24
$\begingroup$

Inspired by the user @Integrals, I thought I'd find some nice integrals! Especially interesting are those involving $\log \pi$. From Borwein and Devlin's "The Computer as Crucible", pg. 58 - show that

$$\displaystyle \int_0^{\infty} \frac{\log x}{\cosh^2x} \ \mathrm{d}x = \log\frac {\pi}4 - \gamma,$$

where $\gamma$ is the Euler-Mascheroni constant.

$\endgroup$
  • 3
    $\begingroup$ This integral appears in the calculation of the transition temperature in a superconductor. For instance, it's evaluated in Fetter and Walecka Book appendix. $\endgroup$ – Felix Marin Jun 29 '14 at 22:13
29
$\begingroup$

Let the considered integral be denoted by $I$. We have $$\eqalign{I&=\int_0^{1}\frac{\ln x}{\cosh^2x}dx+\int_{1}^\infty\frac{\ln x}{\cosh^2x}dx\cr &=\Big[(\ln x)\tanh x \Big]_0^1-\int_0^1\frac{\tanh x}{x}dx+ \Big[(\ln x)(\tanh x-1) \Big]_1^\infty-\int_1^\infty\frac{\tanh x-1}{x}dx\cr &=2\int_1^\infty\frac{dx}{x(1+e^{2x})}-\int_0^1\frac{1-e^{-2x}}{x(1+e^{-2x})}dx\cr &=2\int_1^\infty\frac{dx}{x(1+e^{2x})}-\int_0^1\frac{1-e^{-2x}}{x}\left(\frac{1}{1+e^{-2x}}-1\right)dx- \int_0^1\frac{1-e^{-2x}}{x}dx\cr &=2\int_1^\infty\frac{dx}{x(1+e^{2x})}+\int_0^1\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx- \int_0^1\frac{1-e^{-2x}}{x}dx\cr &=2\int_1^\infty\frac{dx}{x(1+e^{2x})}-\int_1^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx+\int_0^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx- \int_0^1\frac{1-e^{-2x}}{x}dx\cr &=\int_1^\infty\frac{1+e^{-2x}}{x(1+e^{2x})}dx+\int_0^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx- \int_0^1\frac{1-e^{-2x}}{x}dx\cr &=\int_1^\infty\frac{e^{-2x}}{x}dx- \int_0^1\frac{1-e^{-2x}}{x}dx+\int_0^\infty\frac{e^{-2x}-e^{-4x}}{x(1+e^{-2x})}dx\cr &=\underbrace{\int_2^\infty\frac{e^{-t}}{t}dt- \int_0^2\frac{1-e^{-t}}{t}dt}_A+\underbrace{\int_0^\infty\frac{e^{-t}-e^{-2t}}{t(1+e^{-t})}dt}_B} $$ Now, note that $$\eqalign{ A&=\int_2^\infty\frac{e^{-t}}{t}dt-\int_0^1\frac{1-e^{-t}}{t}dt- \int_1^2\frac{1-e^{-t}}{t}dt\cr &=\int_2^\infty\frac{e^{-t}}{t}dt-\int_0^1\frac{1-e^{-t}}{t}dt-\ln2 +\int_1^2\frac{e^{-t}}{t}dt\cr &=\int_1^\infty\frac{e^{-t}}{t}dt-\int_0^1\frac{1-e^{-t}}{t}dt-\ln2\cr &=-\gamma-\ln2 } $$ To calculate $B$ we can do the following $$\eqalign{ B&=\int_0^\infty\frac{e^{-t}-e^{-2t}}{t}\left(\sum_{n=0}^\infty(-1)^ne^{-nt}\right)dt\cr &=\sum_{n=0}^\infty(-1)^n\int_0^\infty\frac{e^{-(n+1)t}-e^{-(n+2)t}}{t}dt\cr &=\sum_{n=0}^\infty(-1)^n\ln\left(\frac{n+2}{n+1}\right)\cr &=\lim_{m\to\infty}\sum_{n=0}^{2m-2}(-1)^n\ln\left(\frac{n+2}{n+1}\right)\cr &=\lim_{m\to\infty}\ln\left(\prod_{k=1}^m\frac{2k}{2k-1}\prod_{k=1}^{m-1}\frac{2k}{2k+1}\right)\cr &=\lim_{m\to\infty}\ln\left(\frac{(2m)!!(2m-2)!!}{((2m-1)!!)^2}\right)=\ln\left(\frac{\pi}{2}\right)\cr } $$ By Stirling's formula. This yields $I=\ln \pi-2\ln 2-\gamma$. $\qquad\square$

$\endgroup$
  • 2
    $\begingroup$ By Wallis' product we have that $$\prod_{n=1}^{\infty} \frac{4n^2}{4n^2-1}=\frac{\pi}{2}$$ that simplifies things. (+1) for your answer $\endgroup$ – user 1357113 May 16 '14 at 12:45
28
$\begingroup$

Let $ \displaystyle I(a) = \int_{0}^{\infty} \frac{x^{a}}{\cosh^{2} x} \ dx$.

