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$\bf{My\; Try::}$ 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

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    Are you familiar with techniques from complex analysis? – 2017-01-30
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    To Michael Saifert. not to much, elementry knowledge. – 2017-01-31

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

Note that $\ds{H_{b - 1} - H_{-b} = -\pi\cot\pars{\pi b}}$ such that: $$ \bbx{\ds{\int_{0}^{\infty}{x^{a - 1} - x^{b - 1} \over 1 - x}\,\dd x = \pi\cot\pars{\pi a} - \pi\cot\pars{\pi b}}} $$

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    I knew the result MyGlasses derived looked familiar! – 2017-01-30
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    @J.G. Yes. It's true. Thanks. – 2017-01-30
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    Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -\pi \cot(\pi b)$ – 2017-01-31
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    @juantheron $H_{m}$ is [related](https://en.wikipedia.org/wiki/Harmonic_number#Applications) to the Digamma Function $\Psi$. Namely, $H_{m} = \Psi\left(m + 1\right) + \gamma$. $\gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = \Psi\left(b\right) - \Psi\left(-b + 1\right) = -\pi\cot\left(\pi b\right)$. The last one is the [Euler Reflection Formula](https://en.wikipedia.org/wiki/Reflection_formula). – 2017-01-31
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    Thanks Felix Marin got it. – 2017-01-31
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    which gives us what? – 2017-01-30
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    I'm stuck here. maybe diverge! – 2017-01-30
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    no the divergencies of the single terms are all of the same magnitude ($\sim \log(n)$)and therefore cancel out...what is left is most likely a so called digamma function – 2017-01-30
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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.