Difficult integral $\int_0^{\infty}\frac {\arctan x\log(1+x^2)}{x(1+x^2)}dx=\frac {\pi}2\log^22$

Question

Question: Show that$$\int\limits_0^{\infty}\frac {\arctan x\log(1+x^2)}{x(1+x^2)}dx=\frac {\pi}2\log^22.$$

I can't tell if I'm being an idiot, or if this is a lot more difficult than it looks. First, I tried integration by parts using the fact that $$\frac 1{x(1+x^2)}=\frac 1x-\frac x{1+x^2}.$$

But quickly I gave up as I wasn't sure what to do with the result. I then decided to make the substitution $t=\arctan x$ to get rid of the $1+x^2$ term in the denominator. Therefore$$\begin{align*}\mathfrak{I} & =\int\limits_0^{\pi/2}\mathrm dt\,t\cot t\log\sec^2t=-2\int\limits_0^{\pi/2}\mathrm dt\, t\cot t \log\cos t.\end{align*}$$

However, I'm not exactly sure what to do after this. Should I use integration by parts? Differentiation under the integral sign? I'm having trouble getting started with this integral. Any ideas?

Answer 1

We can use differentiation under the integral sign and a trick to evaluate this. First define $$ I(a,b) = \int_0^{\infty} \frac{\arctan{ax}}{x} \frac{\log{(1+b^2 x^2)}}{1+x^2} \, dx , $$ so $I(a,0)=I(0,b)=0$ and $I(1,1)$ is what we want. Differentiating one with respect to $a$ and once wrt $b$ gives $$ \partial_a\partial_b I = \int_0^{\infty} \frac{2bx^2 \, dx}{(1+x^2)(1+a^2x^2)(1+b^2x^2)}, $$ which can be done by using partial fractions and the arctangent integral a few times. When the dust settles, $$ \partial_a\partial_b I = \frac{b\pi}{(1+a)(1+b)(a+b)}, $$ and thus $$ I(1,1) = \int_0^1 \int_0^1 \frac{b\pi}{(1+a)(1+b)(a+b)} \, da \, db $$ But we can swap $a$ and $b$ and will get the same result for this integral by the symmetry of the region of integration, so we also have $$ I(1,1) = \int_0^1 \int_0^1 \frac{a\pi}{(1+a)(1+b)(a+b)} \, da \, db. $$ Adding gives $$ I(1,1) = \frac{\pi}{2}\int_0^1 \int_0^1 \frac{1}{(1+a)(1+b)} \, da \, db, $$ but this splits into a product of two copies of $\int_0^1 dy/(1+y) = \log{2}$, so $$ I(1,1) = \frac{\pi}{2}(\log{2})^2 $$ as desired.

Answer 2

Define $$ f(a,b) = \int \limits_0^\infty \frac{\arctan(a x) \ln (1+ b^2 x^2)}{x (1+x^2)} \, \mathrm{d} x $$ for $0 \leq a , b \leq 1$ . Then $f(1,1) = \mathfrak{I}$ and $f(0,b) = f(a,0) = 0 $ . For $0< a,b<1$ we can differentiate under the integral sign to find $$ \partial_a \partial_b f(a,b) = 2 b \int \limits_0^\infty \frac{x^2}{(1+a^2 x^2)(1+b^2 x^2)(1+x^2)} \, \mathrm{d} t = \frac{\pi b}{(1+a)(1+b)(a+b)} \, . $$ The integral can be evaluated using the residue theorem, for example. Now integrate again and exploit the symmetry of the derivative to obtain \begin{align} \mathfrak{I} &= f(1,1) = \pi \int \limits_0^1 \int \limits_0^1 \frac{b}{(1+a)(1+b)(a+b)} \, \mathrm{d}a \, \mathrm{d}b = \pi \int \limits_0^1 \int \limits_0^1 \frac{a}{(1+a)(1+b)(a+b)} \, \mathrm{d}a \, \mathrm{d}b \\ &= \frac{\pi}{2} \int \limits_0^1 \int \limits_0^1 \frac{a+b}{(1+a)(1+b)(a+b)} \, \mathrm{d}a \, \mathrm{d}b = \frac{\pi}{2} \int \limits_0^1 \int \limits_0^1 \frac{1}{(1+a)(1+b)} \, \mathrm{d}a \, \mathrm{d}b \\ &= \frac{\pi}{2} \ln^2 (2) \, . \end{align}

