Calculating an improper integral involving $\log$ Could somebody help me to solve the integral below?
$$\int\limits_0^1\frac{\ln x}{x^{a}(1-x)^{b}}\,\mathrm{d}x,$$
where $a,b \in (0,1)$.
It may be helpful to know the definition of the Beta function:
$$\textbf{B}(r,s):=\int\limits_0^1x^{r-1}(1-x)^{s-1}\mathrm{d}x, \quad r,s&gt;0.$$

Thanks!
It is straightforward to see by differentiation under the integral that we have

\begin{align}I(a,b)&amp;=\int_0^1\frac{\ln(x)}{x^a(1-x)^b}~\mathrm dx\\&amp;=\frac\partial{\partial a}\int_0^1\frac1{x^a(1-x)^b}~\mathrm dx\\&amp;=\frac\partial{\partial a}\mathbf B(1-a,1-b)\end{align}

which can further be written out using the Gamma function and it's derivatives given in terms of the digamma function.
This sounds like homework, and you did not include what you have done.

Anyway, another way to write this in terms of the beta function and its derivative is to integrate by parts.