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As an exercise for myself I constructed the Integral

$$ \int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx $$

with $n\in \mathbb{N}$. With the help of Mathematica I found the analytical result

$$ \int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx=\frac{1+(-1)^n}{4^{n+1}}\Gamma(n+1)\left[\zeta\left(n+1,\frac{1}{4}\right)-\zeta\left(n+1,\frac{3}{4}\right)\right]. $$

For $n=1$ (and probably $n\in \mathbb{N}$) one can employ the methods of complex analysis and find a result. For $n\in \mathbb{N}$ I encountered a nasty recursion relation. I can provide details if needed. Is there another way how to solve the integral at hand?

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  • $\begingroup$ $\displaystyle \beta(n+1)=\frac{1}{4^{n+1}}\left[\zeta\left(n+1,\frac{1}{4}\right)-\zeta\left(n+1,\frac{3}{4}\right)\right]\enspace$ where $\beta(s)$ is the Dirichlet beta function . $\endgroup$ – user90369 Nov 23 '18 at 16:32
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Referring to this answer, $$I_n=\frac{\pi}2\frac{d^n}{dx^n}\sec\left(\frac{\pi x}2\right)\bigg\vert_{x=0}$$

or equivalently,

$$\int^\infty_0\frac{\log^n(x)}{x^2+1}dx=\left(\frac{\pi}2\right)^{n+1}\sec^{(n)}(0)$$

$$\int^\infty_0\frac{\log^{2n}(x)}{x^2+1}dx=(-1)^nE_{2n}\left( \frac{\pi}2\right)^{2n+1} $$

The integral is zero for odd $n$.

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  • $\begingroup$ @Schnarco : $E_n$ are called Euler numbers . $\endgroup$ – user90369 Nov 23 '18 at 11:14
  • $\begingroup$ Thank you both! $\endgroup$ – Schnarco Nov 23 '18 at 13:04
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Let, for $n$, a natural integer,

\begin{align}J_n=\int_0^\infty \frac{\ln^n x}{1+x^2}\,dx\end{align}

First, observe that if $n$ is odd then $J_n=0$ (perform the change of variable $y=\dfrac{1}{x}$ )

Consider, for $n$, a natural integer,

\begin{align}K_n=\int_0^\infty \int_0^\infty\frac{\ln^n(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\end{align}

Observe that,

\begin{align}K_{2n}&=\sum_{k=0}^{2n}\binom{2n}{k} J_kJ_{2n-k}\\ &=\sum_{k=0}^{n}\binom{2n}{2k} J_{2k}J_{2(n-k)} \end{align}

On the other hand,

Perform the change of variable $u=xy$ ($x$ variable, $y$, a "parameter"),

\begin{align}K_n&=\int_0^\infty \int_0^\infty\frac{y\ln^n u}{(1+y^2)(u^2+y^2)}\,du\,dy\\ &=\frac{1}{2}\int_0^\infty \left[\ln\left(\frac{1+y^2}{u^2+y^2}\right)\right]_{y=0}^{y=\infty}\frac{\ln^{n} u}{u^2-1}\,du\\ &=\int_0^\infty \frac{\ln^{n+1} u}{u^2-1}\,du\\ &=\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du+\int_1^\infty \frac{\ln^{n+1} u}{u^2-1}\,du\\ \end{align}

In the second integral perform the change of variable $v=\dfrac{1}{u}$,

\begin{align}K_n&=\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du+\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du\\ &=\int_0^1 \frac{\left(1+(-1)^n\right)\ln^{n+1} u}{u^2-1}\,du\\ &=\left(1+(-1)^n\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du-\left(1+(-1)^n\right)\int_0^1\frac{u\ln^{n+1} u}{u^2-1}\,du\\ \end{align}

In the latter integral perform the change of variable $y=u^2$,

\begin{align}K_n&=\left(1+(-1)^n\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du-\frac{1+(-1)^n}{2^{n+2}}\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du\\ &=\left(1+(-1)^n\right)\left(1-\frac{1}{2^{n+2}}\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du\\ &=\left(1+(-1)^n\right)\left(1-\frac{1}{2^{n+2}}\right)(-1)^{n+2}(n+1)!\zeta(n+2) \end{align}

Therefore, one obtains a recursion relation,

\begin{align} \boxed{\sum_{k=0}^{n}\binom{2n}{2k} J_{2k}J_{2(n-k)}=2\left(1-\frac{1}{2^{2n+2}}\right)(2n+1)!\zeta(2n+2)} \end{align}

Observe that,

\begin{align}J_0=\frac{\pi}{2}\end{align}

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