Conjecture regarding integrals of the form $\int_0^\infty \frac{(\log{x})^n}{1+x^2}\,\mathrm{d}x$. I have been playing around a bit with integrals of the form $$I(n)=\int_0^\infty \frac{(\log{x})^n}{1+x^2}\,\mathrm{d}x,\,\,n\in\mathbb{Z}^+,$$ and I am trying to obtain a closed form solution for $I(n).$ I believe the special cases $I(1)$ and $I(2)$ are somewhat well-known, but I will go over them. When $n=1,$ we have $$I(1)=\int_0^\infty \frac{\log{x}}{1+x^2}\,\mathrm{d}x=\int_0^1 \frac{\log{x}}{1+x^2}\,\mathrm{d}x+\int_1^\infty \frac{\log{x}}{1+x^2}\,\mathrm{d}x.$$ This can be easily shown to be zero by performing the substitution $x=1/y,$ which will yield $$\int_0^1 \frac{\log{x}}{1+x^2}\,\mathrm{d}x=-\int_1^\infty \frac{\log{x}}{1+x^2}\,\mathrm{d}x.$$ Thus, $I(1)=0.$ Clearly, this can be generalized to all odd integers, and $I(2n+1)=0.$ In the case of $n=2$, first observe through the same substitution as above that $$\int_0^1 \frac{(\log{x})^2}{1+x^2}\,\mathrm{d}x=\int_1^\infty \frac{(\log{x})^2}{1+x^2}\,\mathrm{d}x.$$ This implies that $$I(2)=2\int_0^1 \frac{(\log{x})^2}{1+x^2}\,\mathrm{d}x,$$ which is an easier integral to work with. Performing the substitution $x=e^{-y}$ yields $$I(2)=2\int_0^\infty y^2\left(e^{-y}-e^{-3y}+e^{-5y}-\cdot\cdot\cdot\right)\,\mathrm{d}y.$$ Using the identity $$\int_0^\infty x^2 e^{-ax}=\frac{2}{a^3},$$ we obtain $$I(2)=4\left(\frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3}-\cdot\cdot\cdot\right)=4\cdot\frac{\pi^3}{32}=\frac{\pi^3}{8}.$$
I'm not sure how this infinite series is evaluated, but I found this result in a book. I used Mathematica to check a few more values, and I found that $I(4)=5\pi/32,\,I(6)=61\pi/128,\,$ and $I(8)=1385\pi/512.$ Clearly the pattern is $$I(2n)=A_{2n}\left(\frac{\pi}{2}\right)^{2n+1},$$ where $A_{2n}$ is some constant. It turns out that these constants are the Euler numbers, which are the coefficients $E_k$ corresponding to the series $$\operatorname{sech}x=\sum_{k=0}^\infty\frac{E_k}{k!}x^k.$$
All Euler numbers corresponding to odd $n$ are zero, and the first few even Euler numbers are $E_0=1,\, E_2=-1,\, E_4=5, \,E_6=-61,\,$ and $E_8=1385.$ Thus, I have conjectured that $$I(2n)=(-1)^n E_{2n} \left(\frac{\pi}{2}\right)^{2n+1}$$ for all $n\in\mathbb{Z}^+.$ I suppose this could be extended to $n\in\mathbb{Z}_{\geq 0}$ thusly:
