The following result is well known and is interpreted as a Cauchy Principal Value
$$ P.V. \int_{0}^{\infty}\frac{x^{a-1}}{1-x}dx=\pi \cot \left(a \pi\right) \qquad \text{for}\,\,a \in \left(0,1\right) $$
My question is, once it's established , may I bring inside the integral the differential operator to get the result without having to justify it (in the sense of the Principal Value)?
$$ \frac{d}{da}\left( P.V. \int_{0}^{\infty}\frac{x^{a-1}}{1-x}dx=\pi \cot \left(a \pi\right)\right)$$
$$ P.V. \int_{0}^{\infty}\frac{x^{a-1} \ln x}{1-x}dx=-\pi^2 \csc^2 \left(a \pi\right)$$
You may avoid the subtlety of the Principal Value by considering the corresponding converging integral
$$ I(a)=\frac12 \int_{0}^{\infty}\frac{x^{a-1}-x^{-a}}{1-x}dx=\pi \cot \left(a \pi\right) $$ and evaluate the following integral by performing normal differentiation under the integral sign
$$ \int_{0}^{\infty}\frac{x^{a-1} \ln x}{1-x}dx=I’(a) = -\pi^2 \csc^2 \left(a \pi\right) $$