How to prove the following? $$ \int_0^\infty \log x \operatorname{sech}x\,dx = \frac{\pi}{2} \log\left( \frac{4\pi^3}{\Gamma(1/4)^4} \right) $$ I obtained the right side with CAS. It seems like this function has many poles on the imaginary axis so a simple contour integral cannot be used. I also tried the following $$\begin{align*} \int_0^\infty \log x \operatorname{sech} x\, dx &= \int_0^\infty \operatorname{sech} x \left. \frac{\partial}{\partial s} x^s \right|_{s=0} dx \\ &= \left. \frac{\partial}{\partial s} \int_0^\infty x^s \operatorname{sech} x\,dx \right|_{s=0} \end{align*}$$ However, the last integral is very difficult to evaluate and contains terms of $\zeta$ functions.


Hint. One may write $$\begin{align} \int_0^\infty \log x \operatorname{sech}x\,dx &= 2\int_{0}^{\infty} \frac{\partial_s \left.\left(x^s\right)\right|_{s=0} }{1 + e^{-2x}} \:e^{-x}\: dx \\\\ &=2\: \left.\partial_s \left(\int_{0}^{\infty} \frac{x^s e^{-x}}{1 + e^{-2x}} \; dx \right)\right|_{s=0}\\\\ &= 2\:\left.\partial_s \left(\int_{0}^{\infty} \sum_{n=0}^{\infty} (-1)^n x^s e^{-(2n+1)x} \; dx \right)\right|_{s=0}\\\\ &=2\: \left.\partial_s \left(\sum_{n=0}^{\infty} (-1)^n \, \frac{\Gamma(s+1)}{(2n+1)^{s+1}} \right)\right|_{s=0}\\\\ &=2\: \left.\partial_s \left( \Gamma(s+1)\beta(s+1)\right)\right|_{s=0}\\\\ &= -2\:\gamma \cdot\beta(1)+2\:\beta'(1) \end{align}$$ where $\gamma $ is the Euler-Mascheroni constant, $\beta(\cdot)$ is the Dirichlet beta function $$ \beta(s)=\sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)^{s}} $$ then one may use the special values $$ \beta(1)=\frac{\pi}4,\qquad \beta'(1) = \frac{\pi}{4} \left[ \gamma + 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right], $$ proved here to obtain

$$\int_0^\infty \frac{\log x}{\cosh x}\,dx = \frac{\pi}{2} \:\log\left( \frac{4\pi^3}{\Gamma(1/4)^4} \right) $$

as announced.

  • $\begingroup$ Oh, great, I was just missing that $\beta'(1)$ is well-known. Nicely done, (+1). $\endgroup$ – Jack D'Aurizio Dec 23 '15 at 10:15

The residue theorem gives: $$ \text{sech}(z) = \sum_{n\geq 0}\frac{4\pi(-1)^n (2n+1)}{4z^2+(2n+1)^2}\tag{1}$$ hence:

$$ \int_{0}^{+\infty}x^s \text{sech}(x)\,dx = \pi^{s+1}2^{-s}\sec\left(\frac{\pi s}{2}\right)\sum_{n\geq 0}(-1)^n (2n+1)^s\tag{2} $$ and that can be simplified to: $$\begin{eqnarray*}\int_{0}^{+\infty}x^s \text{sech}(x)\,dx &=& \frac{\Gamma(s+1)}{2^{2s+1}}\left(\zeta(1+s,1/4)-\zeta(1+s,3/4)\right)\\ &=&2\Gamma(s+1)\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{s+1}}\tag{3} \end{eqnarray*}$$ i.e. a product between a value of the $\Gamma$ function and a value of a Dirichlet $L$-function associated with the non-principal character $\!\pmod{4}$. It is not difficult to check that the limit as $s\to 0$ of the LHS of $(3)$ is $\pi/2$. To compute $\frac{d}{ds}$ of $(3)$, we may recall that $f'(s)=f(s)\cdot\frac{d}{ds}\log(f(s))$, and both the $\Gamma$ function and the $L$-function appearing in $(3)$ can be written in a nice way as infinite products: the Weierstrass product for the $\Gamma$ function leads to $\frac{\Gamma'}{\Gamma}(1)=-\gamma$, while the Euler product: $$ \sum_{n\geq 0}\frac{(-1)^{n+1}}{(2n+1)^{s+1}}=\prod_{p\equiv 1\pmod{4}}\left(1-\frac{1}{p^{s+1}}\right)^{-1}\cdot \prod_{p\equiv 3\pmod{4}}\left(1+\frac{1}{p^{s+1}}\right)^{-1}\tag{4}$$ should lead us to the final answer. Still working on this one.

  • $\begingroup$ That would be a viable way to obtain $2\Gamma(s+1)\beta(s+1)$. $\endgroup$ – Henricus V. Dec 23 '15 at 3:34

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