# A closed form for a triple integral with sines and cosines

$$\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z) + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz$$

I saw this integral $I$ posted on a page on Facebook . The author claims that there is a closed form for it.

## My Attempt

This can be rewritten as

$$3\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z)}{xyz(x+y+z)}\,dx\,dy\,dz$$

Now consider

$$F(a) = 3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz(x+y+z)}\,dx\,dy\,dz$$

Taking the derivative

$$F'(a) = -3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz}\,dx\,dy\,dz$$

By symmetry we have

$$F'(a) = -3\left(\int^\infty_0 \frac{\sin^2(x)e^{-ax}}{x}\,dx \right)\left( \int^\infty_0 \frac{\sin(x)\cos(x)e^{-ax}}{x}\,dx\right)^2$$

Using W|A I got

$$F'(a) = -\frac{3}{16} \log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)$$

By integeration we have

$$F(0) = \frac{3}{16} \int^\infty_0\log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)\,da$$

Let $x = 2/a$

$$\tag{1}I = \frac{3}{8} \int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$

## Question

I seem not be able to verify (1) is correct nor find a closed form for it, any ideas ?

• why someone posts integrals on facebook? – tired Feb 4 '17 at 11:01
• a more useful comment: i think your last integral can be done by parts and some more or less boring polylog algebra – tired Feb 4 '17 at 12:05
• a more elegant approach goes along the lines of integrating $f(x) =\log(1+ix)^3/x^2$ around a suitable contour in the complex plane (with a suitable choice of branchcut as well). We then can relate the real part of this integral to the integral in question $I$. since you are furthermore capable of doing $\int_0^{\infty}\frac{\log(1+x^2)^3}{x^2}dx$ you should be able to finsih – tired Feb 4 '17 at 12:55
• $\frac{8}{3}I=\frac{1}{12}(\pi^3+3 \pi\log(4)^2)$ with above technique (seems to be numerically correct) – tired Feb 4 '17 at 13:15
• @tired, I am not that good at contour integration. – Zaid Alyafeai Feb 4 '17 at 19:29

Ok I was able to find the integral

$$\int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$

First note that

$$\int \frac{\log(1+x^2)}{x^2}\,dx = 2 \arctan(x) - \frac{\log(1 + x^2)}{x}+C$$

Using integration by parts

$$I = \frac{\pi^3}{12}+2\int^\infty_0\frac{\arctan(x)\log(1 + x^2)}{(1+x^2)x}\,dx$$

For the integral let

$$F(a) = \int^\infty_0\frac{\arctan(ax)\log(1 + x^2)}{(1+x^2)x}\,dx$$

By differentiation we have

$$F'(a) = \int^\infty_0 \frac{\log(1+x^2)}{(1 + a^2 x^2)(1+x^2)}\,dx$$

Letting $1/a = b$ we get

$$\frac{1}{(1 + a^2 x^2)(1+x^2)} = \frac{1}{a^2} \left\{ \frac{1}{((1/a)^2+x)(1+x^2)}\right\} =\frac{b^2}{1-b^2}\left\{ \frac{1}{b^2+x^2}-\frac{1}{1+x^2} \right\}$$

We conclude that $$\frac{b^2}{1-b^2}\int^\infty_0 \frac{\log(1+x^2)}{b^2+x^2}-\frac{\log(1+x^2)}{1+x^2} \,dx = \frac{b^2}{1-b^2}\left\{ \frac{\pi}{b}\log (1+b)-\pi\log(2)\right\}$$

Where we used that

$$\int^\infty_0 \frac{\log(a^2+b^2x^2)}{c^2+g^2x^2}\,dx = \frac{\pi}{cg}\log \frac{ag+bc}{g}$$

By integration we deduce that

$$\int^1_0 \frac{\pi}{a^2-1}\left\{ a\log \left(1+\frac{1}{a} \right)-\log(2)\right\}\,da = \frac{\pi}{2}\log^2(2)$$

For the last one I used wolfram alpha, however it shouldn't be difficult to prove.

Finally we have

$$\int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx = \frac{\pi^3}{12}+\pi \log^2(2)$$

• nice one (+1)!! – tired Feb 4 '17 at 20:42
• @tired, thanks. – Zaid Alyafeai Feb 4 '17 at 20:54

Another approach to break down the last integral might be to consider the integral of $\displaystyle \frac{\log^3 (1-iz)}{z^2}$ along a positively oriented semi-circular contour $\gamma_R = [-R,R]\cup Re^{i[0,\pi]}$ in the upper half-plane. (We choose the branch of logarithm $\log (1-iz)$ in the lower half-plane along $[-i,-i\infty)$).

