TITLE: Need help with $\int_0^\infty\frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx$
QUESTION [17 upvotes]: I need you help with this integral:
$$\int_0^\infty\frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx.$$
Mathematica says it does not converge, which is apparently false.

REPLY [4 votes]: We can use the parametric integral to evaluate the integral. Let
$$ I(\alpha)=\int_0^\infty\frac{\ln(1+\alpha x)}{(1+x^2)(1+x^3)}dx. $$
Then it is easy to see
\begin{eqnarray*}
I'(\alpha)&=&\int_0^\infty\frac{x}{(1+\alpha x)(1+x^2)(1+x^3)}dx\\
&=&\pi\frac{(9+8\sqrt{3})\alpha^2+(8\sqrt{3}-9)}{(\alpha^2+1)(\alpha^2+\alpha+1)}-\frac{\alpha^3\ln\alpha}{(1+\alpha^2)(1-\alpha^3)}
\end{eqnarray*}
and hence
\begin{eqnarray*}
I(1)&=&\int_0^1I'(\alpha)d\alpha\\
&=&\pi\int_0^1\frac{(9+8\sqrt{3})\alpha^2+(8\sqrt{3}-9)}{(\alpha^2+1)(\alpha^2+\alpha+1)}d\alpha+\int_0^1\frac{\alpha^3\ln\alpha}{(1+\alpha^2)(1-\alpha^3)}d\alpha\\
&=&I_1+I_2.
\end{eqnarray*}
It is not hard to get 
$$ I_1=\frac{\pi(5\pi+54\ln 2)}{432}. $$
Note
\begin{eqnarray*}
\frac{\alpha^3}{(1+\alpha^2)(1-\alpha^3)}
&=&-\frac{1}{2}\frac{1+\alpha}{1+\alpha^2}+\frac12\frac{1+\alpha-\alpha^2}{1-\alpha^3}
\end{eqnarray*}
and hence
\begin{eqnarray*}
I_2&=&\int_0^1\frac{\alpha^3\ln\alpha}{(1-\alpha)(\alpha^2+1)(\alpha^2+\alpha+1)}d\alpha\\
&=&-\frac{1}{2}\int_0^1\frac{(1+\alpha)\ln\alpha}{1+\alpha^2}d\alpha+\frac{1}{2}\int_0^1\frac{(1+\alpha-\alpha^2)\ln\alpha}{1-\alpha^3}d\alpha
\end{eqnarray*}
But
\begin{eqnarray*}
\int_0^1\frac{(1+\alpha)\ln\alpha}{1+\alpha^2}d\alpha&=&\int_0^1\frac{\ln\alpha}{1+\alpha^2}d\alpha+\int_0^1\frac{\alpha\ln\alpha}{1+\alpha^2}d\alpha\\
&=&\sum_{n=0}^\infty\int_0^1(-\alpha^2)^n\ln\alpha d\alpha+\sum_{n=0}^\infty\int_0^1\alpha(-\alpha^2)^n\ln\alpha d\alpha\\
&=&-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}-\sum_{n=0}^\infty\frac{(-1)^n}{4(n+1)^2}=-G-\frac{\pi^2}{48}\\
\int_0^1\frac{(1+\alpha-\alpha^2)\ln\alpha}{1-\alpha^3}d\alpha
&=&\int_0^1\frac{\ln\alpha}{1-\alpha^3}d\alpha+\int_0^1\frac{\alpha\ln\alpha}{1-\alpha^3}d\alpha-\int_0^1\frac{\alpha^2\ln\alpha}{1-\alpha^3}d\alpha\\
&=&\sum_{n=0}^\infty\int_0^\infty\alpha^{3n}\ln\alpha d\alpha+\sum_{n=0}^\infty\int_0^\infty\alpha^{3n+1}\ln\alpha d\alpha-\sum_{n=0}^\infty\int_0^\infty\alpha^{3n+2}\ln\alpha d\alpha\\
&=&-\sum_{n=0}^\infty\frac{1}{(3n+1)^2}-\sum_{n=0}^\infty\frac{1}{(3n+2)^2}+\frac{1}{9}\sum_{n=0}^\infty\frac{1}{(n+1)^2}\\
&=&-\sum_{n=0}^\infty\frac{1}{(3n+1)^2}-\sum_{n=0}^\infty\frac{1}{(3n+2)^2}-\frac{1}{9}\sum_{n=0}^\infty\frac{1}{(n+1)^2}+\frac{2}{9}\sum_{n=0}^\infty\frac{1}{(n+1)^2}\\
&=&-\frac{7\pi^2}{54}.
\end{eqnarray*}
Putting everything together, we have
$$ I(1)=\frac{G}{2}-\frac{37\pi}{864}+\frac{1}{8}\pi\ln 2. $$