Answer 3

Notice:

$$ \Im(\log^2(1+ix))=\Im\left(\left(\frac{1}{2}\log(1+x^2)+i\arctan(x)\right)^2\right)=\frac{1}{2}\log(1+x^2)\arctan(x) $$

The integral in question is therefore (use parity) $$ I=\Im\int_{\mathbb{R}}\underbrace{\frac{\log^2(1+ix)}{x(1+x^2)}}_{f(x)} $$

Integrate around a big semicircle in the lower halfplane (to avoid the branchcut) yields

$$ I=\Im \left(2\pi i\text{Res}(f(z),z=-i)\right)=\frac{\pi}{2}\log^2(2)$$

where the residue is easy to calculate since the pole is simple.

The vansihing of the contrbutions at infinity follows from the fact that $R|f(Re^{i\phi})|\sim \log^2(R)/R^2$ in sector of $\mathbb{C}$ we are interested in.

Answer 4

Utilize the known integrals \begin{align} &\int_0^\infty \frac{\ln y}{(1+y)^2+a^2}{dy} = \frac1{2a}\tan^{-1}a\ \ln(1+a^2)\\ &\int_0^\infty \frac{\ln y}{(1+y)(a+y)}{dy} =\frac{\ln^2a}{2(a-1)} \end{align} to evalute \begin{align} &\int_0^{\infty}\frac {\tan^{-1} x\ln(1+x^2)}{x(1+x^2)}dx \\ =& \int_0^\infty\int_0^\infty \frac{2\ln y}{(1+x^2)[(1+y)^2+x^2]}{dy}\ dx\\ =& \int_0^\infty \frac{\pi\ln y}{(1+y)(2+y)}{dy}=\frac\pi2\ln^22 \end{align}