The integral along the arc is $\displaystyle \mathcal{O}\left(\frac{\log^3 R}{R}\right)$, which vanishes as $R \to +\infty$ we have,

\begin{align*}0 = \lim\limits_{R \to \infty} \int\limits_{\gamma_R} \frac{\log^3 (1-iz)}{z^2}\,dz = \int_{-\infty}^{\infty} \frac{\log^3\left((1+x^2)^{1/2} - i\arctan x\right)}{x^2}\,dx\end{align*}

Comparing the real parts on both sides,

\begin{align*} \int_{-\infty}^{\infty} \frac{\log(1+x^2)(\arctan x)^2}{x^2}\,dx &= \frac{1}{12}\int_{-\infty}^{\infty} \frac{\log^3(1+x^2)}{x^2}\,dx \\&= \frac{1}{6}\int_{0}^{\infty} \frac{\log^3(1+x^2)}{x^2}\,dx\\&\underset{\text{(IBP)}}{=} \int_0^{\infty} \frac{\log^2 (1+x^2)}{1+x^2}\,dx \\&= \int_0^{\pi/2} \log^2 (\cos \theta)\,d\theta \\&= \lim\limits_{b \to \frac{1}{2}}\frac{1}{2}\frac{\partial^2}{\partial b^2} B\left(\frac{1}{2},b\right)\\&= \frac{\pi}{2}\left(4\log^2 2 + \frac{\pi^2}{3}\right)\end{align*}

Hence, $$\displaystyle \int_{0}^{\infty} \frac{\log(1+x^2)(\arctan x)^2}{x^2}\,dx = \frac{\pi^3}{12} + \pi\log^2 2$$

• (+1) I showed the evaluation using the same approach in my blog advancedintegrals.com/2017/03/… – Zaid Alyafeai Jun 8 '17 at 6:40
• nice, that is essentially what i proposed in the comments... – tired Jul 6 '17 at 21:10
• @tired Thanks! :) should have read the comments first .. my bad! – r9m Jul 6 '17 at 21:14
• @r9m you did nothing wrong...good to see this approach written down as a proper anser :) – tired Jul 6 '17 at 21:21

Another solution.

$\displaystyle J=\int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2x}{x^2}\,dx$

Perform integration by parts,

\begin{align} J&=-\left[\frac{\log\left(x^2+1 \right)\arctan^2 x}{x}\right]_0^{\infty}+\int_0^{\infty}\frac{1}{x}\left(\frac{2x\arctan^2 x}{1+x^2}+\frac{2\log\left(x^2+1 \right)\arctan x}{1+x^2}\right)\,dx\\ &=2\int_0^{\infty}\frac{\arctan^2 x}{1+x^2}\,dx+2\int_0^{\infty}\frac{\log\left(x^2+1 \right)\arctan x}{x(1+x^2)}\,dx\\ &=\frac{2}{3}\Big[\arctan^3 x\Big]_0^{\infty}+2\int_0^{\infty}\frac{\log\left(x^2+1 \right)\arctan x}{x(1+x^2)}\,dx\\ &=\frac{\pi^3}{12}+2\int_0^{\infty}\frac{\log\left(x^2+1 \right)\arctan x}{x(1+x^2)}\,dx\\ \end{align}

In the following integral, perform the change of variable $x=\tan \theta$,

\begin{align} K&=\int_0^{\infty}\frac{\log\left(x^2+1 \right)\arctan x}{x(1+x^2)}\,dx\\ &=-2\int_0^{\frac{\pi}{2}}\theta\cot\theta\ln\left(\cos\theta\right)\,d\theta\\ \end{align}

Perform integration by parts,

\begin{align} K&=-2\Big[\ln\left(\sin\theta\right)\theta\ln\left(\cos\theta\right)\Big]_0^{\frac{\pi}{2}}+2\int_0^{\frac{\pi}{2}}\ln\left(\sin\theta\right)\Big(\ln\left(\cos\theta\right)-\theta\tan \theta\Big)\,d\theta\\ &=2\int_0^{\frac{\pi}{2}}\ln\left(\sin\theta\right)\ln\left(\cos\theta\right)\,d\theta-2\int_0^{\frac{\pi}{2}}\theta\tan \theta\ln\left(\sin\theta\right)\,d\theta \end{align}