$$I(n)=i^n E_n \left(\frac{\pi}{2}\right)^{n+1},$$ since $E_n=0$ for odd $n.$ So the question, of course, is how to prove this. I tried generalizing the method I used for $I(2),$ and I found that $$\int_0^\infty x^n e^{-ax}=\frac{n!}{a^{n+1}},\,\,n\in\mathbb{Z}_{\geq 0}.$$ Using this, I obtained $$I(2n)=n!\left(\frac{1}{1^{n+1}}-\frac{1}{3^{n+1}}+\frac{1}{5^{n+1}}-\cdot\cdot\cdot\right)=n!\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^{n+1}}.$$ Mathematica wasn't able to evaluate this sum, even for the case of $n=2.$ It gives some expression involving multiple zeta functions, with which I have no experience. Even if we can't prove this, I would be interested to know why the Euler numbers might appear here. Any help would be greatly appreciated.
Edit:
As Claude Leibovici helped point out, there final series expression should be $$I(2n)=2(2n)!\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}=\frac{(2n)!}{2^{4n+1}}\left[\zeta\left(2n+1, \frac{1}{4}\right)-\zeta\left(2n+1, \frac{3}{4}\right)\right].$$
 A: Integrate $f(z)=\dfrac{z^{s}}{1+z^2}$, with the branch cut placed on the positive real axis and $-1<\operatorname{Re}(s)<1$, along a keyhole contour deformed around $[0,R]$. Along the big arc of radius $R$,
$$0\leq\left|\ \int_{R\exp\left(i[0,2\pi]\right)} f(z)\ dz\ \right|\leq\frac{2\pi R^{\operatorname{Re}(s)+1}}{R^2-1}\to0 \text{ as }R\to\infty$$
and along the small arc of radius $\epsilon$,
$$0\leq\left|\ \int_{\epsilon\exp\left(i[0,2\pi]\right)} f(z)\ dz\ \right|\leq\frac{2\pi \epsilon^{\operatorname{Re}(s)+1}}{1-\epsilon^2}\to0 \text{ as }\epsilon\to 0$$
so taking $R\to\infty$ and $\epsilon\to 0$ and applying the Residue Theorem,
\begin{align}
(1-e^{2\pi i s})\int^\infty_0\frac{x^{s}}{1+x^2}\ dx
&=\pi\left(e^{\pi is/2}-e^{3\pi i s/2}\right)\\
\implies \int^\infty_0\frac{x^{s}}{1+x^2}\ dx
&=\pi\cdot\frac{e^{\pi is/2}-e^{-\pi i s/2}}{e^{\pi is}-e^{-\pi is}}
=\pi\cdot\frac{\sin\left(\frac{\pi s}{2}\right)}{\sin\left(\pi s\right)}
=\frac{\pi}{2}\sec\left(\frac{\pi s}{2}\right)
\end{align}
Therefore,
\begin{align}
I(2n)
&=\left.\frac{\pi}{2}\frac{d^{2n}}{ds^{2n}}\sec\left(\frac{\pi s}{2}\right)\right|_{s\to 0}\\
&=\left.\frac{\pi}{2}\frac{d^{2n}}{ds^{2n}}\sum^\infty_{k=0}\frac{(-1)^kE_{2k}}{(2k)!}\left(\frac{\pi s}{2}\right)^{2k}\right|_{s\to 0}\\
&=\left.\frac{\pi}{2}(2n)![s^{2n}]\sum^\infty_{k=0}\frac{(-1)^kE_{2k}}{(2k)!}\left(\frac{\pi s}{2}\right)^{2k}\right|_{s\to 0}\\
&=\frac{\pi}{2}(2n)!\frac{(-1)^nE_{2n}}{(2n)!}\left(\frac{\pi}{2}\right)^{2n}\\
&=(-1)^nE_{2n}\left(\frac{\pi}{2}\right)^{2n+1}
\end{align}
A: First part. We may note that $$\int_{0}^{\infty}\frac{\log^{n}\left(x\right)}{1+x^{2}}dx\stackrel{x\rightarrow1/x}{=}\left(-1\right)^{n}\int_{0}^{\infty}\frac{\log^{n}\left(x\right)}{1+x^{2}}dx
 $$ so obviously if $n$ is odd the integral is $0$. If $n