Answer 5

\begin{align}J&=\int_0^\infty \frac{\ln(1+x^2)\arctan x}{x(1+x^2)}dx\\ &=\int_0^1 \frac{\ln(1+x^2)\arctan x}{x(1+x^2)}dx+\int_1^\infty \frac{\ln(1+y^2)\arctan y}{y(1+y^2)}dy\\ &\overset{x=\frac{1}{y}}=\int_0^1 \frac{\ln(1+x^2)\arctan x}{x(1+x^2)}dx+\int_0^1 \frac{x\ln\left(\frac{1+x^2}{x^2}\right)\arctan\left( \frac{1}{x}\right)}{1+x^2}dx\\ &=\int_0^1 \frac{\ln(1+x^2)\arctan x}{x}dx-2\int_0^1 \frac{x\ln(1+x^2)\arctan x}{1+x^2}dx+2\int_0^1 \frac{x\ln x\arctan x}{1+x^2}dx+\\ &\frac{\pi}{2}\int_0^1 \frac{x\ln(1+x^2)}{1+x^2}dx-\pi\int_0^1 \frac{x\ln x}{1+x^2}dx\\ &\overset{\text{IBP}}=\int_0^1 \frac{\ln(1+x^2)\arctan x}{x}dx-2\int_0^1 \frac{x\ln(1+x^2)\arctan x}{1+x^2}dx+\\&\Big[\ln(1+x^2)\ln x\arctan x\Big]_0^1-\int_0^1 \frac{\ln(1+x^2)\arctan x}{x}dx-\int_0^1 \frac{\ln(1+x^2)\ln x}{1+x^2}dx+\\&\frac{\pi}{2}\int_0^1 \frac{x\ln(1+x^2)}{1+x^2}dx-\pi\int_0^1 \frac{x\ln x}{1+x^2}dx\\ &=-2\int_0^1 \frac{x\ln(1+x^2)\arctan x}{1+x^2}dx-\int_0^1 \frac{\ln(1+x^2)\ln x}{1+x^2}dx+\frac{\pi}{2}\int_0^1 \frac{x\ln(1+x^2)}{1+x^2}dx-\\&\pi\int_0^1 \frac{x\ln x}{1+x^2}dx\\ K&=\int_0^1 \frac{2x\ln(1+x^2)\arctan x}{1+x^2}dx\\ &\overset{\text{IBP}}=\Big[\ln(1+x^2)^2\arctan x\Big]_0^1-K-\int_0^1 \frac{\ln^2(1+x^2)}{1+x^2}dx\\ K&=\frac{\pi\ln^2 2}{8}-\frac{1}{2}\int_0^1 \frac{\ln^2(1+x^2)}{1+x^2}dx\\ J&=-\frac{\pi\ln^2 2}{8}+\frac{1}{2}\int_0^1 \frac{\ln^2(1+x^2)}{1+x^2}dx-\int_0^1 \frac{\ln(1+x^2)\ln x}{1+x^2}dx+\frac{\pi}{2}\int_0^1 \frac{y\ln(1+y^2)}{1+y^2}dy-\\&\pi\int_0^1 \frac{y\ln y}{1+y^2}dy\\ &\overset{x=y^2}=-\frac{\pi\ln^2 2}{8}+\frac{1}{2}\int_0^1 \frac{\ln^2(1+x^2)}{1+x^2}dx-\int_0^1 \frac{\ln(1+x^2)\ln x}{1+x^2}dx+\frac{\pi}{4}\int_0^1 \frac{\ln(1+x)}{1+x}dx-\\&\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\ &=\frac{1}{2}\int_0^1 \frac{\ln^2(1+y^2)}{1+y^2}dy-\int_0^1 \frac{\ln(1+y^2)\ln y}{1+y^2}dy-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\ &\overset{y=\tan t}=2\int_0^{\frac{\pi}{4}}\ln^2\left(\cos t\right)dt-2\int_0^{\frac{\pi}{4}}\ln^2\left(\cos t\right)dt+2\int_0^{\frac{\pi}{4}}\ln\left(\cos t\right)\ln\left(\sin t\right)dt-\\&\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\ &=2\int_0^{\frac{\pi}{4}}\ln\left(\cos t\right)\ln\left(\sin t\right)dt-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\ &=\frac{1}{2}\left(\int_0^{\frac{\pi}{4}}\ln^2(\sin t\cos t)dt-\int_0^{\frac{\pi}{4}}\ln^2\left(\tan u\right)du\right)-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\ &\overset{u=2t}=\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin u}{2}\right)du-\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan u\right)du-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\ &=\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\sin u\right)du-\frac{\ln 2}{2}\int_0^{\frac{\pi}{2}}\ln\left(\sin u\right)du+\frac{\pi\ln^2 2}{8}-\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan u\right)du-\\&\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx\\ L&=\int_0^{\frac{\pi}{2}}\ln^2\left(\sin u\right)du\\ &=\frac{1}{4}\left(2\int_0^{\frac{\pi}{2}}\ln^2\left(\sin u\right)du+2\int_0^{\frac{\pi}{2}}\ln^2\left(\cos u\right)du\right)\\ &=\frac{1}{4}\left(\int_0^{\frac{\pi}{2}}\ln^2\left(\cos u\sin u\right)du+\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\right)\\ &\overset{t=2u}=\frac{1}{4}\left(\frac{1}{2}\int_0^{\pi}\ln^2\left(\frac{\sin t}{2}\right)dt+\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\right)\\ &=\frac{1}{8}\int_0^{\pi}\ln^2\left(\sin t\right)dt-\frac{\ln 2}{4}\int_0^{\pi}\ln\left(\sin t\right)dt+\frac{\pi\ln^2 2}{8}+\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\ &=\frac{1}{8}\left(L+\int_{\frac{\pi}{2}}^\pi\ln^2\left(\sin v\right)dv\right)-\frac{\ln 2}{4}\int_0^{\pi}\ln\left(\sin t\right)dt+\frac{\pi\ln^2 2}{8}+\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\ &\overset{t=\pi-v}=\frac{1}{4}L-\frac{\ln 2}{4}\int_0^{\pi}\ln\left(\sin t\right)dt+\frac{\pi\ln^2 2}{8}+\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\ L&=-\frac{\ln 2}{3}\int_0^{\pi}\ln\left(\sin t\right)dt+\frac{\pi\ln^2 2}{6}+\frac{1}{3}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\ &=-\frac{\ln 2}{3}\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt-\frac{\ln 2}{3}\int_{\frac{\pi}{2}}^{\pi}\ln\left(\sin v\right)dv+\frac{\pi\ln^2 2}{6}+\frac{1}{3}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\ &\overset{t=\pi-v}=-\frac{2\ln 2}{3}\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt+\frac{\pi\ln^2 2}{6}+\frac{1}{3}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan u\right)du\\ \end{align}

