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
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{\Re\pars{z} > -1}$.
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}
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.
