# 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 $$I(2n)=2(2n)!\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}=\frac{(2n)!}{2^{4n+1}}\left[\zeta\left(2n+1, \frac{1}{4}\right)-\zeta\left(2n+1, \frac{3}{4}\right)\right].$$

• This MSE link shows how to set up a recurrence for these integrals which you may then compare to the conjectured closed form. Oct 20, 2016 at 0:23
• For the record, the Maple code at the link I quoted does indeed produce a multiple of OEIS A000364, namely $$\frac{\pi^p}{2^{p+1}} \times p! [z^p] \frac{1}{\cos(z)},$$ Oct 20, 2016 at 1:03
• With $\displaystyle{x \equiv \left(1 - t \over t\right)^{1/2}}$ the integral is rewritten as the Beta Function $\displaystyle{n}$-th derivative. Start from $\displaystyle{\int_{0}^{\infty}{x^{\mu} \over 1 + x^{2}}\,\mathrm{d}x}$. Oct 20, 2016 at 8:29
• This a classical integral that is found in Gradshteyn and Ryzhik (formula 4.271.6 page 549 of my 2nd printing 1981 edition). Oct 20, 2016 at 12:18
• Why $\displaystyle \beta(3)=\sum_{n=0}^{+\infty} \dfrac{(-1)^n}{(2n+1)^3}=\dfrac{\pi^3}{32}$? You can read vixra.org/abs/1607.0569 for example.
– FDP
Oct 20, 2016 at 14:51

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}

• Thank you. I'm not very familiar with contour integration, but this is quite thorough and it seems to show where the Euler numbers come from. Oct 20, 2016 at 11:51

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.

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

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

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

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

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.

• Ah yes, I see where I made an error in the last part. Thank you for your help. Oct 20, 2016 at 11:48
• @ChristianSerio. You are very welcome ! Interesting problem, for sure. Oct 20, 2016 at 12:25

As shown at the following MSE link we have a recurrence for these integrals being parameterized by $n$:

$$Q_n = -\frac{1}{n+1} \sum_k \alpha_{n+1,k} - \frac{1}{n+1} \sum_{p=0}^{n-1} {n+1\choose p} (2\pi i)^{n-p} Q_p$$

where $$\alpha_{n, k} = \mathrm{Res}_{z=\rho_k} \frac{\log^n z}{z^2+1}.$$

with $\rho_0 = i$ and $\rho_1 = -i.$ We now seek to prove

$$Q_n = \frac{\pi^{n+1}}{2^{n+1}} \times n! [z^n] \frac{1}{\cos(z)}.$$

Observe that this is

$$Q_n = \frac{\pi}{2} \times n! [z^n] \frac{1}{\cos(\pi z/2)}.$$

The proof can be done by induction. Starting with $n=0$ we have

$$Q_0 = \frac{\pi^1}{2^1} \times 1 = \frac{\pi}{2} = \int_0^\infty \frac{1}{x^2+1} dx.$$

This establishes the base case. We have

$$\alpha_{n, 0} = \frac{1}{2i} (\pi i/2)^n \quad\text{and}\quad \alpha_{n, 1} = -\frac{1}{2i} (3\pi i/2)^n.$$

Continuing with the induction step and applying the induction hypothesis we obtain on the RHS

