Calculate $\int_{0}^{1} \frac{(\ln{\frac{1}{x}})^{a}}{(1-x)^{2}} dx $

Question

$$\int_{0}^{1} \frac{(\ln{\frac{1}{x}})^{a}}{(1-x)^{2}} dx $$

What I've got after the substitution $\ln{\frac{1}{x}} = t $ :

$$\int_{0}^{\infty} e^{-t} \frac{t^{a}}{(1-e^{-t})^{2}} dt $$

This should be solved with the Gamma function somehow.

Answer 1

The integral is infinite if $a\le 1$. This is because $$ \frac{t^a}{(1-e^{-t})^2} \approx t^{a-2} \quad (t\to 0)$$ which is not locally integrable in dimension 1.

For $a> 1$, the geometric series formula $\frac1{(1-\lambda)^2} = \sum_1^\infty k\lambda^{k-1} $(and Fubini's theorem) gives $$I(a) = \sum_{k=1}^\infty k\int_0^\infty e^{-kt}t^a dt =\sum_{k=1}^\infty k^{-a}\int_0^\infty e^{-\tau}\tau^a d\tau = a!\zeta(a).$$

Here, $a!:=\Gamma(a+1)$ and $\zeta$ is the Riemann Zeta function. This formula seems numerically correct for $a=2$, giving the value $$I(2) = \frac{\pi^2}3 = 3.289868133696452872944830333292050378437899802413596875471\dots$$ (compare a with b)