I want to know how to solve this integral. For $a>0$, $$I(a)=\int_{0}^{\infty}\frac{dx}{(x+1)(x+2)^a}$$ I tried the substitution $1-u=x+2$, in the hope that I could put the integral in terms of the incomplete Beta function: $$I(a)=-\int_{-\infty}^{-1}\frac{du}{u(1-u)^a}$$ Which didn't really work out, because the bounds are all wrong.
I do not know how to proceed. Please help
On
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\vphantom{\large A}\mrm{I}\pars{a}\right\vert_{\ \Re\pars{a}\ >\ 0} & \equiv \int_{0}^{\infty}{\dd x \over \pars{x + 1}\pars{x + 2}^{a}} \,\,\,\stackrel{x + 1\ \mapsto\ x}{=}\,\,\, \int_{1}^{\infty}x^{-1}\pars{1 + x}^{-a}\,\dd x \\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\, \int_{0}^{1}x^{a - 1}\,\pars{1 - x}^{0}\,\pars{1 + x}^{-a}\,\dd x \\[5mm] & = \bbx{\,\mrm{B}\pars{a,4} {}_{2}\mrm{F}_{1}\pars{a,a;a + 1;-1}} \\[2mm] & \mbox{where}\quad \pars{\begin{array}{l} \mrm{B}:\ Beta\ Function \\ {}_{2}\mrm{F}_{1}:\ Hypergeometric\ Function \end{array}} \end{align} See Euler Type Hypergeometric Function.
By writing $\frac{1}{(x+1)}$ as $\sum_{n\geq 1}\frac{1}{(x+2)^n}$ we have that the original integral equals
$$ \sum_{n\geq 1}\int_{0}^{+\infty}\frac{dx}{(x+2)^{n+a}} = \sum_{m\geq 0}\frac{1}{2^{a+m}(a+m)}$$ i.e. a value of the Lerch trascendent, namely $2^{-a}\,\Phi\left(\frac{1}{2},1,a\right)$.