Parametric integral with logarithm

Let $n \in \mathbb{N} \setminus \{1\}$. Prove that:

$$\int_{0}^{\infty} \frac{n^2 x^n \log x}{1+x^{2n}} \, {\rm d}x = \frac{\pi^2}{4} \frac{\sin \left ( \frac{\pi}{2n} \right )}{\cos^2 \left ( \frac{\pi}{2n} \right )}$$

Solution

We give two solutions. One that is elementary using pure properties of the Beta function and another that is using more heavily tools.

1st solution:

One of the many forms Beta function has , is:

$${\rm B} \left ( x, y \right )= \int_{0}^{\infty} \frac{t^{x-1}}{\left ( 1+t \right )^{x+y}} \, {\rm d}t \tag{1} \label{1}$$

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*}

since \eqref{1} for $x=\frac{1}{2}+\frac{1}{2n}$ and $y=\frac{1}{2}-\frac{1}{2n}$ gives:

$${\rm B}\left ( \frac{1}{2}+ \frac{1}{2n}, \frac{1}{2}- \frac{1}{2n} \right )= \Gamma \left ( \frac{1}{2}+ \frac{1}{2n} \right ) \Gamma \left ( \frac{1}{2}- \frac{1}{2n} \right )= \pi \sec \left ( \frac{\pi}{2n} \right )$$

Differentiating \eqref{1} once with respect to $x$ so that we introduce the logarithm at the numerator and the same time making use of the digamma reflection formula

$$ \psi^{(0)} \left ( \frac{1}{2}+ \frac{1}{2n} \right )- \psi^{(0)} \left ( \frac{1}{2} - \frac{1}{2n} \right )=\pi \tan \left ( \frac{\pi}{2n} \right )$$

we get the result.

2nd solution:

We are stating a lemma:

Lemma: It holds that:

$$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{\frac{\pi}{2}- z + n \pi} = \frac{1}{\cos z} \tag{2} \label{2}$$

Proof: Indeed, this follows from the well known identity $\sum \limits_{n=-\infty}^{\infty} \frac{(-1)^n}{n+z}= \frac{\pi}{\sin \pi z}$ which is a simple application of the digamma reflection formula. Thus:

\begin{align*}
\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n+z}= \frac{\pi}{\sin \pi z} &\Rightarrow  \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n \pi + z } = \frac{1}{\sin z}  \\
 &\Rightarrow \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{ n\pi + \frac{\pi}{2}-z} = \frac{1}{\cos z}
\end{align*}

Differentiating once \eqref{2} we get $\displaystyle \frac{\sin{z}}{\cos^2{z}}=\displaystyle\mathop{\sum}_{m=-\infty}^{+\infty}\frac{(-1)^{m}}{\big(\frac{\pi}{2}-z+m\pi\big)^2}$. Setting $x=\frac{\pi}{2n}$ we deduce:

\begin{align*}
\displaystyle \frac{4n^2}{\pi^2}\sum_{m=-\infty}^{\infty}\frac{(-1)^{m}}{(2mn+n-1)^2}&=\frac{\sin\big(\frac{\pi}{2n}\big)}{\cos^2\big(\frac{\pi}{2n}\big)} \quad&\Rightarrow\\ \displaystyle \sum_{m=-\infty}^{\infty}\frac{(-1)^{m}}{(2mn+n-1)^2}&=\frac{\pi^2}{4n^2}\frac{\sin\big(\frac{\pi}{2n}\big)}{\cos^2\big(\frac{\pi}{2n}\big)}\quad &(3)
\end{align*}

Then for the initial integral we have successively:

\begin{align*}
\displaystyle\int_{0}^{\infty}\frac{n^2\,x^n\log{x}}{1+x^{2n}}\,{\rm{d}}x &=\int_{0}^{1}\frac{n^2\,x^n\log{x}}{1+x^{2n}}\,{\rm{d}}x+\int_{1}^{\infty}\frac{n^2\,x^n\log{x}}{1+x^{2n}}\,{\rm{d}}x\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
    {t\,=\,1/x}\\
    {dx\,=\,-dt/t^2} \\
    \end{subarray}}\,\displaystyle\int_{0}^{1}\frac{n^2\,x^n\log{x}}{1+x^{2n}}\,{\rm{d}}x-\int_{0}^{1}\frac{n^2\,t^{n-2}\log{t}}{1+t^{2n}}\,{\rm{d}}t\\
&=n^2\,\displaystyle\int_{0}^{1}\frac{x^n-x^{n-2}}{1+x^{2n}}\log{x}\,{\rm{d}}x\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
    {x\,=\,{\rm{e}}^{-y}}\\
    {dx\,=\,-{\rm{e}}^{-y}dy} \\
    \end{subarray}}\,n^2\,\displaystyle\int_{0}^{\infty}\frac{{\rm{e}}^{-(n-1)y}-{\rm{e}}^{-(n+1)y}}{1+{\rm{e}}^{-2ny}}\,y\,{\rm{d}}y  \\
&=n^2\displaystyle\int_{0}^{\infty}y\,({\rm{e}}^{-(n-1)y}-{\rm{e}}^{-(n+1)y})\mathop{\sum}_{m=0}^{\infty}(-1)^{m}{\rm{e}}^{-2mny}\,{\rm{d}}y  \\
&=n^2\displaystyle\mathop{\sum}_{m=0}^{\infty}(-1)^{m}\int_{0}^{\infty}y\,({\rm{e}}^{-(2mn+n-1)y}-{\rm{e}}^{-(2mn+n+1)y})\,{\rm{d}}y  \\
&=n^2\displaystyle\mathop{\sum}_{m=0}^{\infty}\frac{(-1)^{m}}{(2mn+n-1)^2}-n^2\mathop{\sum}_{m=0}^{\infty}\frac{(-1)^{m}}{(2mn+n+1)^2} \\
&=n^2\displaystyle\mathop{\sum}_{m=0}^{\infty}\frac{(-1)^{m}}{(2mn+n-1)^2}+n^2\mathop{\sum}_{m=1}^{\infty}\frac{(-1)^{m}}{(2mn-n+1)^2} \\ &=n^2\displaystyle\mathop{\sum}_{m=0}^{\infty}\frac{(-1)^{m}}{(2mn+n-1)^2}+n^2\mathop{\sum}_{m=-\infty}^{-1}\frac{(-1)^{m}}{(-2mn-n+1)^2} \\ &=n^2\displaystyle\mathop{\sum}_{m=-\infty}^{\infty}\frac{(-1)^{m}}{(2mn+n-1)^2}  \\
&\stackrel{(3)}{=\!=}n^2\,\frac{\pi^2}{4n^2}\,\frac{\sin\big(\frac{\pi}{2n}\big)}{\cos^2\big(\frac{\pi}{2n}\big)}\\
&=\frac{\pi^2}{4}\,\frac{\sin\big(\frac{\pi}{2n}\big)}{\cos^2\big(\frac{\pi}{2n}\big)}\,.
\end{align*}


The second solution was taken from mathematica.gr