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 evaluate:

$$ \int^\infty_{-\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}dx $$

Maybe we can evaluate it using the well-known result:$\int_{0}^{\frac{\pi}{2}} \ln{\sin t} \text{d}t=\int_{0}^{\frac{\pi}{2}} \ln{\cos t} \text{d}t=-\frac{\pi}{2}\ln{2}$

But how do I evaluate it, using that ?

share|improve this question

4 Answers 4

up vote 11 down vote accepted

Letting $x=\tan t$ leads to $$-4 \int^{\pi/2}_0 \log(\cos t) dt = 2\pi \log 2.$$

share|improve this answer

Alternatively to Ragib's substitution, you could consider $I(s) = \int_\mathbb{R} \left(1+x^2\right)^s \mathrm{d}x$, and then evaluate $I^\prime(-1)$.

$$ I(s) = \int_\mathbb{R} \left(1+x^2\right)^s \mathrm{d}x = 2 \int_0^\infty \left(1+x^2\right)^s \mathrm{d}x \stackrel{x^2=\frac{u}{1-u}}{=} \int_0^1 \left(1-u\right)^{-\frac{3}{2}-s}\frac{\mathrm{d} u}{\sqrt{u}} = B\left(\frac{1}{2}, -\frac{1}{2}-s\right) $$ Thus we established $I(s) = \sqrt{\pi}\frac{\Gamma\left(-\frac{1}{2}-s\right)}{\Gamma\left(-s\right)}$. We are now ready to compute the derivative: $$ I^\prime(s) = I(s) \left( \psi(-s) - \psi\left(-s-\frac{1}{2}\right) \right) $$ and $$ I^\prime(-1) = I(-1) \left( \psi(1) - \psi\left(\frac{1}{2}\right) \right) = \sqrt{\pi} \frac{\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)} \left( \psi(1) - \psi\left(\frac{1}{2}\right) \right) = 2 \pi \log(2) $$ where $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$ was used, as well as a polygamma duplication identity: $$ \psi(2s) = \log(2) + \frac{1}{2}\left(\psi(s) + \psi\left(s+\frac{1}{2}\right)\right) $$ that evaluated at $s=\frac{1}{2}$ gives $\psi(1) - \psi\left(\frac{1}{2}\right) = 2 \log(2)$.

share|improve this answer

since $\frac{\ln(x^{2}+1)}{x^{2}+1} $ is even function, $$ \int^\infty_{-\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}dx = 2\int^\infty_{0}\frac{\ln(x^{2}+1)}{x^{2}+1}dx $$

From here, you can fallow Ragib Zaman's process.

share|improve this answer

Let $$I(\alpha)=\int_{-\infty}^{\infty}\frac{\ln(\alpha x^2+1)}{x^2+1}dx.$$ Then \begin{eqnarray*} I'(\alpha)&=&\int_{-\infty}^{\infty}\frac{x^2}{(\alpha x^2+1)(x^2+1)}dx\\ &=&\frac{1}{\alpha-1}\int_{-\infty}^{\infty}\left[\frac{\alpha}{\alpha x^2+1}-\frac{1}{x^2+1}\right]dx\\ &=&\frac{\pi}{\sqrt{\alpha}+\alpha} \end{eqnarray*} and hence \begin{eqnarray*} I(\alpha)&=&\int\frac{\pi}{\sqrt{\alpha}+\alpha}d\alpha\\ &=&2\pi\ln(1+\sqrt{\alpha})+C. \end{eqnarray*} Clearly $I(0)=0$ implies $C=0$. Thus $$\int_{-\infty}^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx=I(1)=2\pi\ln 2.$$

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.