Let $\alpha$ be a small positive real number.
How do I obtain
$$ I = \int_0^1 \frac{\log(x)}{x+\alpha}\; dx = -\frac{1}{2}(\log\alpha)^2 – \frac{\pi^2}{6} – \operatorname{Li}_2(-\alpha)$$? Maxima told me the result, but I do not understand how to get it.
The appearance of $\pi^2/6$ suggests the necessity of complex analysis. I tried to relate it to some contour integral through the change of variables $x=e^{-t}$:
$$ I = -\int_{0}^{\infty} \frac{t e^{-t}}{e^{-t}+\alpha}\;dt, $$
but I could not find the next step.
Best Answer
$$\int_0^1\frac{\ln(x)}{a+x}dx=\int_0^{1/a}\frac{\ln(ay)}{1+y}dy$$
$$=\underbrace{\ln(1+y)\ln(ay)|_0^{1/a}}_{0}-\int_0^{1/a}\frac{\ln(1+y)}{y}dx=\text{Li}_2(-y)|_0^{1/a}=\text{Li}_2(-1/a)$$
and the result follows on using the dilogarithm inversion formula:
$$\text{Li}_2(-1/z)=-\frac{\pi^2}{6}-\frac{1}{2}\ln^2(z)-\text{Li}_2(-z).$$
To prove the last identity, differentiate $\text{Li}_2(-1/z)$ then integrate back.
In geeneral we have
$$\int_0^1\frac{\ln^p(x)}{a+x}dx=(-1)^{p+1}\, p!\,\text{Li}_{p+1}(-1/a)$$
which follows from the integral represenation of the polylogarithm function
$$\text{Li}_{p+1}(z)=\frac{(-1)^{p}}{p!}\int_0^1\frac{z\ln^{p}(t)}{1-zt}dt$$
upon replacing $z$ by $-1/a$.