 $ is even we have that $$I(2k)=\int_{0}^{\infty}\frac{\log^{2k}\left(x\right)}{1+x^{2}}dx=\int_{0}^{1}\frac{\log^{2k}\left(x\right)}{1+x^{2}}dx+\int_{1}^{\infty}\frac{\log^{2k}\left(x\right)}{1+x^{2}}dx
 $$ $$\stackrel{x\rightarrow1/x}{=}2\int_{0}^{1}\frac{\log^{2k}\left(x\right)}{1+x^{2}}dx=2\sum_{m\geq0}\left(-1\right)^{k}\int_{0}^{1}x^{2m}\log^{2k}\left(x\right)dx
 $$ and integrating by parts $$I(2k)=2\sum_{m\geq0}\left(-1\right)^{m}\int_{0}^{1}x^{2m}\log^{2k}\left(x\right)dx
 $$ $$=\color{blue}{2\left(2k\right)!\sum_{m\geq0}\frac{\left(-1\right)^{m}}{\left(2m+1\right)^{2k+1}}}
 $$ and since the series is absolutely convergent we have $$I(2k)=2\left(2k\right)!\left(\sum_{m\geq0}\frac{1}{\left(4m+1\right)^{2k+1}}-\sum_{m\geq0}\frac{1}{\left(4m+3\right)^{2k+1}}\right)
 $$ $$=\color{green}{\left(2k\right)!2^{-4k-1}\left(\zeta\left(2k+1,\frac{1}{4}\right)-\zeta\left(2k+1,\frac{3}{4}\right)\right)}
 $$ where $\zeta\left(s,a\right)
 $ is the Hurwitz Zeta function. 
Second part. Consider the function $\sin\left(xy\right)
 $ on $\left[-\pi,\pi\right]
 $, $y<1
 $. It is not difficult to see that $$b_{n}=\frac{2}{\pi}\int_{0}^{\pi}\sin\left(yx\right)\sin\left(nx\right)dx=\left(-1\right)^{n-1}\frac{2}{\pi}\frac{n\sin\left(\pi y\right)}{n^{2}-y^{2}}
 $$ but it is also the $n$-th coefficient of the Fourier series of $$\sin\left(xy\right)=\sum_{n\geq1}b_{n}\sin\left(nx\right)
 $$ hence taking $x=\frac{\pi}{2}
 $ we get $$\sin\left(\frac{\pi}{2}y\right)=\frac{2}{\pi}\sin\left(\pi y\right)\sum_{n\geq1}\left(-1\right)^{n-1}\frac{2n-1}{\left(2n-1\right)^{2}-y^{2}}
 $$ $$=\frac{2}{\pi}\sin\left(\pi y\right)\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{2n-1}\sum_{k\geq0}\frac{y^{2k}}{\left(2n-1\right)^{2k}}
 $$ $$=\frac{2}{\pi}\sin\left(\pi y\right)\sum_{k\geq0}\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{\left(2n-1\right)^{2k+1}}y^{2k}
 $$ hence $$\frac{\pi\sin\left(\frac{\pi}{2}y\right)}{\sin\left(\pi y\right)}=\sum_{k\geq0}\sum_{n\geq1}\frac{2\left(-1\right)^{n-1}}{\left(2n-1\right)^{2k+1}}y^{2k}
 $$ but $$\frac{\pi\sin\left(\frac{\pi}{2}y\right)}{\sin\left(\pi y\right)}=\frac{\pi\sec\left(\frac{\pi}{2}y\right)}{2}
 $$ and it is well knonw that $$\frac{\pi\sec\left(\frac{\pi}{2}y\right)}{2}=\sum_{n\geq0}\frac{\left(-1\right)^{n}E_{2n}}{\left(2n\right)!}\left(\frac{\pi}{2}\right)^{2n+1}y^{2n}
 $$ hence, equating the coefficients, we have $$\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{\left(2n-1\right)^{2k+1}}=\sum_{n\geq0}\frac{\left(-1\right)^{n}}{\left(2n+1\right)^{2k+1}}=\color{red}{\frac{\left(-1\right)^{n}E_{2n}}{2\left(2n\right)!}\left(\frac{\pi}{2}\right)^{2n+1}}.$$
Conclusion. 

