compute integral for $n\ge 3$

Question

I was trying to prove Sobolev inequality and estimate the value of the constant for dimensions greater (or equal) to 3 and these integrals came up $$I_\pm=\int_0^\infty \frac{r^{n\pm1}}{(b^2+r^2)^n} dr$$ I tried substitutions $r=b\tan t$, $r=b\sinh t$ and even Cauchy formula for integrals... but I didn't get any results. Any hint?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I_{\pm} & \equiv \int_{0}^{\infty}{r^{n \pm 1} \over \pars{b^{2} + r^{2}}^{n}} \,\dd r \,\,\,\stackrel{{\large{r/\verts{b}}}\ \mapsto\ r}{=}\,\,\, {1 \over \verts{b}^{n - 1 \mp 1}}\int_{0}^{\infty}{r^{n \pm 1} \over \pars{r^{2} + 1}^{n}}\,\dd r \\[5mm] \stackrel{r^{2}\ \mapsto\ r}{=}\,\,\,& {1 \over 2\verts{b}^{n - 1 \mp 1}}\int_{0}^{\infty}{r^{n/2 \pm 1/2 - 1/2} \over \pars{r + 1}^{n}}\,\dd r \\ = &\ {1 \over 2\verts{b}^{n - 1 \mp 1}}\int_{0}^{\infty}r^{n/2 \pm 1/2 - 1/2}\ \overbrace{\bracks{{1 \over \Gamma\pars{n}}\int_{0}^{\infty}t^{n - 1}\expo{-\pars{r + 1}t}\,\dd t}}^{\ds{1 \over \pars{r + 1}^{n}}}\ \,\dd r\quad\pars{~\Gamma:\ Gamma\ Function~} \\[5mm] = &\ {1 \over 2\verts{b}^{n - 1 \mp 1}\,\Gamma\pars{n}} \int_{0}^{\infty}t^{n - 1}\expo{-t}\ \overbrace{\int_{0}^{\infty}r^{n/2 \pm 1/2 - 1/2}\expo{-tr}\,\dd r} ^{\ds{\Gamma\pars{n/2 \pm 1/2 + 1/2}/t^{n/2 \pm 1/2 + 1/2}}}\,\dd t \\[5mm] = &\ {1 \over 2\verts{b}^{n - 1 \mp 1}}\,{\Gamma\pars{n/2 \pm 1/2 + 1/2} \over \Gamma\pars{n}} \int_{0}^{\infty}t^{n/2 \mp 1/2 - 3/2}\expo{-t}\dd t \\[5mm] = &\ {1 \over 2\verts{b}^{n - 1 \mp 1}}\,{\Gamma\pars{n/2 \pm 1/2 + 1/2} \over \Gamma\pars{n}}\,\,\Gamma\pars{{n \over 2} \mp {1 \over 2} - {1 \over 2}} \\[5mm] = &\ \bbx{{\mrm{B}\pars{n/2 \pm 1/2 + 1/2,n/2 \mp 1/2 - 1/2} \over 2\verts{b}^{n - 1 \mp 1}}} \qquad\pars{~\mrm{B}:\ Beta\ Function~} \end{align}

Answer 2

If you consider the problem of the antiderivative $$J=\int \frac{r^{a}}{(b^2+r^2)^n}\, dr$$ the result is given in terms of hypergeometric function $$J=\frac{r^{a+1} \, _2F_1\left(1,\frac{1}{2} (a-2 n+3);\frac{a+3}{2};-\frac{r^2}{b^2}\right)}{(a+1) b^2\left(b^2+r^2\right)^{n-1}}$$ making $$I_a=\int_0^\infty \frac{r^{a}}{(b^2+r^2)^n}\, dr=\frac 12\, b^{(a+1-2 n)}\,\,\frac{\Gamma \left(\frac{a+1}{2}\right)\, \Gamma \left(n-\frac{a+1}{2}\right)}{\Gamma (n)}$$ provided $$\Re(b)\neq 0\land \Re(a-2 n)<-1\land (a+1) \arg \left(\frac{1}{b^2}\right)\leq 2 \pi \land \Re(a)>-1$$

Edit

Otherwise, use $$r=b \tan(x) \qquad dr=\frac{dx} {\cos^2(x)}\qquad 1+\tan^2(x)=\frac{1} {\cos^2(x)}$$ to make $$I_a=\int_0^\infty \frac{r^{a}}{(b^2+r^2)^n}\, dr=b^{(a+1-2n)} \int_0^{\frac \pi 2} \sin^a(x) \cos^{(2n-a-2)}(x) \,dx$$ which is a classical one (since, morover, in your case $a$ is an integer).