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A parametric logarithmic integral

Let n \in \mathbb{N} \mid n>1. Prove that

    \[\int_{0}^{\infty} \frac{n^2 x^n \ln x}{1+x^{2n}} \, \mathrm{d}x = \frac{\pi^2}{4} \frac{\sin \frac{\pi}{2n}}{\cos^2 \frac{\pi}{2n}}\]

Solution

We are basing the whole solution on the Beta function and its derivative. We recall that

(1)   \begin{equation*} \mathrm{B}(x, y) = \int_{0}^{\infty} \frac{t^{x-1}}{\left ( 1+t \right )^{x+y}} \, \mathrm{d}t \end{equation*}

Setting x=\frac{1}{2} + \frac{1}{2n} and y=\frac{1}{2} - \frac{1}{2n} back at (1) we get that

    \begin{align*} \mathrm{B}\left ( \frac{1}{2}+ \frac{1}{2n}, \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 )\\ &= \pi \sec \left ( \frac{\pi}{2n} \right ) \end{align*}

Differentiating (1) with respect to x 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*}

where we made use of the reflection formulae of both the Gamma and the digamma function; \Gamma (x) \Gamma(1-x) = \pi \csc \pi x and \psi^{(0)} (1-x)- \psi^{(0)}(x) = \pi \cot \pi x.

Now for our integral we have successively:

    \begin{align*} \int_{0}^{\infty} \frac{ n^2 x^n \log x}{1+x^{2n}}\, {\rm d}x &\overset{u=x^n}{=\! =\! =\!} \int_{0}^{\infty} \frac{u^{1/n} \log u}{1+u^2} \, {\rm d}u \\ &\overset{y=u^2}{=\! =\! =\! =\!} \frac{1}{4}\int_{0}^{\infty} \frac{y^{1/2n -1/2} \log y}{1+y} \, {\rm d}y \\ &=\frac{1}{4}\pi^2 \sec \left ( \frac{\pi}{2n} \right ) \tan \left ( \frac{\pi}{2n} \right ) \\ &= \frac{\pi^2}{4} \frac{\sin \left ( \frac{\pi}{2n} \right )}{\cos^2 \left ( \frac{\pi}{2n} \right )} \end{align*}

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