How to evaluate this integral $\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$ How to evaluate this integral
$$\mathcal{I}=\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$$
Mathematica gave me the answer below
$$\mathcal{I}=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left ( 2+\sqrt{3} \right )-\frac{\mathbf{G}}{3}$$
where $\mathbf{G}$ is Catalan's constant.
 A: It is possible to calculate the integral via the usual tool of differentiating with respect to a parameter. I don't claim the calculations to be especially nice, but it is nice to have as a comparison to the residue approach which is shorter, and nicer (but needs someone with a very good feeling about what contour to integrate along). We will use the fact that
$$
\int_0^1\frac{\log x}{1+x^2}\,dx=-\mathrm G.
$$
where $\mathrm G$ denotes Catalan's constant (it must show up somehow).
I will give some details below, but I cannot motivate myself to write everything explicitly.
First, let
$$
f(s)=\int_0^{+\infty}\frac{\log(s+x^3)}{1+x^2}\,dx
$$
Note that (just split the integral into $\int_0^1+\int_1^{+\infty}$ and do $y=1/x$ in the latter)
$$
f(0)=\int_0^{+\infty}\frac{3\log x}{1+x^2}\,dx=0.
$$
The integral we want to calculate becomes
$$
f(1)=f(0)+\int_0^1 f'(s)\,ds=\int_0^1 f'(s)\,ds.
$$
We calculate $f'(s)$ below. Differentiating, making a partial fraction decomposition, and calculating elementary but horrible primitives, we find that
$$
\begin{aligned}
f'(s)&=\int_0^{+\infty}\frac{1}{(s+x^3)(1+x^2)}\,dx\\
&=\frac{1}{1+s^2}\int_0^{+\infty}\frac{s+x}{1+x^2}+\frac{1-sx-x^2}{s+x^3}\,dx\\
&=\cdots\\
&=\frac{1}{18(1+s^2)}\Bigl(\frac{4\sqrt{3}\pi}{s^{2/3}}-4\sqrt{3}\pi s^{2/3}+9\pi s+6\log s\Bigr)
\end{aligned}
$$
Next, we calculate more elementary, but horrible, primitives, (let $u=s^{1/3}$)
$$
\int \frac{1}{18(1+s^2)}\Bigl(\frac{4\sqrt{3}\pi}{s^{2/3}}-4\sqrt{3}\pi s^{2/3}\Bigr)\,ds=\frac{\pi}{3}\log\Bigl(\frac{1+\sqrt{3}s^{1/3}+s^{2/3}}{1-\sqrt{3}s^{1/3}+s^{2/3}}\Bigr).
$$
Hence,
$$
\begin{aligned}
f(1)&=\int_0^1 f'(s)\,ds\\
&=\biggl[\frac{\pi}{3}\log\Bigl(\frac{1+\sqrt{3}s^{1/3}+s^{2/3}}{1-\sqrt{3}s^{1/3}+s^{2/3}}\Bigr)+\frac{\pi}{4}\log(1+s^2)\biggr]_0^1+\frac{1}{3}\int_0^1\frac{\log s}{1+s^2}\,ds\\
&=\frac{\pi}{4}\log 2+\frac{2\pi}{3}\log(2+\sqrt{3})-\frac{\mathrm G}{3}.
\end{aligned}
$$
A: Lemma 1::$$\int_{0}^{\infty}\dfrac{\ln{(x^2-x+1)}}{x^2+1}=\dfrac{2\pi}{3}\ln{(2+\sqrt{3})}-\dfrac{4}{3}G$$
Use this well known
$$\int_{0}^{+\infty}\dfrac{\ln{(x^2+2\sin{a}\cdot x+1)}}{1+x^2}dx=\pi\ln{2\cos{\dfrac{a}{2}}}+a\ln{|\tan{\dfrac{a}{2}}|}+2\sum_{k=0}^{+\infty}\dfrac{\sin{(2k+1)a}}{(2k+1)^2}$$
this indentity proof is very easy consider $\ln{(x^2+2\sin{a}\cdot x+1)}$ Fourier expansions(possion fourier).
then you can take
$a=-\dfrac{\pi}{6}$
then we have
$$\pi\ln{2\cos{\dfrac{\pi}{12}}}=\dfrac{\pi}{2}\ln{(2+\sqrt{3})}$$
$$-\dfrac{\pi}{6}\ln{\tan{\dfrac{\pi}{12}}}=\dfrac{\pi}{6}\ln{(2+\sqrt{3})}$$
and 
$$2\sum_{k=0}^{3N}\dfrac{\sin{(2k+1)-\pi/6}}{(2k+1)^2}=-\sum_{k=0}^{3N}\dfrac{(-1)^k}{(2k+1)^2}-3\sum_{k=0}^{N-1}\dfrac{(-1)^k}{(6k+3)^2}\to -G-\dfrac{G}{3}=-\dfrac{4}{3}G$$
so
$$\int_{0}^{\infty}\dfrac{\ln{(x^2-x+1)}}{x^2+1}=\dfrac{2\pi}{3}\ln{(2+\sqrt{3})}-\dfrac{4}{3}G$$
By done!
Lemma 2:$$\int_{0}^{+\infty}\dfrac{\ln{(1+x)}}{1+x^2}dx=\dfrac{\pi}{4}\ln{2}+G$$
\begin{align*} \int_{0}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx
&= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{1}^{\infty} \frac{\log (x + 1)}{x^2 + 1} \, dx \\
&= \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx + \int_{0}^{1} \frac{\log (x^{-1} + 1)}{x^2 + 1} \, dx \quad (x \mapsto x^{-1}) \\
&= 2 \int_{0}^{1} \frac{\log (x + 1)}{x^2 + 1} \, dx - \int_{0}^{1} \frac{\log x}{x^2 + 1} \, dx\\
&=\dfrac{\pi}{4}\ln{2}+G
\end{align*}
so 
$$\int_{0}^{+\infty}\dfrac{\ln{(1+x^3)}}{1+x^2}dx=\int_{0}^{+\infty}\dfrac{\ln{(1+x)}}{1+x^2}dx+\int_{0}^{+\infty}\dfrac{\ln{(x^2-x+1)}}{1+x^2}dx=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left ( 2+\sqrt{3} \right )-\frac{\mathbf{G}}{3}$$
A: We can attack this integral
$$I = \int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2}$$
by considering the complex contour integral
$$\oint_C dz \frac{\log{(1+z^3)} \log{z}}{1+z^2}$$
where $C$ is the following contour

