What is the closed form solution for this integral or series?

Question

Here is the question: $$F(n)=\int_0^1\frac{\ln\left(1-x\right)}{x^n}dx ;\ \ \ \ \forall \ n < 2$$

find the explicit expression of $F(n)$.

I find this relation:

$$\int_0^1\frac{\ln\left(1-x\right)}{x^{n}}dx=-\sum_{k=1}^\infty\frac{1}{k^2-\left(n-1\right)k}$$

I try to solve using this identity:

$$\frac{\pi\tan\left(\pi x\right)}{2x}=\sum_{k=n}^\infty\frac{1}{\left(k-\frac{\left(2n-1\right)}{2}\right)^2-x^2} \ \ ; \forall \ n\in \mathbb{N} $$

but when I did this evaluation $x=\frac{\left(2n-1\right)}{2}$ the equality in the result is no longer met.

$$\frac{\pi\tan\left(\frac{\pi}{2}\left(2n-1\right)\right)}{2n-1}=\sum_{k=n}^\infty\frac{1}{k^2-\left(2n-1\right)k}$$

Does anyone know why equality is no longer fulfilled?

Answer 1

Call the integral $I$. After partial fraction expansion of the sum that is mentioned, we see that $$I=\frac{1}{n-1}\sum_{k=0}^{+\infty} \frac{1}{k+1}-\frac{1}{k+2-n}$$ From the series expansion of the Digamma function: $$\psi^{(0)}(z)=-\gamma+\sum_{k=0}^{+\infty} \frac{1}{k+1}-\frac{1}{k+z}$$ We can find that $$\boxed{\int_0^1 \frac{\log(1-x)}{x^n}\,dx=\frac{\psi^{(0)}(2-n)+\gamma}{n-1}}$$

This result seems to agree with the numerical values that @marty cohen found.

EDIT: If we take the limit as $n\to 1$, we can use l'Hopital's rule to conclude that $$\int_0^1\frac{\log(1-x)}{x}\,dx=\psi^{(1)}(1)=\frac{\pi^2}{6}$$

Answer 2

This is more a comment.

Wolfy says:

$F(1) = -\dfrac{\pi^2}{6} \approx -1.64493 $

$F(\frac12) = \log(16)-4 \approx -1.22741 $

$F(\frac32) = -\log(16)≈-2.7726 $

$F(\frac23) =\frac12 (\sqrt{3}\pi + 9 (\log(3) - 2))≈-1.3355 $

$F(\frac43) = \frac12 (\sqrt{3}\pi- 9 \log(3))≈-2.2231 $

$F(\frac53) = \frac14 (-\sqrt{3}\pi - 9 \log(3))≈-3.8322 $

So something is going on but I'm not sure what.