Integral $ \int_a^\infty \frac{dr}{r^5 \sqrt{r-a}}$

Question

$$ \int_a^\infty \frac{dr}{r^5 \sqrt{r-a}}$$

By taking $r-a=t^2$ we are getting $$ \int_0^\infty \frac{2dt}{(a+t^2)^5}$$ Now I am stuck after this. Thanks.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} 2\int_{0}^{\infty}{\dd t \over \pars{t^{2} + a}^{5}} & = {1 \over 12}\,\totald[4]{}{a}\int_{0}^{\infty}{\dd t \over t^{2} + a} = {\pi \over 24}\,\totald[4]{\pars{a^{-1/2}}}{a} = {\pi \over 24}\,{105 \over 16\,a^{9/2}} = \bbx{{35\pi \over 128}\,{1 \over a^{9/2}}} \end{align}

Answer 2

Let $t = \sqrt{a}\tan(\theta), dt = \sqrt{a}\sec^2(\theta)$. Then your integral becomes $$2\int \frac{\sqrt{a}\sec^2(\theta)}{(a+a\tan^2(\theta))^5} d \theta = \frac{2\sqrt a}{a^5} \int \cos^8(\theta)\, d\theta$$ $$ = \frac{2}{a^{9/2}} \int \left( \frac{1+\cos(2\theta)}{2}\right)^4 d\theta$$$$ = \frac{1}{8a^{9/2}}\int 1+4\cos(2\theta)+6\cos^2(2\theta)+4\cos^3(2\theta)+\cos^4(2\theta) d\theta$$