Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to Prove: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$

share|improve this question
1  
Please avoid titles that are entirely in $\TeX$. –  J. M. Apr 20 '12 at 17:35
2  
@J.M. I couldn't think of a better title. –  peter Apr 20 '12 at 17:45

3 Answers 3

up vote 6 down vote accepted

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*}

share|improve this answer
1  
With regards to the log sine integral, see this article. –  J. M. Apr 20 '12 at 17:50
    
@J.M. Yeah i think $\log{\sin{\theta}}$ was asked here sometime before. –  user9413 Apr 20 '12 at 17:53
1  
@Chandrasekhar : You might consider up-voting the question. –  Michael Hardy Apr 20 '12 at 19:10
    
@Hardy: Why do you say like that –  user9413 Apr 20 '12 at 20:00

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 $s=1$: $$ \psi(s) = \log(2) + \frac{1}{2} \left( \psi\left( \frac{s}{2} \right) + \psi\left( \frac{s+1}{2} \right) \right) \stackrel{s=1}{\implies } \psi(1) - \psi(1/2) = 2 \log(2) $$

share|improve this answer

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 $$ I=\pi \log 2.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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