Residue Theorem for Improper Integral

Question

$$\int^\infty_0 log(1+x^2)\frac{dx}{x^{1+\alpha}}(0<\alpha<2)$$

How to do with the log and it says try integration by parts.

Answer 1

You shouldn't fear following the book hint.

$$ I=\int_{0}^{+\infty}\log(1+x^2)\frac{dx}{x^{1+\alpha}} \stackrel{\text{IBP}}{=} \frac{1}{\alpha}\int_{0}^{+\infty}\frac{2x}{1+x^2}\cdot \frac{dx}{x^\alpha} $$ and by setting $\frac{1}{1+x^2}=u$ the last integral turns into a value for a Beta function:

$$ I = \frac{1}{\alpha}\int_{0}^{1} u^{\frac{\alpha}{2}-1}(1-u)^{-\frac{\alpha}{2}}\,du = \frac{\Gamma\left(\frac{\alpha}{2}\right)\,\Gamma\left(1-\frac{\alpha}{2}\right)}{\alpha}=\frac{\pi}{\alpha \sin\frac{\pi\alpha}{2}} $$ where the last equality follows from the reflection formula for the $\Gamma$ function.
The same result can be achieved by computing $$\int_{0}^{+\infty}\frac{x^{1-\alpha}}{1+x^2}\,dx $$ through suitable contours in the complex plane.