Then if $\text{Re} (a) >0$,

$$ \begin{align} I(a) &= 4 \int_{0}^{\infty} \frac{x^{a}}{(e^{x}+e^{-x})^{2}} \ dx \\ &= 4 \int_{0}^{\infty}x^{a} \frac{e^{-2x}}{(1+e^{-2x})^{2}} \ dx \\ &=4\int_{0}^{\infty} x^{a} \sum_{n=1}^{\infty} (-1)^{n-1} n e^{-2nx} \ dx \\ &=4 \sum_{n=1}^{\infty} (-1)^{n-1}n \int_{0}^{\infty} x^{a} e^{-2nx} \ dx \\ &= 4 \sum_{n=1}^{\infty} (-1)^{n-1} n \frac{\Gamma(a+1)}{(2n)^{a+1}} \\ &= \frac{2 \Gamma(a+1)}{2^{a}}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{a}} \\ &= \frac{2 \Gamma(a+1) \eta(a)}{2^{a}} \end{align}$$

where $\eta(a)$ is the Dirichlet eta function.

But by analytic continuation, the above formula is valid if $\text{Re}(a) >- 1$.

Then differentiating under the integral sign we get

$$ \begin{align} I'(a) &= \int_{0}^{\infty} \frac{x^{a} \log(x)}{\cosh^{2} x} \ dx \\ &= 2 \ \frac{[\Gamma'(a+1) \eta(a) + \Gamma(a+1) \eta'(a)]2^{a}- \Gamma(a+1) \eta(a) \log (2) 2^{a}}{2^{{2a}}} \ , \end{align}$$

which means

$$ \int_{0}^{\infty}\frac{\log x}{\cosh^{2}x} \ dx = I'(0) = 2 \Big( \Gamma'(1) \eta(0)+\eta'(0) - \eta(0) \log(2) \Big) .$$

The eta values can be determined using the relation

$$\eta(s) = (1-2^{1-s})\zeta(s) $$

and the known Riemann zeta values $$\zeta(0) = -\frac{1}{2} $$ and $$ \zeta'(0) = - \frac{1}{2} \log (2 \pi) . $$

Therefore,

$$ \begin{align} \int_{0}^{\infty}\frac{\log x}{\cosh^{2}x} \ dx &= 2 \left[-\gamma \left(\frac{1}{2}\right)+ \frac{1}{2} \log \left(\frac{\pi}{2}\right) -\frac{1}{2}\log(2) \right] \\ &= \log \left( \frac{\pi}{4}\right) - \gamma . \end{align}$$

$\endgroup$
  • 2
    $\begingroup$ Excellent method! $\endgroup$ – Bennett Gardiner May 8 '14 at 22:17
  • 1
    $\begingroup$ +1. That is the traditional proof. Pretty fine. This integral appears in the evaluation of the transition temperature of a superconductor. We arrive to that expression from the integration by parts of $\displaystyle\int_{0}^{\Lambda}{\tanh\left(\, x\,\right) \over x}\,{\rm d}x$. In that way we skip the divergence when $\displaystyle\Lambda \to \infty$. $\endgroup$ – Felix Marin Jan 16 '15 at 22:54
14
$\begingroup$

Largely speaking, we have

$$\begin{align} \int_0^{\infty} \frac{\log x}{\cosh^2x}\ \mathrm{d}x &=\lim_{a\to0} \lim_{b\to2}\frac{\partial}{ \partial a \partial b}\left(-4\int_0^{ \infty}\frac{x^{a-1}}{1+e^{b x}} \ dx\right)\\&= \lim_{a\to0} \lim_{b\to2}\frac{\partial}{\partial a \partial b}\left(b^{-a} (2^{3-a}-4) \Gamma(a) \zeta(a)\right)\\&=\log\left(\frac{\pi}{4}\right)-\gamma\end{align}$$

Done.

$\endgroup$
  • $\begingroup$ Obviously you find this solution in less 10 secondes right ? $\endgroup$ – Hexacoordinate-C Jul 13 '15 at 12:04
  • $\begingroup$ @Shadock I don't remember how fast I was, but now I only edited the answer because of the latex rendering issue. $\endgroup$ – user 1357113 Jul 13 '15 at 12:07
  • $\begingroup$ So use \mathrm{d}x in you integral it looks better :) $\endgroup$ – Hexacoordinate-C Jul 13 '15 at 12:08

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.