Concerning this integral $\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{x^s\over 2}\right){\mathrm dx\over 1-x}$

Question

Motivation from this question $(1)$

$$\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{1\over 2}\right){x^s\over 1-x}\mathrm dx=F(s)\tag1$$

setting $s=0$ then $F(0)=-{1\over 2}+{1\over 2}\ln(2\pi)-{1\over 2}\gamma$

by a slight variation of $(1)$, we have

$$\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{x^s\over 2}\right){\mathrm dx\over 1-x}={1\over 2}H_s-{1\over 2}+{1\over 2}\ln(2\pi)-{1\over 2}\gamma\tag2$$

$H_s$ is the harmonic number, $H_0=0$

How do we go about to prove $(2)?$

Answer 1

Use \begin{eqnarray*} H_v= \int_0^1 \frac{1-x^v}{1-x} dx. \end{eqnarray*} We have \begin{eqnarray*} \int_0^1 \left( \frac{1}{\ln x}+ \frac{1}{1-x}-\frac{x^s}{2}\right)\frac{ dx}{1-x} &=& \int_0^1 \left( \frac{1}{\ln x}+ \frac{1}{1-x}-\frac{1}{2} +\frac{1-x^s}{2}\right)\frac{ dx}{1-x} \\ &=& F(0)+ \int_0^1 \frac{1-x^s}{2(1-x)}dx \\ &=& -\frac{1}{2}+ \frac{1}{2} \ln( 2 \pi) -\frac{1}{2} \gamma + \frac{1}{2} H_s. \end{eqnarray*}