Therefore,

\begin{align}J&=\frac{1}{6}\pi\ln^2 2-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx-\frac{2\ln 2}{3}\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt+\frac{1}{12}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan t\right)dt-\\&\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan t\right)dt\\ &=\frac{1}{6}\pi\ln^2 2-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx-\frac{2\ln 2}{3}\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt+\\&\frac{1}{12}\left(\int_0^{\frac{\pi}{4}}\ln^2\left(\tan t\right)dt+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln\left(\tan u\right)du\right)-\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan t\right)dt\\ &\overset{t=\frac{\pi}{2}-v}=\frac{1}{6}\pi\ln^2 2-\frac{\pi}{4}\int_0^1 \frac{\ln x}{1+x}dx-\frac{2\ln 2}{3}\int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt-\frac{1}{3}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan t\right)dt\\ &=\frac{1}{6}\pi\ln^2 2-\frac{\pi}{4}\times -\frac{\pi^2}{12}-\frac{2\ln 2}{3}\times -\frac{\pi\ln 2}{2}-\frac{1}{3}\times \frac{\pi^3}{16}\\ &=\boxed{\frac{\pi\ln^2 2}{2}} \end{align}

I assume that:

\begin{align}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan t\right)dt&=\frac{\pi^3}{16}\\ \int_0^1 \frac{\ln x}{1-x}dx&=-\zeta(2)=-\frac{\pi^2}{6}\\ \int_0^{\frac{\pi}{2}}\ln\left(\sin t\right)dt&=-\frac{\pi\ln 2}{2}\\ \int_0^1 \frac{\ln x}{1+x}dx&=\int_0^1 \frac{\ln x}{1-x}dx-\int_0^1 \frac{2y\ln y}{1-y^2}dy\\ &\overset{y=x^2}=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx=-\frac{\pi^2}{12} \end{align}