In the second integral perform the change of variable $x=\frac{\pi}{2}-\theta$,

\begin{align} K&=2\int_0^{\frac{\pi}{2}}\ln\left(\sin\theta\right)\ln\left(\cos\theta\right)\,d\theta-2\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-\theta\right)\cot \theta\ln\left(\cos\theta\right)\,d\theta\\ &=2\int_0^{\frac{\pi}{2}}\ln\left(\sin\theta\right)\ln\left(\cos\theta\right)\,d\theta-\pi\int_0^{\frac{\pi}{2}}\cot \theta\ln\left(\cos\theta\right)\,d\theta+2\int_0^{\frac{\pi}{2}}\theta\cot \theta\ln\left(\cos\theta\right)\,d\theta\\ &=2\int_0^{\frac{\pi}{2}}\ln\left(\sin\theta\right)\ln\left(\cos\theta\right)\,d\theta-\pi\int_0^{\frac{\pi}{2}}\cot \theta\ln\left(\cos\theta\right)\,d\theta-K \end{align}

Therefore,

\begin{align} K&=\int_0^{\frac{\pi}{2}}\ln\left(\sin\theta\right)\ln\left(\cos\theta\right)\,d\theta-\frac{\pi}{2}\int_0^{\frac{\pi}{2}}\cot \theta\ln\left(\cos\theta\right)\,d\theta \end{align}

Moreover,

\begin{align} \int_0^{\frac{\pi}{2}}\cot \theta\ln\left(\cos\theta\right)\,d\theta&=\int_0^{\frac{\pi}{4}}\cot \theta\ln\left(\cos\theta\right)\,d\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cot \theta\ln\left(\cos\theta\right)\,d\theta \end{align}

Perform the change of variable $x=\frac{\pi}{2}-\theta$ in the second integral,

\begin{align} \int_0^{\frac{\pi}{2}}\cot \theta\ln\left(\cos\theta\right)\,d\theta&=\int_0^{\frac{\pi}{4}}\cot \theta\ln\left(\cos\theta\right)\,d\theta+\int_0^{\frac{\pi}{4}}\tan \theta\ln\left(\sin\theta\right)\,d\theta\\ &=\left(\Big[\ln\left(\sin\theta\right)\ln\left(\cos\theta\right)\Big]_0^{\frac{\pi}{4}}+\int_0^{\frac{\pi}{4}}\tan \theta\ln\left(\sin\theta\right)\,d\theta\right)+\int_0^{\frac{\pi}{4}}\tan \theta\ln\left(\sin\theta\right)\,d\theta\\ &=\frac{1}{4}\ln^2 2+2\int_0^{\frac{\pi}{4}}\tan \theta\ln\left(\sin\theta\right)\,d\theta\\ &=\frac{1}{4}\ln^2 2+2\int_0^{\frac{\pi}{4}}\tan \theta\ln\left(\cos\theta\tan\theta\right)\,d\theta\\ &=\frac{1}{4}\ln^2 2+2\int_0^{\frac{\pi}{4}}\tan \theta\ln\left(\cos\theta\right)\,d\theta+2\int_0^{\frac{\pi}{4}}\tan \theta\ln\left(\tan\theta\right)\,d\theta\\ &=\frac{1}{4}\ln^2 2-\Big[\ln^2\left(\cos\theta\right)\Big]_0^{\frac{\pi}{4}}+2\int_0^{\frac{\pi}{4}}\tan \theta\ln\left(\tan\theta\right)\,d\theta\\ &=2\int_0^{\frac{\pi}{4}}\tan \theta\ln\left(\tan\theta\right)\,d\theta \end{align}

Perform the change of variable $x=\tan \theta$,

\begin{align} \int_0^{\frac{\pi}{2}}\cot \theta\ln\left(\cos\theta\right)\,d\theta&=\int_0^1 \frac{2x\ln x}{1+x^2}\,dx\\ &=\frac{1}{2}\int_0^1 \frac{2x\ln(x^2)}{1+x^2}\,dx\\ \end{align}

Perform the change of variable $y=x^2$,

\begin{align} \int_0^{\frac{\pi}{2}}\cot \theta\ln\left(\cos\theta\right)\,d\theta&=\frac{1}{2}\int_0^1 \frac{\ln x}{1+x}\,dx\\ &=-\frac{1}{24}\pi^2 \end{align}

\begin{align} L&=\int_0^{\frac{\pi}{2}}\ln\left(\sin\theta\right)\ln\left(\cos\theta\right)\,d\theta\\ &=\frac{1}{4}\int_0^{\frac{\pi}{2}}\left(\left(\ln\left(\sin\theta\right)+\ln\left(\cos\theta\right)\right)^2-\left(\ln\left(\sin\theta\right)-\ln\left(\cos\theta\right)\right)^2\right)\,d\theta\\ &=\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\sin\theta\cos\theta\right)\,d\theta-\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta\\ &=\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{1}{2}\sin(2\theta)\right)\,d\theta-\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta \end{align}

