5
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.