# $\int_0^\infty \log\left(1 - \frac{x}{\sinh x}\right)\,dx$ and generalisations

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.

• 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. Apr 22, 2021 at 15:27
• $$\theta(2) := -\frac{1}{2\pi^2}\int_0^\infty \log\left(1 - \frac{x}{\sinh x}\right)\,dx.$$ to make it equivalent to the log gamma integral. Apr 30, 2021 at 15:02
• @JamesArathoon: Thanks, fixed. May 1, 2021 at 8:55
• A couple of equivalent integrals for $\theta(2)$ $$\theta(2) :=\frac{1}{2 \pi ^2}\int_0^{\infty } \frac{z\, (1-z \coth (z)) \text{csch}(z)}{z\, \text{csch}(z)-1} \, dz$$ $$\theta(2) :=\frac{1}{2 \pi ^2}\int_0^1 \frac{2 \tanh ^{-1}(t) \left(\left(t^2+1\right) \tanh ^{-1}(t)-t\right)}{t \left(\left(t^2-1\right) \tanh ^{-1}(t)+t\right)} \, dt$$ May 1, 2021 at 14:45
• 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. May 1, 2021 at 16:50

I will illustrate why it is unlikely a closed form in elementary functions exists.

Let $$I=\int_0^\infty\log\left(1-\frac x{\sinh x}\right)\,dx=\int_0^\infty\frac{x(x\coth x-1)}{x-\sinh x}\,dx$$ on integration by parts. In exponential notation, we have $$I=-\int_{-\infty}^\infty\frac{xe^x((x-1)e^{2x}+x+1)}{(e^{2x}-1)(e^{2x}-2xe^x-1)}\,dx$$ where we use the fact that the integrand $$f(x)$$ is even.

Taking a semicircular contour on the upper half plane, we have $$I=2\pi i\left(\sum_{k>0}\underbrace{\operatorname{Res}(f,ik\pi)}_{e^{2z}-1\,\text{contribution}}+\sum_j\underbrace{\operatorname{Res}(f,w_j)}_{\text{where}\,e^{2w_j}-2w_je^{w_j}=1}\right)$$ since the arc contribution $$\lim_{R\to+\infty}\int_0^\pi iRe^{i\theta}f(Re^{i\theta})\,d\theta=\lim_{R\to+\infty}\int_0^\pi iRe^{i\theta}\frac{Re^{i\theta}(Re^{i\theta}-1)+O(1)}{e^{Re^{i\theta}}-2Re^{i\theta}+O(1)}\,d\theta=0$$ vanishes due to the dominant exponential term on the denominator.

By direct computation we can show that $$\operatorname{Res}(f,ik\pi)=-ik\pi/2$$ for a positive integer $$k$$. However, this cannot be said for the roots $$w_j$$ of $$e^{2z}-2ze^z-1$$. Upon writing $$w_j=x+iy$$, we obtain a system of highly transcendental equations \begin{align}e^{2x}\cos2y-2e^x(x\cos y-y\sin y)&=1\\(e^x\sin y+x\tan y+y)\cos y&=0\end{align} which appears to have infinitely many solutions; note that theoretically there must be infinitely many since they need to cancel out with the divergent sum $$\sum\limits_{k>0}\operatorname{Res}(f,ik\pi)$$. So unless there is a miraculous simplification of this system, it is extremely unlikely we can obtain a closed form for $$\sum\limits_j\operatorname{Res}(f,w_j)$$ in elementary functions, and hence for $$I$$.

• (+1) The integral is a sum of roots of a transcendental equation. Is there a way to show this result without the residue theorem nor contour integrals? Nov 8, 2022 at 18:47
• One can show the integral is a series of values of the form $\zeta(2n)$ over $\mathbb{Q}$, where the series is not absolutely convergent, see math.stackexchange.com/a/4081308/5051. The coefficients involved are the so-called central factorial numbers. Nov 9, 2022 at 14:48

Instead of trying to solve the question for $$\ell = 2$$ first it might be easier to solve the question for $$\ell = 3,5,7,9,...$$ as all of these result in integrals involving $$\cosh(x)$$ and for $$l=1$$ one can solve it.

For $$\ell=3$$ one can do the following: $$\left| \frac{\Gamma(\frac 3 2 + i x)}{\Gamma(\frac 3 2)} \right|^2 = \left| \frac{\Gamma(\frac 1 2 + i x)}{\Gamma(\frac 1 2)} \right|^2 (1+ 4 x^2) = \frac 1 {\cosh(\pi x)} (1 + 4 x^2).$$ Using this and expaning the logarithm, we get $$-\int_0^\infty dx \log \left( 1 - \left| \frac{\Gamma(\frac 3 2 + i x)}{\Gamma(\frac 3 2)} \right|^2\right) =\sum_{n=1}^\infty \frac 1 n \int_0^\infty dx (1 + 4x^2)^n \frac 1 {\cosh^n(\pi x)}.$$ Now we use the following identity (Gradshtein/Ryzhik eq. 3.512, which is also true for $$b=0$$) $$\int_0^\infty dx \frac{\cosh(2 b x)}{\cosh^{2v}(ax)} = \frac {4^{v-1}} a B\left(v+ \frac b a, v - \frac b a\right)$$ where $$B(\ ,\ )$$ is the Beta function. We further note that $$\frac {d^{2k}}{d b^{2k}} \cosh(2bx)\big|_{b=0} = (4x)^{2k}.$$ Not bothering about any convergence and expanding $$(1+4x^2)^n$$, this implies \begin{align} \int_0^\infty dx (1 + 4x^2)^n \frac 1 {\cosh^n(\pi x)} &= \sum_{k=0}^n \binom{n}{k}\int_0^\infty dx \frac { \frac {d^{2k}}{d b^{2k}} \cosh(2bx)\big|_{b=0}} {\cosh^n(\pi x)} \\ &= \sum_{k=0}^n \binom{n}{k} \frac{4^{n/2-1}} \pi \frac {d^{2k}}{d b^{2k}} B\left(\frac n 2 + \frac b \pi, \frac n 2 - \frac b \pi \right)\big|_{b=0}. \end{align} Writing the Beta function in terms of the $$\Gamma$$ function and rearranging the summations a little, we end up with \begin{align} &-\int_0^\infty dx \log \left( 1 - \left| \frac{\Gamma(\frac 3 2 + i x)}{\Gamma(\frac 3 2)} \right|^2\right) \\ =& \frac 1 \pi \sum_{k=0}^\infty \frac{ d^{2k}}{d b^{2k}} \left(\sum_{n: n\geq \max\{k,1\}} \frac {4^{\frac n 2 -1}}{k! (n-k)!} \Gamma\left(\frac n 2 +\frac b \pi\right) \Gamma\left(\frac n 2 -\frac b \pi\right)\right)\Big|_{b = 0} \\ =& \frac 1 \pi \sum_{k=0}^\infty \frac {4^{k/2}}{k!}\frac{ d^{2k}}{d b^{2k}} \left(\sum_{n = 0}^\infty \frac {4^{\frac n 2 -1}}{n!} \Gamma\left(\frac {n+k} 2 +\frac b \pi\right) \Gamma\left(\frac {n + k} 2 -\frac b \pi\right)\right)\Big|_{b = 0} . \end{align} This is not a complete answer but maybe a step in the right direction? Similar calculations work for all odd $$\ell$$.