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align*} & \color{#44f}{\int_{0}^{\infty}{\arctan\pars{x}\ln\pars{1 + x^{2}} \over x\pars{1 + x^{2}}}\dd x} \\ = & \ {1 \over 2}\int_{-\infty}^{\infty}\ \overbrace{{\ln\pars{1 + x\ic} - \ln\pars{1 - x\ic} \over 2\ic}}^{\ds{\arctan\pars{x}}}\ \times \\ & \phantom{{1 \over 2}\int_{-\infty}^{\infty}}\underbrace{\bracks{\rule{0pt}{5mm}\ln\pars{1 + x\ic} + \ln\pars{1 - x\ic}}}_{\ds{\ln\pars{1 + x^{2}}}}\,\, {\dd x \over x\pars{1 + x^{2}}} \\[5mm] = & \ {1 \over 2}\,\Im\int_{-\infty}^{\infty} {\ln^{2}\pars{1 + x\ic} \over x\pars{1 + x^{2}}}\dd x \\[5mm] \sr{1\ +\ x\ic\ =\ s}{=} & \ {1 \over 2}\,\Im\int_{1\ -\ \infty\ic}^{1\ +\ \infty\ic} {\ln^{2}\pars{s} \over \bracks{-\pars{s - 1}\ic}\bracks{-s\pars{s - 2}}}\pars{-\ic}\dd s \\[5mm] = & \ -{1 \over 2}\,\Im\int_{1\ -\ \infty\ic}^{1\ +\ \infty\ic} {\ln^{2}\pars{s} \over s\pars{s - 1}\pars{s - 2}}\dd s \\[5mm] = & \ -{1 \over 2}\,\Im\braces{-2\pi\ic\,\on{Res}\bracks{{\ln^{2}\pars{s} \over s\pars{s - 1}\pars{s - 2}}, s = 2}} \\[5mm] = & \ -{1 \over 2}\pars{-2\pi}{\ln^{2}\pars{2} \over 2\pars{2 - 1}} = \bbx{\color{#44f}{\pi\ln^{2}\pars{2} \over 2}} \approx 0.7547 \\[5mm] & \begin{array}{l} \mbox{The integral along}\ \pars{1 - \infty\ic,1 + \infty\ic} \\ \mbox{is "closed" with the arc}\ \left.R\!\expo{\pars{\pi\ic/2,-\pi\ic/2}}\,\,\right\vert_{\,R\ \to\ \infty} \\ \mbox{which does not yield any contribution to} \\ \mbox{the final result.} \end{array} \end{align*}

Answer 7

Today, I used real methods to calculate an integral. The integral to be evaluated is as follows:

$$I = \int_{0}^{\infty} \frac{\arctan x \ln \left(1+x^2 \right)}{x\left(1+x^2\right)} \, dx.$$

Using the idea of parameter integration, let

$$I(b) = \int_{0}^{\infty} \frac{\arctan bx \ln \left(1+x^2 \right)}{x\left(1+x^2\right)} \, dx, \quad (b > 0).$$

Then,

\begin{align} I'(b) &= \int_{0}^{\infty} \frac{\ln \left(1+x^2 \right)}{\left(1+x^2 \right) \left(1+b^2 x^2 \right)} \, dx \\ &= \frac{1}{1-b^2} \int_{0}^{\infty} \left[\frac{\ln \left(1+x^2\right)}{1+x^2} - \frac{b^2 \ln \left(1+x^2\right)}{1+b^2 x^2}\right] \, dx \\ &= \frac{1}{1-b^2} J - \frac{b^2}{1-b^2} K, \end{align}

where

\begin{align} J &= \int_{0}^{\infty} \frac{\ln \left(1+x^2\right)}{1+x^2} \, dx \quad \text{(let } x \rightarrow \tan x) \\ &= -2 \int_{0}^{\frac{\pi}{2}} \ln \cos x \, dx \quad \text{(let } x \rightarrow \frac{\pi}{2} - x) \\ &= -2 \int_{0}^{\frac{\pi}{2}} \ln \sin x \, dx \\ &= -\int_{0}^{\frac{\pi}{2}} \ln (\sin x \cos x) \, dx \quad \text{(average of the two expressions)} \\ &= -\int_{0}^{\frac{\pi}{2}} \ln \sin 2x \, dx \quad \text{(let } 2x \rightarrow x) + \int_{0}^{\frac{\pi}{2}} \ln 2 \, dx \\ &= \frac{\pi \ln 2}{2} - \frac{1}{2} \int_{0}^{\pi} \ln \sin x \, dx \\ &= \frac{\pi \ln 2}{2} - \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} \ln \sin x \, dx + \int_{\frac{\pi}{2}}^{\pi} \ln \sin x \, dx \quad \text{(let } x \rightarrow \pi - x) \right) \\ &= \frac{\pi \ln 2}{2} - \int_{0}^{\frac{\pi}{2}} \ln \sin x \, dx \\ &= \pi \ln 2. \end{align}

