How to evaluate the integral
$$
I_p=\int_{0}^{\infty} \frac{\log x}{x^p (x-1)}\, dx\,?$$
Attempt:
My first idea was to find the area of convergence, which gives me $p \in (0,1)$. Now let's consider the function $f(z) = \frac{\log^2 z}{z^p (z-1)}$. Then we can use Cauchy theorem and residuals.
But I'm stuck here. Because there is no term to describe $I_1$. Any ideas?
Best Answer
We can take from here the following representation of the digamma function: $$\psi(s+1)=-\gamma+\int_0^1 \frac{1-x^s}{1-x}dx\Rightarrow \psi_1(s)=\int_0^1 \frac{x^{s-1}\ln x}{x-1}dx$$ Now we are just one step away to solve it by splitting the integral into two pieces because we have: $$\int_1^\infty \frac{\ln x}{x^p(x-1)}dx\overset{\large x=\frac{1}{t}}=\int_0^1 \frac{t^{p-1}\ln t}{t-1}dt$$
$$\int_0^\infty \frac{\ln x}{x^p(x-1)}dx=\int_0^1 \frac{x^{-p}\ln x}{x-1}dx+\int_0^1\frac{x^{p-1}\ln x}{x-1}dx=\psi_1(1-p)+\psi_1(p)=\boxed{\frac{\pi^2}{\sin^2 (\pi p)}}$$ Above follows by the the trigamma's reflection formula.