$$I\left(2n\right)=\left(-1\right)^{n}E_{2n}\left(\frac{\pi}{2}\right)^{2n+1}.$$

A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{I}\pars{n} \equiv
\int_{0}^{\infty}{\ln^{n}\pars{x} \over 1 + x^2}\,\dd x:\ ?.\qquad
n \in \mathbb{Z}^{+}}$.

\begin{align}
\mrm{I}\pars{n} & \equiv
\int_{0}^{\infty}{\ln^{n}\pars{x} \over 1 + x^2}\,\dd x =
\int_{0}^{1}{\ln^{n}\pars{x} \over 1 + x^2}\,\dd x +
\int_{1}^{\infty}{\ln^{n}\pars{x} \over 1 + x^2}\,\dd x =
\bracks{1 + \pars{-1}^{n}}\int_{0}^{1}{\ln^{n}\pars{x} \over 1 + x^2}\,\dd x
\end{align}

Then,


*

*$\ds{{\large n\ \underline{odd}} \implies \mrm{I}\pars{n} \equiv
\int_{0}^{\infty}{\ln^{n}\pars{x} \over 1 + x^2}\,\dd x = 0}$

*$\ds{{\large n\ \underline{even}} \implies
\mrm{I}\pars{n} \equiv
\int_{0}^{\infty}{\ln^{n}\pars{x} \over 1 + x^2}\,\dd x =
2\int_{0}^{1}{\ln^{n}\pars{x} \over 1+x^2}\,\dd x =
\left.2\,\partiald[n]{}{\mu}
\int_{0}^{1}{x^{\mu} \over 1 + x^2}\,\dd x
\,\right\vert_{\ \mu\ =\ 0}}$

\begin{align}
\int_{0}^{1}{x^{\mu} \over 1 + x^2}\,\dd x & =
\int_{0}^{1}{x^{\mu} - x^{\mu + 2} \over 1 - x^4}\,\dd x =
\int_{0}^{1}{x^{\mu/4} - x^{\mu/4 + 1/2} \over 1 - x}
\,{1 \over 4}\,x^{-3/4}\,\dd x
\\[5mm] & =
{1 \over 4}\pars{\int_{0}^{\infty}{1 - x^{\mu/4 - 1/4} \over 1 - x}\,\dd x -
\int_{0}^{\infty}{1 - x^{\mu/4 - 3/4} \over 1 - x}\,\dd x}
\\[5mm] & =
{1 \over 4}\bracks{\Psi\pars{\mu + 3 \over 4} - \Psi\pars{\mu + 1 \over 4}}
\qquad\pars{~\Psi:\ Digamma\ Function~}
\end{align}

$$\bbox[15px,#ffe,border:1px groove navy]{%
\mrm{I}\pars{n} \equiv
\int_{0}^{\infty}{\ln^{n}\pars{x} \over 1 + x^{2}}\,\dd x =
\left\{\begin{array}{lcl}
\ds{0} & \mbox{if} & \ds{n \in \mathbb{Z}^{+}\ \mbox{is}\ odd}
\\[3mm]
\ds{{1 \over 2^{2n +1}}\bracks{\Psi^{\mrm{\pars{n}}}\pars{3 \over 4} -
\Psi^{\mrm{\pars{n}}}\pars{1 \over 4}}} & \mbox{if} &
\ds{n \in \mathbb{Z}^{+}\ \mbox{is}\ even}
\end{array}\right.}
$$


Note that
  $\ds{\pars{~Euler\ Reflection\ Formula~}}$
  $$
\bracks{\Psi^{\mrm{\pars{n}}}\pars{3 \over 4} -
\Psi^{\mrm{\pars{n}}}\pars{1 \over 4}}_{\ n\ \in\ \mathbb{Z}^{+}\ even} =
\left.\pars{-1}^{n}\,\pi^{n + 1}\,\totald[n]{\cot\pars{z}}{z}
\right\vert_{\ z\ =\ \pi/4}
$$

