I'm interested in the value of

$$ \theta(2) := -\frac{1}{2\pi^2}\int_0^\infty \log\left(1 - \frac{x}{\sinh x}\right)\,dx. $$

I have some hopes this admits a closed-form solution. Expanding the logarithm yields the integral $I_n := \int_0^\infty(\frac{x}{\sinh x})^n\,dx$, which has recently been asked about here on math.sa.

More generally, I'd love to know a closed-form solution for

$$\theta(\ell) := -\frac{1}{2\pi}\int_0^\infty \log\left(1 - \left\lvert \frac{\Gamma(\frac{\ell}{2}+ix)}{\Gamma(\frac{\ell}{2})}\right\rvert^2\right)\,dx$$

with $\ell\in\mathbb{N}$. For $\ell=2$ this is the same integral as above; see below for details.

The integral $\theta(\ell)$ has some relation to the eigenvalues of truncated random orthogonal matrices; see this paper, especially equation (7.1).

Some partial results below.

For $\ell=1$, the integral $\theta(1)$ is solvable by expanding the $\log$ and its value is

$$\theta(1) = \frac{3}{16}.$$

Note that for $b\in\mathbb{R}$,

$$ \left|\Gamma\left(\tfrac{1}{2}+bi\right)\right|^2 = \frac{\pi}{\cosh \pi b} \qquad\text{and}\qquad \left|\Gamma\left(1+bi\right)\right|^2 = \frac{\pi b}{\sinh \pi b}, $$

or more generally for $n\in\mathbb{N}$,

$$\begin{align} \left|\Gamma\left(1+n+bi\right)\right|^2 & = \frac{\pi b}{\sinh \pi b} \prod_{k=1}^n \left(k^2 + b^2 \right)\\ \left|\Gamma\left(\tfrac{1}{2} + n+bi\right)\right|^2 & = \frac{\pi}{\cosh \pi b} \prod_{k=1}^n \left(\left( k-\tfrac{1}{2}\right)^2 + b^2 \right) \end{align}$$

(for a derivation, see Wikipedia). Hence, when expanding the $\log$ the integral turns into the sum of the integral of powers of $\frac{1}{\sinh}$ and $\frac{1}{\cosh}$ times a polynomial.

For instance for $\ell=2$, one has

$$2\pi^2\theta(2) = \sum_{n=1}^\infty\frac{1}{n}\int_0^\infty \Bigl(\frac{x}{\sinh x}\Bigr)^n\,dx.$$

The integral $I_n := \int_0^\infty(\frac{x}{\sinh x})^n\,dx$ has recently been asked about here on math.sa. The answers prove a recurrence relation and indicate that $I_n$ is a linear combination of $\zeta(2), \zeta(4), \ldots, \zeta(2⌈n/2⌉)$ over $\mathbb{Q}$, where the coefficients are the "central factorial numbers", see this answer. (The central factorial numbers (OEIS A008955 and A008956) turn out to also show up in the power series expansion of $x\mapsto(\frac{x}{\sinh x})^n$, $x\mapsto(\arcsin x)^n$ and powers of various other trigonometric functions; see this paper for a thorough treatment.)

I'd be interested to know if $\sum_{n=1}^\infty\frac{1}{n}I_n$ admits a more succinct representation.

For larger values of $\ell$, the expansion produces terms of the form $\int_0^\infty x^n(\frac{1}{\sinh x})^m \, dx$ and $\int_0^\infty x^n(\frac{1}{\cosh x})^m \, dx$. In the question mentioned before, user Quanto has given a recurrence relation for such integrals. The polynomials in question make it likely central factorial numbers play a role in the linear combinations there too.

  • 4
    $\begingroup$ An idea for the case $\ell=2$ might be to compute $$ \int_0^\infty \log\big(1-\frac x {\sinh(x)}\big) dx = \int_0^\infty \int_0^1 \frac x {\sinh(x) -\lambda x} d\lambda dx. $$ Now $\sinh(x) -\lambda x\geq 0$ and we write $$ \int_0^\infty \int_0^1 \frac x {\sinh(x) -\lambda x} d\lambda dx = \int_0^\infty \int_0^1 \int_0^\infty x e^{-t(\sinh(x) -\lambda x)} d\lambda dx dt. $$ Performing the $\lambda$ integration $$\int_0^\infty \log\big(1-\frac x {\sinh(x)}\big) dx =\int_0^\infty \int_0^\infty e^{-t\sinh(x)}\frac {e^{x t} -1}t dx dt.$$ How to continue form here, I don't know. $\endgroup$
    – Martin
    Apr 22, 2021 at 15:27
  • 1
    $\begingroup$ $$\theta(2) := -\frac{1}{2\pi^2}\int_0^\infty \log\left(1 - \frac{x}{\sinh x}\right)\,dx.$$ to make it equivalent to the log gamma integral. $\endgroup$ Apr 30, 2021 at 15:02
  • $\begingroup$ @JamesArathoon: Thanks, fixed. $\endgroup$
    – heiner
    May 1, 2021 at 8:55
  • 1
    $\begingroup$ A couple of equivalent integrals for $\theta(2)$ $$\theta(2) :=\frac{1}{2 \pi ^2}\int_0^{\infty } \frac{z\, (1-z \coth (z)) \text{csch}(z)}{z\, \text{csch}(z)-1} \, dz$$ $$\theta(2) :=\frac{1}{2 \pi ^2}\int_0^1 \frac{2 \tanh ^{-1}(t) \left(\left(t^2+1\right) \tanh ^{-1}(t)-t\right)}{t \left(\left(t^2-1\right) \tanh ^{-1}(t)+t\right)} \, dt$$ $\endgroup$ May 1, 2021 at 14:45
  • $\begingroup$ A deleted answer suggested starting from $\int_0^\infty \ln(1-e^{-2x})\,dx = -\frac{\pi^2}{12}$, but didn't lead to a solution either. $\endgroup$
    – heiner
    May 1, 2021 at 16:50


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.