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 $|\sigma \omega^{-1/2}| &lt; 1$ and $|\sigma \omega^{-1/2}| &gt; 1.$  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 $0 &lt; s &lt; 1$,

$$ I = I(s,\alpha,\beta) := \int_{0}^{\infty} \frac{\log(1+\alpha x)\log(1+\beta x)}{x^{2+s}} \, dx, $$

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 $\beta &gt; 0$). 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 $\alpha, \beta &gt; 0$. Taking real parts of the identity above and taking $s \to 1^{-}$, it follows that

\begin{align*}
\tilde{I}(\alpha, \beta) 
&amp;:= \int_{0}^{\infty} \frac{\log(1+\alpha x)\log(1+\beta^{2}x^{2})}{x^{3}} \, dx \\
&amp;= \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.
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 :

$|\sigma \omega^{-1/2}| &lt; 1$ or $|\sigma \omega^{-1/2}| &gt; 1$