Conjecture regarding integrals of the form $\int_0^\infty \frac{(\log{x})^n}{1+x^2}\,\mathrm{d}x$.

Solution 1:

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}

Solution 2:

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.

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

Solution 3:

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

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

\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}

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

  ---

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

Solution 4:

This is not an answer but too long for a comment.

$$K_n=\int \frac{\log ^n(x)}{x^2+1} \, dx$$ does not show any closed form but (using a CAS) $$J_n=\int_0^\infty \frac{\log ^n(x)}{x^2+1} \, dx=\frac{n!}{2^{2(n+1)}} \left(1+(-1)^n\right) \left(\zeta \left(n+1,\frac{1}{4}\right)-\zeta \left(n+1,\frac{3}{4}\right)\right) $$ Clearly, as you showed, $J_{2n+1}=0$ and $$J_{2n}=\frac{(2n)!}{2^{(4 n+1)}} \left(\zeta \left(2 n+1,\frac{1}{4}\right)-\zeta \left(2 n+1,\frac{3}{4}\right)\right) $$ while, the last expression of the post write $$I_{2m}=m!\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^{m+1}}=\frac{m!}{2^{2( m+1)}} \left(\zeta \left(m+1,\frac{1}{4}\right)-\zeta \left(m+1,\frac{3}{4}\right)\right)$$ which are not the same even if $m=2n$ (they will differ by a factor $2$).

I give below a table of $J_{2n}$ as a function of $n$ $$\left( \begin{array}{cc} n & J_{2n} \\ 1 & \frac{\pi ^3}{8} \\ 2 & \frac{5 \pi ^5}{32} \\ 3 & \frac{61 \pi ^7}{128} \\ 4 & \frac{1385 \pi ^9}{512} \\ 5 & \frac{50521 \pi ^{11}}{2048} \\ 6 & \frac{2702765 \pi ^{13}}{8192} \\ 7 & \frac{199360981 \pi ^{15}}{32768} \\ 8 & \frac{19391512145 \pi ^{17}}{131072} \\ 9 & \frac{2404879675441 \pi ^{19}}{524288} \\ 10 & \frac{370371188237525 \pi ^{21}}{2097152} \end{array} \right)$$

Looking at $OEIS$, the numerators correspond to sequence $A000364$ and they effectively are Euler numbers (also named "Zig" and "secant" numbers).

So, as you conjectured, $$J_{2n}=\left(\frac \pi 2 \right)^{2n+1}E_n$$

As said in the $OEIS$ page, Euler numbers appear in the Taylor expansion of $\sec (x)$ (this explains probably one of the names) as well as in the expansion of $2 \tanh ^{-1}(\csc (x)-\cot (x))$.

Doing the same as you did with the integral, changing $x=e^{-y}$, we heve $$\int_0^\infty-\frac{e^{-y} (-y)^n}{e^{-2 y}+1}dy=\sum_{k=0}^\infty (-1)^{k+n}\int_0^\infty y^ne^{-(2k+1)y}dy=\sum_{k=0}^\infty (-1)^{k+n}\frac{ n!}{ (2 k+1)^{n+1}}$$ that is to say $$\int_0^\infty-\frac{e^{-y} (-y)^n}{e^{-2 y}+1}dy=(-1)^n\frac{ n!}{2^{2(n+1)}} \left(\zeta \left(n+1,\frac{1}{4}\right)-\zeta \left(n+1,\frac{3}{4}\right)\right)$$

Edit

I already totally agree that this does not explain why Euler numbers appear here in the same manner as I could not explain why Bernoulli numbers appear appear in the Taylor series expansion of the tangent and the hyperbolic tangent functions.