How would one evaluate the integral $$\int_0^\infty \frac{\arctan(x) }{x(1+x^2)}\,dx$$?
I was told it had a nice closed form and could have been solved with differentiation under the integral sign; however, I tried to set $$I(\alpha) = \int_0^\infty \frac{\arctan(\alpha x)}{x(1+x^2)}~dx$$ and got nowhere (the resulting integral was very messy). Is there a much more clever substitution that can be used to tackle the integral?
Rewrite
$$\frac{\arctan{x}}{x} = \int_0^1 \frac{du}{1+x^2 u^2} $$
Then plugging this in and reversing the order of integration, we get the integral; value as
$$\begin{align}\int_0^1 \frac{du}{u^2} \, \int_0^{\infty} dx \, \left (\frac1{\frac1{u^2}+x^2} \frac1{1+x^2} \right ) &= \int_0^1 \frac{du}{1-u^2} \, \int_0^{\infty} dx \left ( \frac1{1+x^2}-\frac1{\frac1{u^2}+x^2} \right )\\ &= \int_0^1 \frac{du}{1-u^2} \, \frac{\pi}{2} (1-u) \\ &= \frac{\pi}{2} \log{2}\end{align}$$
Let $x=\tan u$ then $$\int_0^\infty \frac{\arctan(x) \,dx}{x(1+x^2)}=\int_0^{\pi/2} \frac{u}{\tan u}\ du =\int_0^{\pi/2} u\cot u\ du $$ now use $$\int_0^{\pi/2} u\cot u \ du=\dfrac{\pi}{2}\ln2$$
Let $x=\tan(u)$, then $$ \begin{align} \int_0^\infty\frac{\arctan(x)\,\mathrm{d}x}{x\left(1+x^2\right)} &=\int_0^{\pi/2}u\cot(u)\,\mathrm{d}u\\ &=\int_0^{\pi/2}u\,\mathrm{d}\log(\sin(u))\\ &=-\int_0^{\pi/2}\log(\sin(u))\,\mathrm{d}u\\[6pt] &=\frac\pi2\log(2) \end{align} $$ where the last step uses $(2)$ from this answer.
Let
$$ I=\int_0^\infty \frac{\arctan(x) \,dx}{x(1+x^2)}.$$
The change of variable $\arctan x=t$ give us
$$I=\int_0^{\pi/2} \frac{t\cos t \,dt}{\sin t}=\frac{1}{2}\pi\log{2}$$
A special case of Ramanujan's master theorem for integrating even functions over the positive reals was found by Glaisher in the late 19th century. If $f(x)$ is an even function with a series expansion around zero of the form:
$$f(x) = \sum_{n=0}^{\infty}(-1)^n c_n x^{2n}$$
then
$$\int_0^{\infty}f(x) dx = \frac{\pi}{2}c_{-\frac{1}{2}}$$
if the integral converges. Here the rigorous definition of $c_{-\frac{1}{2}}$ follows from the rigorous statement of Ramanujan's master theorem, but in practice one can simply analytically continue the $c_k$ to fractional values of $k$ in a natural way, e.g. by replacing factorials by gamma functions etc.
In this case, we have:
$$c_n = \sum_{k=0}^n\frac{1}{2k+1}\tag{1}$$
The natural way to analytically continue integer limits of summations to the reals or the complex plane is given here. To evaluate $c_{-\frac{1}{2}}$ one can consider the large $n$ asymptotic expansion of $c_n$, treating $n$ as a continuous variable there allows one to shift lower limit of the summation to $\frac{1}{2}$, allowing one to extract the value of the summation from 0 to $-\frac{1}{2}$.
