Integral with logarithm

• Mar 5th 2010, 02:05 PM
Krizalid
Integral with logarithm
Compute $\displaystyle \int_0^\infty\frac{\ln\big(1+x^2\big)}{x^{1+\alpha }}\,dx$ for $\displaystyle 0<\alpha<2.$
• Mar 5th 2010, 06:35 PM
TheEmptySet
Quote:

Originally Posted by Krizalid
Compute $\displaystyle \int_0^\infty\frac{\ln\big(1+x^2\big)}{x^{1+\alpha }}\,dx$ for $\displaystyle 0<\alpha<2.$

First integrate by parts to get with
$\displaystyle u=\ln(x^2+1) \implies du=\frac{2x}{x^2+1}$ and
$\displaystyle dv=x^{-(\alpha+1)} \implies v=-\frac{1}{\alpha}x^{-\alpha}$

This Formally gives

$\displaystyle \int_0^\infty\frac{\ln\big(1+x^2\big)}{x^{1+\alpha }}\,dx=-\frac{1}{\alpha}x^{-\alpha}\ln(x^2+1) \bigg|_{0}^{\infty}+\int_{0}^{\infty}\frac{1}{\alp ha}x^{-\alpha}\frac{2x}{x^2+1}dx$

Now justifying the limits

$\displaystyle \lim_{x\to \infty}\frac{\ln(x^2+1)}{x^\alpha}=\lim_{x \to \infty}\frac{\frac{2x}{x^2+1}}{\alpha x^{\alpha-1}} =\lim_{x \to \infty}\frac{2x^2}{(x^2+1)}\frac{1}{x^\alpha }=1\cdot 0$

$\displaystyle \lim_{x\to 0}\frac{\ln(x^2+1)}{x^\alpha}=\lim_{x \to 0}\frac{\frac{2x}{x^2+1}}{\alpha x^{\alpha-1}} =\lim_{x\to 0}\frac{2x^{2-\alpha}}{1}\frac{1}{ (x^2+1)}=0 \cdot 1=0$

Both of the above limits only exsist when $\displaystyle 0 < \alpha < 2$

This gives

$\displaystyle \int_0^\infty\frac{\ln\big(1+x^2\big)}{x^{1+\alpha }}\,dx=\frac{1}{\alpha}\int_{0}^{\infty}x^{-\alpha}\frac{2x}{x^2+1}dx$

Now let $\displaystyle t=x^2 \implies t^{\frac{1}{2}}=x \implies dt=2xdx$

This turns the right hand side into a form of the Beta function!

$\displaystyle \frac{1}{\alpha}\int_{0}^{\infty}x^{-\alpha}\frac{2x}{x^2+1}dx=\frac{1}{\alpha}\int_{0} ^{\infty}\frac{t^{-\frac{\alpha}{2}}}{(t+1)}dt=\frac{1}{\alpha}\beta \left( \frac{2-\alpha}{2},\frac{\alpha}{2}\right)=\frac{1}{\alpha }\Gamma\left( \frac{2-\alpha}{2}\right)\Gamma\left( \frac{\alpha}{2}\right)$

Now using Euler reflection formula the product of the two Gamma functions can be written as

$\displaystyle \frac{1}{\alpha}\Gamma\left( \frac{2-\alpha}{2}\right)\Gamma\left( \frac{\alpha}{2}\right)=\frac{1}{\alpha}\Gamma\lef t( 1-\frac{\alpha}{2}\right)\Gamma\left( \frac{\alpha}{2}\right) =\frac{\pi}{\alpha \sin\left( \frac{\pi \alpha}{2}\right)}$
• Mar 5th 2010, 10:13 PM
simplependulum
Are you expecting a beautiful solution ?

Intuitively , I think there must be another method which doesn't require any knowledges of Gamma function .
• Mar 6th 2010, 11:11 AM
Krizalid
no, i'm not looking for a beautiful solution, i just wanted to share this problem.

here's my solution:

$\displaystyle \int_{0}^{\infty }{\frac{x^{1-\alpha}}{1+x^{2}t}\,dx}=\frac{1}{2}t^{\frac{\alpha }{2}-1}\int_{0}^{\infty }{\frac{x^{-\frac{\alpha}{2}}}{1+x}\,dx}=\frac{1}{2}t^{\frac{\ alpha}{2}-1}\beta \left( 1-\frac{\alpha}{2},\frac{\alpha}{2} \right),$ thus the integral equals $\displaystyle \frac{1}{2}t^{\frac{\alpha}{2}-1}\Gamma \left( 1-\frac{\alpha}{2} \right)\Gamma \left( \frac{\alpha}{2} \right)=\frac{\pi }{2}t^{\frac{\alpha}{2}-1}\csc \left( \frac{\alpha\pi }{2} \right).$

hence $\displaystyle \int_{0}^{\infty }{\frac{\ln \left( 1+x^{2} \right)}{x^{1+\alpha }}}\,dx=\int_{0}^{\infty }{\int_{0}^{1}{\frac{x^{1-\alpha }}{1+x^{2}t}\,dt}\,dx}=\frac{\pi }{2}\csc \left( \frac{a\pi }{2} \right)\int_{0}^{1}{t^{\frac{a}{2}-1}\,dt}=\frac{\pi }{a}\csc \left( \frac{a\pi }{2} \right).$