Help solving $\int_0^\infty\frac{\ln{(x)}\ln{(1+ix)}}{x^2+1}dx$ I want to integrate the following integral using a variety of methods. It came up while I was working out a solution for $\int_0^\pi\frac{x(\pi-x)}{\sin{(x)}}$.
$$\int_0^\infty\frac{\ln{(x)}\ln{(1+ix)}}{1+x^2}dx$$
Using the Taylor series, I have tried to expand $\ln{(1+ix)}$. I tried to use $\int_0^\infty\frac{x^n\ln{(1+ix)}}{x^2+1}dx$ to solve the above problem, but I couldn't solve this one either. I tried integration by parts by setting $u=\ln{(x)}\ln{(1+ix)}$ and $dv=\frac{1}{x^2+1}$ but it seemed to complicate it further. I'm not acquainted enough with complex analysis to solve it that way, but I would appreciate a complex analysis answer regardless. I'm not sure what else I can do.
Thank you in advance
 A: Note
\begin{align}
&\int_0^\infty\frac{\ln{x}\ln{(1+ix)}}{1+x^2}dx\\
=&\ \frac12 \int_0^\infty\frac{\ln{x}\ln{(1+x^2)}}{1+x^2}dx
+ i\int_0^\infty\frac{\ln{x}\tan^{-1}x}{1+x^2}dx
\end{align}
where
\begin{align}
\int_0^\infty\frac{\ln{x}\ln{(1+x^2)}}{1+x^2}{dx}& \overset{x\to \frac1x}=\int_0^\infty\frac{\ln^2{x}}{1+x^2}dx
=\frac{\pi^3}8\\
\\
\int_0^\infty \frac{\ln x\tan^{-1}x}{1+x^2}dx
=& \int_0^\infty \int_0^1 \frac{x\ln x}{(1+x^2)(1+y^2x^2)} \overset{x\to \frac1{xy}}{dx}dy\\
 = & \ 
 \frac1{2}\int_0^1\int_0^\infty \frac{-x\ln y}{(1+x^2)(1+{y^2}x^2)} {dx}\ dy\\
=& \ \frac12\int_0^1\frac{\ln^2 y}{1-y^2}dy
=\frac78\zeta(3)
\end{align}
A: Using the fact that $$
\begin{aligned}
\ln (1+i x) &=\ln \left(\sqrt{1+x^2} \cdot e^{i \tan ^{-1} x}\right) \\
&=\frac{1}{2} \ln \left(1+x^2\right)+i \tan ^{-1}x,
\end{aligned}
$$
we splits the integral into two as
$$
I=\frac{1}{2} \underbrace{\int_0^{\infty} \frac{\ln x \ln \left(1+x^2\right)}{1+x^2}}_J d x+i \underbrace{\int_0^{\infty} \frac{\ln x \tan ^{-1} x}{1+x^2} d x}_K
$$
Letting $x\mapsto\tan x$ yields
$$
\begin{aligned}
J &=-2 \int_0^{\frac{\pi}{2}} \ln (\tan x) \ln (\cos x) d x \\
&=2 \int_0^{\frac{\pi}{2}} \ln ^2(\cos x)dx-2 \int_0^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x \\
&=2 \cdot \frac{1}{24}\left(\pi^3+3 \pi \ln ^2 4\right)-2\left(-\frac{\pi^3}{48}+\frac{\pi}{2} \ln ^2 2\right)\cdots (*) \\
&=\frac{\pi^3}{8}
\end{aligned}
$$
where $(*)$ comes from my post .
$$
\begin{aligned}
K &=\int_0^{\infty} \frac{\ln x \tan ^{-1} x}{1+x^2} d x=\int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\frac{7}{8} \zeta(3),
\end{aligned}
$$
where the last result comes from my post.
We can now conclude that
$$
\boxed{I=\frac{1}{16}\left(\pi^3+14 i\zeta(3)\right)}
$$