From the asymptotic formula:
$$\sum_{k=1}^{n}\frac{1}{k} = \log(n) +\gamma +\mathcal{O}\left(n^{-1}\right)\tag{2}$$
we can derive the large $n$ asymptotics of $c_n$ by considering the summation over even $k$:
$$\sum_{k=1}^{n}\frac{1}{2k} = \frac{1}{2}\sum_{k=1}^{n}\frac{1}{k} =\frac{1}{2}\log(n) +\frac{1}{2}\gamma +\mathcal{O}\left(n^{-1}\right)\tag{3}$$
Using(2) and (3) we can then write:
$$c_n = \sum_{k=1}^{2n+2}\frac{1}{k} - \sum_{k=1}^{n+1}\frac{1}{2k} = \log(2) + \frac{1}{2}\log(n+1)+\frac{1}{2}\gamma +\mathcal{O}\left(n^{-1}\right)\tag{4}$$
Then the summation to $n$ in (1) can also be written as a summation to some arbitrary $u$ plus the summation from $u+1$ to $n$. This rule continues to hold for fractional summations whren $u$ and $n$ are arbitrary real or complex numbers. We thus have:
$$c_n = c_{-\frac{1}{2}} + \sum_{k=\frac{1}{2}}^n \frac{1}{2k+1}\tag{5}$$
We can write:
$$\sum_{k=\frac{1}{2}}^n \frac{1}{2k+1} = \sum_{k=1}^{n+\frac{1}{2}} \frac{1}{2k} = \frac{1}{2}\log\left(n+\frac{1}{2}\right) +\frac{1}{2}\gamma +\mathcal{O}\left(n^{-1}\right)$$
Inserting this in (5) and using (4), we then find:
$$c_{-\frac{1}{2}} = \log(2) + \frac{1}{2}\log(n+1) - \frac{1}{2}\log\left(n+\frac{1}{2}\right) + \mathcal{O}\left(n^{-1}\right)$$
Since there cannot be any dependence on $n$, the r.h.s. is actually a constant but this is not visible to us as we're not keeping track of any $\mathcal{O}\left(n^{-1}\right)$ terms. But taking the limit of $n\to\infty$ makes it clear that $c_{-\frac{1}{2}} = \log(2)$, therefore:
$$\int_0^{\infty}\frac{\arctan(x)dx}{x(1+x^2)}=\frac{\pi}{2}\log(2)$$
The approach you were pointed to isn't that messy; we have $$I'=\int_0^\infty\frac{1}{1-\alpha^2}\bigg(\frac{1}{1+x^2}-\frac{\alpha^2}{1+\alpha^2 x^2}\bigg)dx=\frac{\pi/2}{1+\alpha},$$so from $I(0)=0$ we get $$I(\alpha)=\frac{\pi}{2}\ln|1+\alpha|,\,I(1)=\frac{\pi}{2}\ln 2.$$However, I think @RonGordon's use of Feynman's trick is more interesting than this use of Feynman's trick.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[15px,#ffd]{\int_{0}^{\infty}{\arctan\pars{x} \over x\pars{1 + x^{2}}}\,\dd x} \,\,\,\stackrel{x\ =\ \tan\pars{\theta}}{=}\,\,\, \int_{0}^{\pi/2}{\theta \over \tan\pars{\theta}}\,\dd\theta \\[5mm] = &\ \left.\Re\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2}{-\ic\ln\pars{z} \over -\ic\pars{z^{2} - 1}/\pars{z^{2} + 1}}\,{\dd z \over \ic z} \,\right\vert_{\ z\ =\ \exp\pars{\ic\theta}} \\[5mm] = &\ \left.-\,\Im\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2} {\pars{1 + z^{2}}\ln\pars{z} \over 1 - z^{2}}\,{\dd z \over z} \,\right\vert_{\ z\ =\ \exp\pars{\ic\theta}} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\, &\ \Im\int_{1}^{\epsilon} {\pars{1 - y^{2}}\bracks{\ln\pars{y} + \ic\pi/2} \over 1 + y^{2}} \,{\ic\,\dd y \over \ic y} + \Im\int_{\pi/2}^{0}\bracks{\ln\pars{\epsilon} + \ic\theta}\, {\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \epsilon\expo{\ic\theta}} \\[5mm] = &\ -\,{\pi \over 2}\int_{\epsilon}^{1} {1 - y^{2} \over 1 + y^{2}}\,{\dd y \over y} - {\pi \over 2}\ln\pars{\epsilon} \\[5mm] = &\ -\,{\pi \over 2}\int_{\epsilon}^{1} \pars{{1 - y^{2} \over 1 + y^{2}} - 1}\,{\dd y \over y} - {\pi \over 2}\int_{\epsilon}^{1}{\dd y \over y} - {\pi \over 2}\ln\pars{\epsilon} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\,& \pi\int_{0}^{1}{y \over y^{2} + 1}\,\dd y = \bbx{{1 \over 2}\,\pi\ln\pars{2}} \end{align}
Note $$\int_0^\infty \frac{\arctan(x) }{x(1+x^2)}\,dx=\frac12\int_0^\infty \frac1x d\arctan^2(x)=\frac12\int_0^\infty \frac{\arctan^2(x)}{x^2} \,dx.$$ Now define $$ I(a,b)=\int_0^\infty \frac{\arctan(ax)\arctan(bx)}{x^2} \,dx $$ and then \begin{eqnarray} \frac{\partial^2I(a,b)}{\partial a\partial b}&=&\int_0^\infty \frac{1}{(1+a^2x^2)(1+b^2x^2)} \,dx\\ &=&\frac{1}{a^2-b^2}\int_0^\infty\bigg(\frac{a^2}{1+a^2x^2}-\frac{b^2}{1+b^2x^2}\bigg)\;dx\\ &=&\frac{\pi}{2(a+b)}. \end{eqnarray} So $$ \int_0^\infty \frac{\arctan(x) }{x(1+x^2)}\,dx=\frac12\int_0^\infty \frac{\arctan^2(x)}{x^2} \,dx=\frac\pi4\int_0^1\int_0^1\frac{1}{a+b}=\frac{\pi}{2}\ln2. $$