In the first integral perform the change of variable $x=2\theta$,

\begin{align} L&=\frac{1}{8}\int_0^{\pi}\ln^2\left(\frac{1}{2}\sin\theta\right)\,d\theta-\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta\\ &=\frac{1}{8}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{1}{2}\sin\theta\right)\,d\theta+\frac{1}{8}\int_{\frac{\pi}{2}}^\pi\ln^2\left(\frac{1}{2}\sin\theta\right)\,d\theta-\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta\\ \end{align}

In the second integral perform the change of variable $x=\pi-\theta$,

\begin{align} L&=\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{1}{2}\sin\theta\right)\,d\theta-\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta\\ &=\frac{1}{8}\pi\ln^2 2+\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2(\sin \theta)\,d\theta-\frac{1}{2}\ln 2\int_0^{\frac{\pi}{2}}\ln(\sin \theta)\,d\theta-\frac{1}{4}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta\\ \end{align}

\begin{align} M&=\int_0^{\frac{\pi}{2}}\ln^2(\sin \theta)\,d\theta\\ &=\int_0^{\frac{\pi}{2}}\ln^2(\tan\theta\cos \theta)\,d\theta\\ &=\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta+\int_0^{\frac{\pi}{2}}\ln^2\left(\cos\theta\right)\,d\theta+2\int_0^{\frac{\pi}{2}}\ln\left(\tan\theta\right)\ln\left(\cos\theta\right)\,d\theta\\ &=\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta+\int_0^{\frac{\pi}{2}}\ln^2\left(\cos\theta\right)\,d\theta+2\int_0^{\frac{\pi}{2}}\ln\left(\sin \theta\right)\ln\left(\cos \theta\right)\,d\theta-2\int_0^{\frac{\pi}{2}}\ln^2\left(\cos \theta\right)\,d\theta\\ &=\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta+2L-\int_0^{\frac{\pi}{2}}\ln^2\left(\cos\theta\right)\,d\theta \end{align}

In the second integral perform the change of variable $x=\frac{\pi}{2}-\theta$,

\begin{align} M&=\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta+2L-M \end{align}

Therefore,

\begin{align} M&=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta+L \end{align}

Therefore,

\begin{align} L&=\frac{1}{8}\pi\ln^2 2-\frac{1}{8}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta-\frac{1}{2}\ln 2\int_0^{\frac{\pi}{2}}\ln(\sin \theta)\,d\theta+\frac{1}{4}L \end{align}

Thus,

\begin{align} L&=\frac{1}{6}\pi\ln^2 2-\frac{1}{6}\int_0^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta-\frac{2}{3}\ln 2\int_0^{\frac{\pi}{2}}\ln(\sin \theta)\,d\theta\\ &=\frac{1}{6}\pi\ln^2 2-\frac{1}{6}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan\theta\right)\,d\theta-\frac{1}{6}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln^2\left(\tan\theta\right)\,d\theta-\frac{2}{3}\ln 2\int_0^{\frac{\pi}{2}}\ln(\sin \theta)\,d\theta\\ \end{align}

In the second integral perform the change of variable $x=\frac{\pi}{2}-\theta$,

\begin{align} L&=\frac{1}{6}\pi\ln^2 2-\frac{1}{3}\int_0^{\frac{\pi}{4}}\ln^2\left(\tan\theta\right)\,d\theta-\frac{2}{3}\ln 2\int_0^{\frac{\pi}{2}}\ln(\sin \theta)\,d\theta\\ \end{align}

In the second integral perform the change of variable $x=\tan\theta$,

\begin{align} L&=\frac{1}{6}\pi\ln^2 2-\frac{1}{3}\int_0^1\frac{\ln^2 x}{1+x^2}\,dx-\frac{2}{3}\ln 2\int_0^{\frac{\pi}{2}}\ln(\sin \theta)\,d\theta\\ \end{align}

It is well known that,

$\displaystyle \int_0^1\frac{\ln^2 x}{1+x^2}\,dx=\dfrac{1}{16}\pi^3$

and,

$\displaystyle \int_0^{\frac{\pi}{2}}\ln(\sin \theta)\,d\theta=-\frac{1}{2}\pi\ln 2$

