t} \right)\frac{d t}{t}$$ Then take $\frac{1+t}{1 - t} = e^{x}$, to get: $$ \int\limits_{0}^{\infty}x^{n + 1}\frac{4e^{x}dx}{e^{2x} - 1}$$

which is known to be equal to $2^{-n}(2^{n + 2} - 1)(n+1)!\zeta(n + 2)$, which gives $$\mathcal{G}_5 = \frac{(2^7 - 1)\cdot 6!}{2^{5}}\zeta(7) = \frac{5715}{2}\zeta(7)$$


  • 5
    $\begingroup$ To make it entirely self-contained, we can prove that integral representation by expanding $e^x / (e^{2x}-1)$ in a geometric series $e^{-x} + e^{-3x} + e^{-5x} + \cdots$ and integrating termwise, $\endgroup$ Jul 24, 2019 at 18:26

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