Integral $\int_0^1\frac{\ln x}{x^2+1}\cdot\ln\left(\frac{3\,x^2+1}{x^2+3}\right)dx$

Question

I need to evaluate the following integral: $$\int_0^1\frac{\ln x}{x^2+1}\cdot\ln\left(\frac{3\,x^2+1}{x^2+3}\right)dx.$$ Could you suggest how to find a closed form for it? I am not sure if there is one, but the integrand seems simple enough, so I hope it might exist.

Answer 1

My calculation shows that

$$ \int_{0}^{1} \frac{\log x}{x^{2} + 1} \log \left( \frac{t^{2}x^{2} + 1}{x^{2} + t^{2}} \right) \, dx = - \pi \chi_{2} \left( \frac{1 - t}{1 + t} \right), \tag{1} $$

where $\chi_{2}$ stands for the Legendre chi function. I will post a detailed solution later, but I should note that the idea is very simple: denote this integral as $I(t)$ and differentiate it to obtain

$$ I'(t) = -\frac{\pi \log t}{1 - t^{2}}. $$

Restricting $|t| < 1$ temporarily, this gives

$$ I(t) = I(0) + \int_{0}^{t} I'(s) \, ds = -\pi \left\{ \frac{\pi^{2}}{8} + \log t \operatorname{artanh} t - \chi_{2}(t) \right\}. $$

Then $\text{(1)}$ restricted to $|t| < 1$ follows from the following identity:

$$ \chi_{2}\left( \frac{1-t}{1+t} \right) + \chi_{2}(t) = \frac{\pi^{2}}{8} + \log t \operatorname{artanh} t. $$

Then the equality for the general $t$ follows by analytic continuation. Here is a Mathematica code for testing this:

LegendreChi[n_, z_] := z/2^n LerchPhi[z^2, n, 1/2];

t = Sqrt[3];
NIntegrate[(Log[x] Log[(t^2 x^2 + 1)/(x^2 + t^2)])/(1 + x^2), {x, 0, 1},
    WorkingPrecision -> 100]
N[-Pi LegendreChi[2, (1 - t)/(1 + t)], 100]
Clear[t];