And,

$$ K = \int_{0}^{\infty} \frac{\ln \left(1+x^2\right)}{1+b^2 x^2} \, dx. $$

For $K$, consider parameter integration, let

$$ K(a) = \int_{0}^{\infty} \frac{\ln \left(1+a^2x^2\right)}{1+b^2 x^2} \, dx, \quad (a > 0). $$

Then,

\begin{align} K'(a) &= \int_{0}^{\infty} \frac{2ax^2}{\left(1+a^2x^2\right)\left(1+b^2x^2\right)} \, dx \\ &= \frac{2a}{a^2 - b^2} \int_{0}^{\infty} \left( \frac{1}{1+b^2x^2} - \frac{1}{1+a^2x^2} \right) \, dx \\ &= \frac{2a}{a^2 - b^2} \left( \frac{\pi}{2b} - \frac{\pi}{2a} \right) \\ &= \frac{\pi}{b(a + b)}. \end{align}

Thus, we have

$$ K(a) = \frac{\pi \ln (a + b)}{b} + C, $$

where $C$ is the integration constant. From

\begin{align} K(0) &= \int_{0}^{\infty} 0 \, dx \\ &= 0 = \frac{\pi \ln b}{b} + C, \end{align}

we get

$$ C = -\frac{\pi \ln b}{b}. $$

Therefore,

$$ \begin{align} K(a) &= \frac{\pi \ln (a + b)}{b} - \frac{\pi \ln b}{b} \\ &= \frac{\pi}{b} \ln \left(1 + \frac{a}{b}\right). \end{align} $$

Or, we can write

$$ K(a, b) = \frac{\pi}{b} \ln \left(1 + \frac{a}{b}\right). $$

It is easy to see that the integral $J$ is actually $K(1, 1) = \pi \ln 2$, so the integral $J$ can also be derived from the integral $K$. The reason for calculating the integral $J$ is to obtain the by-product integrals $\int_{0}^{\frac{\pi}{2}} \ln \cos x \, dx$ and $\int_{0}^{\frac{\pi}{2}} \ln \sin x \, dx$.

From the above, the integral

$$ K = K(1) = \frac{\pi}{b} \ln \left(1 + \frac{1}{b}\right). $$

Thus, $I'(b)$ is

\begin{align} I'(b) &= \frac{1}{1 - b^2} J + \frac{b^2}{1 - b^2} K \\ &= \frac{\ln 2 - b \ln \left(1 + \frac{1}{b}\right)}{1 - b^2} \pi. \end{align}

Therefore, the desired integral

$$ I = I(1) = \int_{0}^{1} I'(b) \, db + I(0). $$

And,

$$ I(0) = \int_{0}^{\infty} 0 \, dx = 0. $$

So, the desired integral

\begin{align} I &= \int_{0}^{1} \frac{\ln 2 - b \ln \left(1 + \frac{1}{b}\right)}{1 - b^2} \pi \, db \\ &= \pi \int_{0}^{1} \frac{\ln 2 - b \ln (1 + b) + b \ln b}{1 - b^2} \, db \\ &= \pi \left[ \int_{0}^{1} \frac{\ln 2 - b \ln (1 + b)}{1 - b^2} \, db + \int_{0}^{1} \frac{b \ln b}{1 - b^2} \, db \right] \\ &= \pi (L + M), \end{align}