A: This isn't really an answer, just an evaluation of the integral.
Let
$$J(a)=\int_0^\infty \frac{x^a}{x^2+1}\mathrm dx$$
Hence
$$I(n)=\left[\left(\frac{d}{da}\right)^nJ(a)\right]_{a=0}=J^{(n)}(0)$$
Anyway,
$$J(a)=\int_0^\infty \frac{x^a}{x^2+1}\mathrm dx$$
$x=\tan t$:
$$J(a)=\int_0^{\pi/2}\tan(t)^a\mathrm dt$$
$$J(a)=\int_0^{\pi/2}\sin(t)^a\cos(t)^{-a}\mathrm dt$$
Because $$\int_0^1 t^{a-1}(1-t)^{b-1}\mathrm dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
with the substitution $t=\sin(x)^2$ we have that
$$\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$
So 
$$J(a)=\frac12\Gamma\left(\frac{1+a}2\right)\Gamma\left(\frac{1-a}2\right)$$
And since $$\Gamma(s)\Gamma(1-s)=\frac\pi{\sin\pi s}$$
We have that 
$$J(a)=\frac\pi2\sec\frac{\pi a}2$$
So 
$$I(n)=\frac\pi2\left[\left(\frac{d}{da}\right)^n\sec\frac{\pi a}2\right]_{a=0}$$
A: One more
A. When the power is odd.
Obviously, letting $x\mapsto \frac{1}{x} $ yields
$$
\begin{aligned}
\therefore \int_{0}^{\infty} \frac{(\ln x)^{2 n+1}}{1+x^{2}} d x &=\int_{0}^{\infty} \frac{-(\ln x)^{2 n+1}}{1+x^{2}} d x \\
&=-\int_{0}^{\infty} \frac{(\ln x)^{2 n+1}}{1+x^{2}} d x \\
\therefore \int_{0}^{\infty} \frac{(\ln x)^{2 n+1}}{1+x^{2}} d x &=0
\end{aligned}
$$
B.When the power is even.
Noticing that
$$
\int_{0}^{\infty} \frac{(\ln x)^{2n}}{x^{2}+1} d x=\left.\frac{d^{2n}}{d a^{2n}} \int_{0}^{\infty} \frac{x^{a}}{x^{2}+1} d x\right|_{a=0}
$$
By my post, we get $$
\begin{aligned}
I&= \left.\frac{\partial ^{2n}}{\partial a^{2n}} \Gamma\left(\frac{a+1}{2}\right) \Gamma\left(\frac{1-a}{2}\right)\right|_{a=0}\\&=\left.\pi \frac{\partial^{2n}}{\partial a^{2n}} \csc \frac{(a+1) \pi}{2}\right|_{a=0} \quad \textrm{ (By the Reflection Property )}\\
&=\left.\frac{\pi^{2n+1}}{2^{2n}} \frac{\partial^{2n}}{\partial x^{2n}}(\csc x)\right|_{x=\frac{\pi}{2}}
\end{aligned}
$$
By the post,
$$\lim_{x\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\csc(x)=|E_{2n}|$$
$$
\boxed{\int_{0}^{\infty} \frac{(\ln x)^{2 n}}{1+x^{2}} d x =\frac{\pi^{2 n+1}}{2^{2 n}}\left|E_{2 n}\right|}
$$
A: \begin{align*}J_{n}&=\int_0^\infty \frac{\ln^{2n}  x}{1+x^2}dx\\
J_0&=\int_0^1 \frac{1}{1+x^2}dx=\frac{\pi}{2}\\ 
K_{n}&=\int_0^\infty\int_0^\infty\frac{\ln^{2n}(xy)}{(1+x^2)(1+y^2)}dxdy\\
&=\sum_{k=0}^{n}\binom{2n}{2k}J_kJ_{n-k}\\
J_n=&\overset{u(x)=yx}=\int_0^\infty\int_0^\infty\frac{y\ln^{2n}u}{(y^2+u^2)(1+y^2)}dudy \\
&=\frac{1}{2}\int_0^\infty \frac{\ln^{2n} u}{1-u^2}\left[\ln\left(\frac{y^2+u^2}{y^2+1}\right)\right]_0^\infty du\\
&=-\int_0^\infty \frac{\ln^{2n+1} u}{1-u^2}du\\
&=-2\int_0^1 \frac{\ln^{2n+1} u}{1-u^2}du\\
&=\underbrace{\int_0^1 \frac{2u\ln^{2n+1} u}{1-u^2}du}_{z=u^2}-2\int_0^1 \frac{\ln^{2n+1} u}{1-u}du\\
&=\left(\frac{1}{2^{2n+1}}-2\right)\int_0^1 \frac{\ln^{2n+1} u}{1-u}du\\
&=2(2n+1)!\left(1-\frac{1}{2^{2n+2}}\right)\zeta(2n+2)\\
\end{align*}
Therefore,
\begin{align}\boxed{J_n=\frac{1}{\pi}\left(2(2n+1)!\left(1-\frac{1}{2^{2n+2}}\right)\zeta(2n+2)-\sum_{k=1}^{n-1}\binom{2n}{2k}J_kJ_{n-k}\right)}\end{align}
