How can we solve this integral $$\int_{0}^{\infty} \frac{x^p \ln(x)}{x^2 + 1}dx\;?$$
The result is given to be $$\frac{\pi^2}{4} \frac{\sin^2(\pi p/2)}{\cos(\pi p /2)}.$$
But I have no idea how to come at this integral solution. Any help?
At least I see method of substitution is not going to help. The denominator some how reminds me of $\tan^{-1}(x)$ but still alas. The improper integral hints me to use complex analysis stuff.
Consider the integral$$\mathfrak{I}(p)=\int\limits_0^{\infty}\mathrm dx\,\frac {x^p}{1+x^2}$$Make the transformation $x\mapsto\tan x$ to get$$\mathfrak{I}(p)=\int\limits_0^{\pi/2}\mathrm dx\,\tan^px=\frac 12\Gamma\left(\frac {1+p}2\right)\Gamma\left(\frac {1-p}2\right)=\frac {\pi}2\sec\left(\frac {\pi p}2\right)$$Now differentiate with respect to $p$ to get the answer.