Evaluate the integral $\int_0^\infty\frac{x^a\ln{x}}{x+b} \, dx$

Question

Integrate $\dfrac{x^a\ln{x}}{x+b}$ from 0 to infinity where $b > 0$ and $-1 < a < 0 $

I'm having trouble deciding how to approach the problem! Thank you!

Answer 1

This integral is the derivative of $$ \int_0^{\infty} \frac{x^a}{x+b} \, dx $$ with respect to $a$. We can remove $b$ by setting $bu=x$, so $dx=b \, du$, $$ b^{a}\int_0^{\infty} \frac{u^a}{1+u} \, du. $$ I evaluated this integral here, so we have $$ \int_0^{\infty} \frac{x^a}{x+b} \, dx = -b^{a} \pi \csc{\pi a}. $$ Now we differentiate this with respect to $a$, to obtain the answer $$ b^a \pi \csc{\pi a} \left( \pi \cot{\pi a}-\log{b} \right). $$