Evaluation of $\int_{0}^{\infty}\frac{x^{a-1}-x^{b-1}}{1-x}dx$ 
For $0<a,b<1.$
  Evaluation of $$\int_{0}^{\infty}\frac{x^{a-1}-x^{b-1}}{1-x}dx$$

$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{x^{a-1}-x^{b-1}}{1-x}dx+\int_{1}^{\infty}\frac{x^{a-1}-x^{b-1}}{1-x}dx$$
Now how can i proceed further, Help required, Thanks
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\int_{0}^{\infty}{x^{a - 1} - x^{b - 1} \over 1 - x}\,\dd x & =
\int_{0}^{1}{x^{a - 1} - x^{b - 1} \over 1 - x}\,\dd x +
\int_{1}^{\infty}{x^{a - 1} - x^{b - 1} \over 1 - x}\,\dd x
\end{align}
In the RHS second integral I'll perform the change $\ds{x\ \mapsto\ 1/x}$:
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{a - 1} - x^{b - 1} \over 1 - x}\,\dd x}
\\[5mm] = &\
\int_{0}^{1}{x^{a - 1} - x^{b - 1} \over 1 - x}\,\dd x +
\int_{1}^{0}{\pars{1/x}^{a - 1} - \pars{1/x}^{1 - b} \over 1 - 1/x}
\,{\dd x \over -x^{2}}
\\[5mm] = &\
\int_{0}^{1}{x^{a - 1} - x^{b - 1} \over 1 - x}\,\dd x -
\int_{0}^{1}{x^{-a} - x^{-b} \over 1 - x}\,\dd x
\\[5mm] & =
\bbx{\ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
\quad\mbox{with}\ \Re\pars{a}\,,\ \Re\pars{b} \in \pars{0,1}
\end{align}
$\ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $\ds{H_{z} = \int_{0}^{1}{1 - t^{z} \over 1 - t}\,\dd t}$ with  $\ds{\Re\pars{z} > -1}$.

Note that $\ds{H_{b - 1} - H_{-b} = -\pi\cot\pars{\pi b}}$ such that:
$$
\bbx{\ds{\int_{0}^{\infty}{x^{a - 1} - x^{b - 1} \over 1 - x}\,\dd x  =
\pi\cot\pars{\pi a} - \pi\cot\pars{\pi b}}}
$$
A: For $0<a,b<1$, with $x=e^u$
\begin{eqnarray}
\int_{0}^{\infty}\frac{x^{a-1}-x^{b-1}}{1-x}dx
&=&
\int_{-\infty}^{\infty}\frac{e^{au}-e^{bu}}{1-e^u}dx\\
&=&
\int_{0}^{\infty}\frac{e^{au}-e^{bu}}{1-e^u}dx+\int_{0}^{\infty}\frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\\
&=&
\int_{0}^{\infty}\frac{-e^{-u}}{1-e^{-u}}\Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}\Big)dx\\
&=&
\int_{0}^{\infty}\sum_{n=1}^\infty e^{-nu}\Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}\Big)dx\\
&=&
\sum_{n=1}^\infty\int_{0}^{\infty}\Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}\Big)dx\\
&=&
\sum_{n=1}^\infty\Big(\frac{1}{n-a}-\frac{1}{n-b}+\frac{1}{n+a-1}-\frac{1}{n+b-1}\Big)
\end{eqnarray}
A: You have to study what happens near $x=1$ and when $x\to \infty$, as I suppose you have notice. 
Defining $f(x)=\frac{x^{a-1}-x^{b-1}}{1-x}$, $\lim_{x\to 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $\int_0^1 f(x)\ dx$ exists and it's a number.
I do not know how to proceed from here. Any help would be appreciated.