Therefore,

\begin{align} L&=\frac{1}{2}\pi\ln^2 2-\frac{1}{48}\pi^3 \end{align}

Therefore,

\begin{align} K&=\frac{1}{2}\pi\ln^2 2 \end{align}

Therefore,

$\boxed{\displaystyle J=\frac{1}{12}\pi^3+\pi\ln^2 2}$

• Nice approach , thanks ! – Zaid Alyafeai Aug 3 '17 at 23:11

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Given the 'cyclic symmetry' of your integral $\ds{\color{#f00}{\mc{J}}}$, it's equivalent to \begin{align} \color{#f00}{\mc{J}} & \equiv 3\int_{0}^{\infty}\!\!\int_{0}^{\infty}\!\!\int_{0}^{\infty}\!\! \mrm{sinc}\pars{x}\mrm{sinc}\pars{y}\mrm{sinc}\pars{z}\sin\pars{x}\cos\pars{y} \cos\pars{z}\ \times \\[3mm] & \phantom{\equiv 3} \underbrace{\bracks{\int_{0}^{\infty}\expo{-\pars{x + y + z}t}\,\dd t}} _{\ds{1 \over x + y + z}}\dd x\,\dd y\,\dd z = 3\int_{0}^{\infty}\mrm{f}\pars{t}\mrm{g}^{2}\pars{t}\,\dd t \\[5mm] & \mbox{where}\qquad \left\{\begin{array}{rcl} \ds{\mrm{f}\pars{t}} & \ds{=} & \ds{\Im\mc{I}\pars{t}} \\[2mm] \ds{\mrm{g}\pars{t}} & \ds{=} & \ds{\Re\mc{I}\pars{t}} \\[2mm] \ds{\mc{I}\pars{t}} & \ds{\equiv} & \ds{\int_{0}^{\infty}\mrm{sinc}\pars{x}\expo{-\pars{t - \ic}x}\dd x} \end{array}\right.\label{1}\tag{1} \end{align}

Then \begin{align} \left.\mc{I}\pars{t}\right\vert_{\large\ \color{#f00}{t\ >\ 0}} & \equiv \int_{0}^{\infty}\mrm{sinc}\pars{x}\expo{-\pars{t - \ic}x}\dd x = \int_{0}^{\infty} \overbrace{\pars{{1 \over 2}\int_{-1}^{1}\expo{\ic kx}\,\dd k}} ^{\ds{\mrm{sinc}\pars{x}}}\ \expo{-\pars{t - \ic}x}\dd x \\[5mm] & = {1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}\expo{-\pars{t - \ic - \ic k}x} \dd x\,\dd k = {1 \over 2}\int_{-1}^{1}{\dd k \over t - \ic - \ic k} = {1 \over 2}\int_{-1}^{1}{t + \pars{k + 1}\ic \over \pars{k + 1}^{2} + t^{2}}\,\dd k \\[5mm] & = {1 \over 2}\int_{0}^{2}{t + k\ic \over k^{2} + t^{2}}\,\dd k = {1 \over 2}\int_{0}^{2/t}{1 + k\ic \over k^{2} + 1}\,\dd k = {1 \over 2}\arctan\pars{2 \over t} + {1 \over 4}\ln\pars{{4 \over t^{2}} + 1 }\ic \end{align}
$\ds{\color{#f00}{\mc{J}}}$ becomes ( see \eqref{1} ): \begin{align} \color{#f00}{\mc{J}} & = 3\int_{0}^{\infty}\bracks{{1 \over 4}\ln\pars{{4 \over t^{2}} + 1}} \bracks{{1 \over 2}\arctan\pars{2 \over t}}^{2}\dd t \\[5mm] & \stackrel{2/t\ \mapsto\ t}{=}\,\,\, {3 \over 16}\int_{\infty}^{0}\ln\pars{t^{2} + 1} \arctan^{2}\pars{t}\pars{-\,{2\,\dd t \over t^{2}}} = {3 \over 8}\ \underbrace{\int_{0}^{\infty} {\ln\pars{t^{2} + 1}\arctan^{2}\pars{t} \over t^{2}}\,\dd t} _{\ds{{\Large\color{#f00}{\S}}: {\pi^{3} \over 12} + \pi\ln^{2}\pars{2}}} \end{align}

$\ds{{\Large\color{#f00}{\S}}}$: The integral was already evaluated in the $\texttt{@Zaid Alyafeai}$ fine answer.

Finally, the answer to the proposed OP integral is given by $$\bbox[15px,#ffe,border:1px dotted navy]{\ds{{\color{#f00}{\mc{J}} = {\pi^{3} \over 32} + {3 \over 8}\,\pi\ln^{2}\pars{2}}}} \approx 1.5350$$

• @ZaidAlyafeai Thanks. $\left)\substack{\bullet\quad\bullet\\ /\setminus\\ \smile}\right\{$ – Felix Marin Aug 3 '17 at 23:29