Let 
. Prove that
![\[\int_{0}^{\infty} \frac{n^2 x^n \ln x}{1+x^{2n}} \, \mathrm{d}x = \frac{\pi^2}{4} \frac{\sin \frac{\pi}{2n}}{\cos^2 \frac{\pi}{2n}}\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-c77cae5570d4f884213d00d3fe133109_l3.png "Rendered by QuickLaTeX.com")
Solution
We are basing the whole solution on the Beta function and its derivative. We recall that
(1) 
Setting 
and 
back at 
we get that
Differentiating with respect to 
we get that
![\begin{align*} \mathrm{B} '\left ( \frac{1}{2n}+\frac{1}{2}, \frac{1}{2} - \frac{1}{2n} \right ) & = \Gamma \left ( \frac{1}{2}+ \frac{1}{2n} \right ) \Gamma \left ( \frac{1}{2}- \frac{1}{2n} \right ) \bigg [ \psi^{(0)} \left ( \frac{1}{2}+ \frac{1}{2n} \right )- \\ & \quad \quad \quad - \psi^{(0)} \left ( \frac{1}{2} - \frac{1}{2n} \right ) \bigg ]\\ &=\pi^2 \sec \left ( \frac{\pi}{2n} \right) \tan \left ( \frac{\pi}{2n} \right ) \end{align*}](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-f460216518bae099ecfe7e0be3e8fdc4_l3.png "Rendered by QuickLaTeX.com")
where we made use of the reflection formulae of both the Gamma and the digamma function; 
and 
.
Now for our integral we have successively:
