I'm trying to prove that
$$ \int_{2}^{x-2}\frac{t}{\log(t)}\frac{1}{\log(x-t)}\mathrm{dt} \sim \frac{1}{2}\frac{x^2}{\log(x)^2}, $$ that is,
$$\lim_{x\to +\infty} \frac{\int_{2}^{x-2}\frac{t}{\log(t)}\frac{1}{\log(x-t)}\mathrm{dt}}{x^2/\log(x)^2} = \frac{1}{2}.$$
Numerically it seems to be right, as for $x=10^{20}$ the above expression gives $\approx 0.523$, and for $x=10^{100}$ it is $\approx 0.504$. (I calculated using Python.)
However, I didn't find a way to effectively tackle this problem. Substituting $u:=t/x$ in the integral, as $\mathrm{d}u = 1/x~ \mathrm{d}t$ one obtain
\begin{align} \int_{2}^{x-2}\frac{t}{\log(t)}\frac{1}{\log(x-t)}\mathrm{dt} &= \int_{2/x}^{1-2/x}\frac{xu}{\log(x)+\log(u)}\frac{x}{\log(x) + \log(1-u)}\mathrm{du} \\ &= \frac{x^2}{\log(x)}\int_{2/x}^{1-2/x}\frac{u}{\left(1+\frac{\log(u)}{\log(x)}\right)\left(1+\frac{\log(1-u)}{\log(x)}\right)}\mathrm{du}, \end{align}
but here I'm basically stuck. The Leibniz rule doesn't seem to work either.
Any tips would be helpful. Thanks!
There is a clever symmetry trick: $$I(x)=\int_{2}^{x-2}\frac{t}{\log(t)\log(x-t)}\,dt = \int_{2}^{x-2}\frac{x-t}{\log(t)\log(x-t)}\,dt \tag{1}$$ giving: $$ I(x) = \frac{x}{2}\int_{2}^{x-2}\frac{dt}{\log(t)\log(x-t)}= x\int_{2}^{\frac{x}{2}}\frac{dt}{\log(t)\log(x-t)}\tag{2}$$ where the last integral in concentrated around the right endpoint of the integration range, hence it can be approximated pretty well by: $$ \frac{x}{\log(x-2)}\int_{2}^{x/2}\frac{dt}{\log(t)} \tag{3}$$ where the logarithmic integral has a well-known asymptotic expansion.