where

\begin{align} M &= \int_{0}^{1} \frac{b \ln b}{1 - b^2} \, db \\ &= -\frac{1}{2} \int_{0}^{1} \ln b \, d \ln \left(1 - b^2\right) \\ &= -\frac{1}{2} \ln b \ln \left(1 - b^2\right) \bigg|_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{\ln \left(1 - b^2\right)}{b} \, db \\ &= \lim_{b \to 1} -\frac{1}{2} \ln b \ln \left(1 - b^2\right) + \lim_{b \to 0} \frac{1}{2} \ln b \ln \left(1 - b^2\right) - \frac{1}{2} \int_{0}^{1} \frac{1}{b} \sum_{n=1}^{\infty} \frac{b^{2n}}{n} \, db \\ &= -\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} b^{2n-1} \, db \\ &= -\frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} \\ &= -\frac{\pi^2}{24}. \end{align}

And,

\begin{align} L &= \int_{0}^{1} \frac{\ln 2 - b \ln (1 + b)}{1 - b^2} \, db \\ &= \frac{1}{2} \int_{0}^{1} \left[ \ln 2 - b \ln (1 + b) \right] \, d \ln \left( \frac{1 + b}{1 - b} \right) \\ &= \frac{1}{2} \left[ \ln 2 - b \ln (1 + b) \right] \ln \left( \frac{1 + b}{1 - b} \right) \bigg|_{0}^{1} + \frac{1}{2} \int_{0}^{1} \ln \left( \frac{1 + b}{1 - b} \right) \left[ \ln (1 + b) + \frac{b}{1 + b} \right] \, db \\ &= \frac{1}{2} \left[ \int_{0}^{1} \ln^2 (1 + b) \, db + \int_{0}^{1} \frac{b \ln (1 + b)}{1 + b} \, db - \int_{0}^{1} \ln (1 - b) \ln (1 + b) \, db - \int_{0}^{1} \frac{b \ln (1 - b)}{1 + b} \, db \right] \\ &= \frac{1}{2} \left[ \int_{0}^{1} \ln^2 (1 + b) \, db + \int_{0}^{1} \ln (1 + b) \, db - \int_{0}^{1} \frac{\ln (1 + b)}{1 + b} \, db - \int_{0}^{1} \ln (1 - b) \ln (1 + b) \, db - \int_{0}^{1} \ln (1 - b) \, db + \int_{0}^{1} \frac{\ln (1 - b)}{1 + b} \, db \right] \\ &= \frac{1}{2} (A + B - C - D - E + F), \end{align}

where

\begin{align} A &= \int_{0}^{1} \ln^2 (1 + b) \, db \\ &= \int_{1}^{2} \ln^2 b \, db \\ &= b \ln^2 b \bigg|_{1}^{2} - 2 \int_{1}^{2} \ln b \, db \\ &= 2 \ln^2 2 - 4 \ln 2 + 2. \\ B &= \int_{0}^{1} \ln (1 + b) \, db \\ &= b \ln (1 + b) \bigg|_{0}^{1} - \int_{0}^{1} \frac{b}{1 + b} \, db \\ &= 2 \ln 2 - 1. \\ C &= \int_{0}^{1} \frac{\ln (1 + b)}{1 + b} \, db \\ &= \frac{1}{2} \ln^2 (1 + b) \bigg|_{0}^{1} \\ &= \frac{1}{2} \ln^2 2. \\ E &= \int_{0}^{1} \ln (1 - b) \, db \\ &= b \ln (1 - b) \bigg|_{0}^{1} - \int_{0}^{1} \frac{-b}{1 - b} \, db \\ &= -1. \end{align}

