# Integral with several parameters

Let $r>0$. Find $(p,q) \in \mathbb{R}^{2}$ such that the integral: $$\int_{1}^{\infty}{\frac{(x^{r}-1)^{p}}{x^{q}}} ~dx$$ converges and for those values calculate it.

I've already calculated the values for which it is convergent. The integral above converges iff $rp-q <-1$ or equivalently when $p<\frac{q-1}{r}$.

I get stuck when trying to calculate it. I've derived the integrand with respect to every parameter and then tried to substitute the order of integration but in all cases I get a non-elementary integral. Any ideas how to calculate this? Any help would be appreciated.

• maybe limit test with $1/x^q$? That may take care of some cases, but not all – GPerez Jan 12 '14 at 11:37

Carry out the substitution $y=x^{1-q}$ and then use Newton's generalized binomial theorem.
• It leads to a calculation of the integral. From Wikipedia, binomial theorem, $(a+b)^p=\sum_{k=0}^\infty {p\choose k} a^k b^{p-k}$. In your case $a$ is a power of $x$ and hence so is $a^k$. – JPi Jan 12 '14 at 14:32
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \color{#00f}{\large\int_{1}^{\infty}{\pars{x^{r} - 1}^{p} \over x^{q}}\,\dd x}&= \int_{1}^{0}{\pars{x^{-r} - 1}^{p} \over x^{-q}}\,\pars{-\,{\dd x \over x^{2}}} =\int_{0}^{1}{\pars{1 - x^{r}}^{p} \over x^{pr - q + 2}}\,\dd x \\[3mm]&=\int_{0}^{1}{\pars{1 - x}^{p} \over \pars{x^{1/r}}^{pr - q + 2}} \,{1 \over r}\,x^{1/r - 1}\dd x ={1 \over r}\int_{0}^{1}\pars{1 - x}^{p}x^{-p + \pars{q - 1}/r -1}\,\dd x \\[3mm]&=\color{#00f}{\large{1 \over r}\,{\rm B}\pars{p + 1,-p + {q - 1 \over r}}} \\[5mm]&\quad \Re\pars{{q - 1 \over r}} > \Re p > -1 \end{align} where ${\rm B}\pars{x,y}$ is the Beta function.