$$-\frac{1}{n+1} \frac{1}{2i} (\pi i/2)^{n+1} + \frac{1}{n+1} \frac{1}{2i} (3\pi i/2)^{n+1} \\ - \frac{\pi}{2} \frac{1}{n+1} \sum_{p=0}^{n-1} {n+1\choose p} (2\pi i)^{n-p} \times p! [z^p] \frac{1}{\cos(\pi z/2)} \\ = -\frac{1}{n+1} \frac{1}{2i} (\pi i/2)^{n+1} + \frac{1}{n+1} \frac{1}{2i} (3\pi i/2)^{n+1} \\ - \frac{1}{4i} \frac{1}{n+1} \sum_{p=0}^{n-1} {n+1\choose p} (2\pi i)^{n+1-p} \times p! [z^p] \frac{1}{\cos(\pi z/2)} \\ = -\frac{1}{n+1} \frac{1}{2i} (\pi i/2)^{n+1} + \frac{1}{n+1} \frac{1}{2i} (3\pi i/2)^{n+1} \\ + \frac{1}{4i} \frac{1}{n+1} (n+1) (2\pi i) n! [z^n] \frac{1}{\cos(\pi z/2)} + \frac{1}{4i} \frac{1}{n+1} (n+1)! [z^{n+1}] \frac{1}{\cos(\pi z/2)} \\ - \frac{1}{4i} \frac{1}{n+1} \sum_{p=0}^{n+1} {n+1\choose p} (2\pi i)^{n+1-p} \times p! [z^p] \frac{1}{\cos(\pi z/2)}.$$

What we have here is a convolution of two exponential generating functions and we obtain

$$-\frac{1}{n+1} \frac{1}{2i} (\pi i/2)^{n+1} + \frac{1}{n+1} \frac{1}{2i} (3\pi i/2)^{n+1} \\ + \frac{\pi}{2} n! [z^n] \frac{1}{\cos(\pi z/2)} + \frac{1}{4i} n! [z^{n+1}] \frac{1}{\cos(\pi z/2)} \\ - \frac{1}{4i} \frac{1}{n+1} (n+1)! [z^{n+1}] \frac{\exp(2\pi i z)}{\cos(\pi z/2)}.$$

Simplifying the last two terms we get

$$\frac{1}{4i} n! [z^{n+1}] \frac{\exp(2\pi i z)-1}{\cos(\pi z/2)} \\ = \frac{1}{4i} n! [z^{n+1}] \exp(\pi i z) \frac{\exp(\pi i z)-\exp(-\pi i z)}{\cos(\pi z/2)} \\ = \frac{1}{2} n! [z^{n+1}] \exp(\pi i z) \frac{\sin(\pi z)}{\cos(\pi z/2)} \\ = n! [z^{n+1}] \exp(\pi i z) \sin(\pi z/2) \\ = \frac{1}{2i} n! [z^{n+1}] (\exp(3\pi i z/2) - \exp(\pi i z/2)) \\ = \frac{1}{n+1} \frac{1}{2i} (3\pi i/2)^{n+1} -\frac{1}{n+1} \frac{1}{2i} (\pi i/2)^{n+1}.$$

The contribution from the last two terms cancels the first two terms, leaving just the middle term which is

$$\bbox[5px,border:2px solid #00A000]{ \frac{\pi}{2} n! [z^n] \frac{1}{\cos(\pi z/2)}.}$$

This concludes the proof by induction. The EGF of the Euler numbers is

$$\frac{1}{\cosh(z)} = \frac{2}{\exp(z)+\exp(-z)} = \sum_{n\ge 0} E_n \frac{z^n}{n!}$$

so that

$$\frac{1}{\cos(\pi z/2)} = \sum_{n\ge 0} E_n (\pi i/2)^n \frac{z^n}{n!}$$

and we have the alternate closed form

$$\bbox[5px,border:2px solid #00A000]{ E_n i^n \left(\frac{\pi}{2}\right)^{n+1} = E_{2m} (-1)^m \left(\frac{\pi}{2}\right)^{2m+1}}$$

where we put $n=2m$ as the odd Euler numbers vanish.

The logarithm here refers to the branch with argument from $0$ to $2\pi$ and branch cut on the positive real axis.

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 $$I(n)=\frac\pi2\left[\left(\frac{d}{da}\right)^n\sec\frac{\pi a}2\right]_{a=0}$$

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, $$\lim_{x\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\csc(x)=|E_{2n}|$$

$$\boxed{\int_{0}^{\infty} \frac{(\ln x)^{2 n}}{1+x^{2}} d x =\frac{\pi^{2 n+1}}{2^{2 n}}\left|E_{2 n}\right|}$$

\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}