This is a keyhole contour about the positive real axis, but with additional keyholes about the branch points at $z=e^{i \pi/3}$, $z=-1$, and $z=e^{i 5 \pi/3}$.  There are simple poles at $z=\pm i$.
I will outline the procedure for evaluation.  The integral about the circular arcs, large and small, go to zero as the radii go to $\infty$ and $0$, respectively.  Each of the branch points introduces a jump of $i 2 \pi$ due to the logarithm in the integrand.  By the residue theorem, we have
$$-i 2 \pi \int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2} - i 2 \pi \int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{1+t^2} \\ - i 2 \pi \int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} - i 2 \pi \int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{1+t^2} = \\ i 2 \pi \sum_{\pm} \left[\frac{\log{(1+z^3)} \log{z}}{2 z} \right]_{z=\pm i} $$
Without going into too much detail, I will illustrate how the integrals are done by evaluating one of them.  Consider
$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} = -\int_1^{\infty} dy \frac{\log{y}+i \pi}{1+y^2}$$
Now, 
$$\int_1^{\infty} \frac{dy}{1+y^2} = \int_{\pi/4}^{\pi/2} d\theta  = \frac{\pi}{4}$$
$$\begin{align}\int_1^{\infty} dy\frac{\log{y}}{1+y^2} &= G\end{align}$$
so that
$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} = -G - i \frac{\pi^2}{4} $$
Along similar lines,
$$\int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{1+t^2} =  \frac{2}{3} G + \frac{\pi}{6} \log{(2+\sqrt{3})}$$
$$\int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{1+t^2} = \frac{2}{3} G - \frac{5 \pi}{6} \log{(2+\sqrt{3})} + i \frac{\pi^2}{2}$$
Combining the integrals, I get
$$\frac{G}{3} - \frac{2 \pi}{3} \log{(2+\sqrt{3})} + i \frac{\pi^2}{4}$$
The sum of the residues on the RHS is relatively simple to evaluate; I get
$$\sum_{\pm} \left[\frac{\log{(1+z^3)} \log{z}}{2 z} \right]_{z=\pm i} = \frac{(1/2 \log{2} -i \pi/4)(i \pi/2)}{2 i} + \frac{(1/2 \log{2} + i \pi/4)(i 3 \pi/2)}{-2 i}\\ = -\frac{\pi}{4} \log{2}-i \frac{\pi^2}{4}$$
The integral we seek is then the negative of the sum of the combined integrals and the sum of the residues, which gives us

$$\int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2} = -\frac{G}{3} + \frac{\pi}{4} \log{2} +\frac{2 \pi}{3} \log{(2+\sqrt{3})} $$

which agrees with Mathematica.  
