How to Prove: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$
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1Please avoid titles that are entirely in $\TeX$. – J. M. ain't a mathematician Apr 20 '12 at 17:35
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3@J.M. I couldn't think of a better title. – peter Apr 20 '12 at 17:45
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Put $x = \tan{\theta}$. Then you have
\begin{align*} I &= \int_{0}^{\pi/2} \log(\tan\theta + \cot\theta) \ d\theta \\\ &= \int_{0}^{\pi/2} \log\frac{2}{\sin{2\theta}} \ d\theta \\\ &= - \int_{0}^{\pi/2} (\log{\sin\theta} +\log{\cos\theta}) d\theta \\\ &= -2 \int\limits_{0}^{\pi/2} \log\: {\sin\theta} \ d\theta = \pi \: \log{2} \end{align*}
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1With regards to the log sine integral, see this article. – J. M. ain't a mathematician Apr 20 '12 at 17:50
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Rewrite it as $$ I =\int_0^\infty \frac{\log\left(x+\frac{1}{x}\right)}{x+\frac{1}{x}} \frac{\mathrm{d} x}{x} = \left.\frac{\mathrm{d}}{\mathrm{d}s} \mathcal{I}(s)\right|_{s=-1} $$ where $$ \mathcal{I}(s) = \int_0^\infty \left( x+\frac{1}{x}\right)^s \frac{\mathrm{d} x}{x} \stackrel{x=\sqrt{\frac{u}{1-u}}}{=} \frac{1}{2} \int_0^{1} (u(1-u))^{-s/2-1} \mathrm{d} x $$ Thus $$ \mathcal{I}(s) = \frac{1}{2} \operatorname{B}\left(-\frac{s}{2},-\frac{s}{2} \right) = \frac{1}{2} \frac{\Gamma^2\left(-\frac{s}{2}\right)}{\Gamma(-s)} $$ It remains to evaluate the derivative at $s=-1$: $$ I = \left.\frac{\mathrm{d}}{\mathrm{d}s} \mathcal{I}(s)\right|_{s=-1} = \mathcal{I}(-1)\left( \psi(1) - \psi\left(\frac{1}{2}\right) \right) = \pi \log(2) $$ The result follows from $\mathcal{I}(-1) = \frac{1}{2} B(1/2,1/2) = \frac{\pi}{2}$, and from duplication identity for digamma, evaluated as $s=1$: $$ \psi(s) = \log(2) + \frac{1}{2} \left( \psi\left( \frac{s}{2} \right) + \psi\left( \frac{s+1}{2} \right) \right) \stackrel{s=1}{\implies } \psi(1) - \psi(1/2) = 2 \log(2) $$
Sasha
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Here I give another way to solve the problem. We rewrite the integral as \begin{eqnarray} I&=&\int_0^\infty\log(x+\frac{1}{x})\frac{1}{1+x^2}dx\\ &=&\int_0^\infty\frac{\log(1+x^2)}{1+x^2}dx-\int_0^\infty\frac{\log x}{1+x^2}dx \end{eqnarray} Note $$ \int_0^\infty\frac{\log(1+x^2)}{1+x^2}dx=\pi\log 2$$ from Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$ and $$ \int_0^\infty\frac{\log x}{1+x^2}dx=0. $$ Thus we have $$ I=\pi \log 2.$$
