Integral $ \int_0^\infty \frac{\ln(1+\sigma x)\ln(1+\omega x^2)}{x^3}dx$ Hello there I am trying to calculate
$$
\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+\omega x^2)}{x^3}dx
$$
NOT using mathematica, matlab, etc.  We are given that $\sigma, \omega$ are complex.  Note, the integral should have different values for $|\sigma \omega^{-1/2}| < 1$ and $|\sigma \omega^{-1/2}| > 1.$  I am stuck now and not sure how to approach it.
Note this integral is useful since in the limit $\sigma \to \sqrt{\omega}$ and using $Li_2(-1)=-\pi^2/12$ we obtain
$$
\int_0^\infty \frac{\ln(1+x)\ln(1+x^2)}{x^3}dx=\frac{\pi}{2}.
$$
We also know that
$$
\ln(1+x)=-\sum_{n=1}^\infty \frac{(-1)^nx^n}{n}, \ |x|\leq 1.
$$
Thanks
 A: One may adopt the approach as in Pranav Arora's comment. But this approach involves a double integral whose calculation seems painful. So here is an indirect approach that makes calculation slightly easier (at least to me):
Let us consider the following integral: for $\alpha, \beta \in \Bbb{C}\setminus(-\infty, 0]$ and $0 < s < 1$,
$$ I = I(s,\alpha,\beta) := \int_{0}^{\infty} \frac{\log(1+\alpha x)\log(1+\beta x)}{x^{2+s}} \, dx, $$
Differentiating w.r.t. $\alpha$ and $\beta$, we have
$$ \frac{\partial^{2}I}{\partial\alpha\partial\beta} = \int_{0}^{\infty} \frac{dx}{x^{s}(1+\alpha x)(1+\beta x)}. $$
Using standard complex analysis technique (you man use keyhole contour), it follows that
$$ \frac{\partial^{2}I}{\partial\alpha\partial\beta} = \frac{\pi}{\sin \pi s} \frac{\beta^{s} - \alpha^{s}}{\beta - \alpha}
\quad \Longrightarrow \quad
I = \frac{\pi}{\sin \pi s} \int_{0}^{\alpha}\int_{0}^{\beta} \frac{x^{s} - y^{s}}{x - y} \, dxdy. \tag{1} $$
Replace $\beta$ by $i\beta$ (with $\beta > 0$). Then (1) yields
$$ 2I(s, \alpha, i\beta) = \frac{2\pi}{\sin \pi s} \int_{0}^{\alpha}\int_{0}^{\beta} \frac{i^{s}x^{s} - y^{s}}{x + iy} \, dxdy. $$
Now assume that $\alpha, \beta > 0$. Taking real parts of the identity above and taking $s \to 1^{-}$, it follows that
\begin{align*}
\tilde{I}(\alpha, \beta) 
&:= \int_{0}^{\infty} \frac{\log(1+\alpha x)\log(1+\beta^{2}x^{2})}{x^{3}} \, dx \\
&= \int_{0}^{\alpha}\int_{0}^{\beta} \frac{2xy \log(y/x) + \pi x^{2}}{x^{2}+y^{2}} \, dxdy. \tag{2}
\end{align*}
In particular, when $\beta = \alpha$, by symmetry we retrieve the following formula
$$ \tilde{I}(\alpha, \alpha)
= \pi \int_{0}^{\alpha}\int_{0}^{\alpha} \frac{x^{2}}{x^{2}+y^{2}} \, dxdy
= \frac{\pi}{2} \int_{0}^{\alpha}\int_{0}^{\alpha} dxdy
= \frac{\pi}{2}\alpha^{2}. $$
which also follows from the formula in OP's posting. In general, using polar coordinates shows that we have
$$ \tilde{I}(\alpha, \beta) = \beta^{2}J(\alpha/\beta) - \alpha^{2}J(\beta/\alpha) + \frac{\pi \alpha \beta}{2} + \frac{\pi^{2}\beta^{2}}{4} - \frac{\pi(\alpha^{2}+\beta^{2})}{2}\arctan(\beta/\alpha), \tag{3} $$
where $J$ is defined by
$$ J(x) =  \int_{0}^{x} \frac{t \log t}{1+t^{2}} \, dt. $$
This function can be written in terms of elementary functions and dilogarithm.
Remark. Though we have derived this formula for positive $\alpha, \beta$, by the principle of analytic continuation (3) continues to hold on the region containing $(0, \infty)^{2}$ where both sides of (3) are holomorphic.
A: I really do not know how much this could help you but, using a CAS, I obtained for the integral the following result
$$\frac{1}{24} \left(6 \left(\sigma ^2+\omega \right) \text{Li}_2\left(-\frac{\sigma
   ^2}{\omega }\right)+6 \left(\sigma ^2+\omega \right) (2 \log (\sigma )-\log
   (\omega )) \log \left(\frac{\sigma ^2+\omega }{\omega }\right)+12 \pi 
   \left(\sigma ^2+\omega \right) \tan ^{-1}\left(\frac{\sigma }{\sqrt{\omega
   }}\right)+\sigma  \left(\pi  \left(12 \sqrt{\omega }-5 \pi  \sigma \right)-3
   \sigma  (\log (\omega )-2 \log (\sigma ))^2\right)\right)$$ I suppose that there are restrictions but I have not been able to find the ensemble of them.
A: why not use the $\Gamma$ function ? it seems that this question is the special case!
$\int_0^\infty \frac{\ln(1+ x)\ln(1+x^2)}{x^3}dx$$=\frac{1}{2}(\int_0^\infty \frac{e^{-x}}{\sqrt{x}}dx)^{2}$
$
\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+\omega x^2)}{x^3}dx
$$=$
$\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+(\sigma x)^2)}{x^3}dx
$
$\Longrightarrow$$
\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+(\sigma x)^2)}{(\sigma{x})^3}dx$$=$$\frac{\sigma^{2}}{2}(\int_0^\infty \frac{e^{-\sigma x}}{\sqrt{\sigma x}}dx)^{2}$
$\Longrightarrow$
$
\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+(\sigma x)^2)}{{x}^3}dx$$=$$\frac{\sigma^{5}}{2}(\int_0^\infty \frac{e^{-\sigma x}}{\sqrt{\sigma x}}dx)^{2}$
$=\frac{1}{2}(\int_0^\infty \frac{e^{-\sigma x}}{\sqrt{x}}dx)^{2}$$\cdot{\sigma^{4}}$
therefore, it seems that you just need to multiply $\sigma^{3}$
and the positive and negative of your equation depends on :
$|\sigma \omega^{-1/2}| < 1$ or $|\sigma \omega^{-1/2}| > 1$
