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

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.