And,

\begin{align} D &= \int_{0}^{1} \ln (1 - b) \ln (1 + b) \, db \\ &= b \ln (1 - b) \ln (1 + b) \bigg|_{0}^{1} - \int_{0}^{1} b \left[ \frac{-\ln (1 + b)}{1 - b} + \frac{\ln (1 - b)}{1 + b} \right] \, db \\ &= b \ln (1 - b) \ln (1 + b) \bigg|_{0}^{1} - \int_{0}^{1} \left[ \ln (1 + b) - \frac{\ln (1 + b)}{1 - b} + \ln (1 - b) - \frac{\ln (1 - b)}{1 + b} \right] \, db \\ &= b \ln (1 - b) \ln (1 + b) \bigg|_{0}^{1} - B + \int_{0}^{1} \frac{\ln (1 + b)}{1 - b} \, db - E + F \\ &= b \ln (1 - b) \ln (1 + b) \bigg|_{0}^{1} - B - E + F - \int_{0}^{1} \ln (1 + b) \, d \ln (1 - b) \\ &= b \ln (1 - b) \ln (1 + b) \bigg|_{0}^{1} - B - E + F - \ln (1 + b) \ln (1 - b) \bigg|_{0}^{1} + \int_{0}^{1} \frac{\ln (1 - b)}{1 + b} \, db \\ &= (b - 1) \ln (1 - b) \ln (1 + b) \bigg|_{0}^{1} - B - E + 2F \\ &= 2F - 2 \ln 2 + 2. \end{align}

Next, solve for $F$. Continue using the idea of parameter integration, let

$$ F(g) = \int_{0}^{1} \frac{\ln (1 - gb)}{1 + b} \, db. $$

Then,

\begin{align} F'(g) &= \int_{0}^{1} \frac{-b}{(1 - gb)(1 + b)} \, db \\ &= \frac{1}{1 + g} \int_{0}^{1} \left( \frac{1}{1 + b} - \frac{1}{1 - gb} \right) \, db \\ &= \frac{\ln 2}{1 + g} - \frac{\ln (1 - g)}{1 + g} + \frac{\ln (1 - g)}{g}. \end{align}

Similarly, from $F(0) = 0$, we have

$$ F = F(1) = \int_{0}^{1} F'(g) \, dg. $$

So,

\begin{align} F &= \int_{0}^{1} \left[ \frac{\ln 2}{1 + g} - \frac{\ln (1 - g)}{1 + g} + \frac{\ln (1 - g)}{g} \right] \, dg \\ &= \ln 2 \int_{0}^{1} \frac{1}{1 + g} \, dg - \int_{0}^{1} \frac{\ln (1 - g)}{1 + g} \, dg + \int_{0}^{1} \frac{\ln (1 - g)}{g} \, dg \\ &= \ln^2 2 - \int_{0}^{1} \frac{\ln (1 - b)}{1 + b} \, db + G \\ &= \ln^2 2 - F + G, \end{align}

where

\begin{align} G &= \int_{0}^{1} \frac{\ln (1 - g)}{g} \, dg \\ &= -\int_{0}^{1} \frac{1}{g} \sum_{n=1}^{\infty} \frac{g^n}{n} \, dg \\ &= -\sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} g^{n-1} \, dg \\ &= -\sum_{n=1}^{\infty} \frac{1}{n^2} \\ &= -\frac{\pi^2}{6}. \end{align}

Therefore,

\begin{align} F &= \frac{1}{2} \left( \ln^2 2 + G \right) \\ &= \frac{1}{2} \ln^2 2 - \frac{\pi^2}{12}. \\ D &= 2F - 2 \ln 2 - 2 \\ &= \ln^2 2 - \frac{\pi^2}{6} - 2 \ln 2 + 2. \\ L &= \frac{1}{2} (A + B - C - D - E + F) \\ &= \frac{1}{2} \ln^2 2 + \frac{\pi^2}{24}. \\ I &= \pi (L + M) = \frac{1}{2} \pi \ln^2 2. \end{align}

Finally, the value of the desired integral is

\begin{align} I &= \int_{0}^{\infty} \frac{\arctan x \ln \left(1 + x^2 \right)}{x \left(1 + x^2\right)} \, dx \\ &= \frac{\pi \ln^2 2}{2}. \end{align}

Additionally, some by-products such as $D$, $F$, $G$, $L$, and $M$ were obtained.