While reviewing an old calculus book the following integral was assigned: \begin{align} \int_{0}^{1} \left( x^{a-1} - x^{n-a-1} \right) \, \frac{\ln^{2}x \, dx}{1-x^{n}} = \frac{2 \, \pi^{3} \, \cos\left(\frac{a \pi}{n}\right)}{n^{3} \, \sin^{3}\left(\frac{a \pi}{n}\right)} \end{align} for $a \neq n$. The proposed questions here are:
- What are some methods to prove the given integral?
- What are the changes if the power of the logarithm is lowered to one or raised to 3?
The old calculus book is: Ralph A Roberts, "A treatise on the integral calculus; part 1", 1887. The propoblem presented here is found on page 354 as example 36. The book can be found in the Google Books collection.
That is not terribly difficult. One just has to prove that: $$\int_{0}^{1}\frac{x^{a}-x^{n+a}}{1-x^n}\cdot\frac{dx}{x}=\frac{1}{a},\qquad \int_{0}^{1}\frac{x^{a}-x^{n-a}}{1-x^n}\cdot\frac{dx}{x} = \frac{\pi}{n}\,\cot\left(\frac{\pi a}{n}\right)\tag{1}$$ that follows, for instance, from the reflection formula for the digamma function or from the residue theorem, then differentiate the above line with respect to $a$ the right number of times. Another chance is given by exploiting:
$$ \int_{0}^{1}x^{\alpha-1} \log(x)^k\,dx = \frac{(-1)^k k!}{\alpha^{k+1}}\tag{2}$$ together with: $$ \psi'(z)=\sum_{n\geq 0}\frac{1}{(z+n)^2},\quad \sum_{n\geq 0}\frac{1}{(n+z)^k} = \frac{(-1)^k}{k!}\, \psi^{(k-1)}(z)\tag{3}$$ from which: $$\int_{0}^{1}\frac{x^{a/n}}{1-x}\cdot\frac{\log(x)^k}{x}\,dx = (-1)^k k!\sum_{m\geq 0}\frac{1}{\left(m+\frac{a}{n}\right)^{k+1}}=\psi^{(k)}\left(\frac{a}{n}\right).\tag{4}$$