Austin Rangel

2022-10-02

How to compute ${\int }_{a}^{b}\left(b-x{\right)}^{\frac{n-1}{2}}\left(x-a{\right)}^{-1/2}dx$

Demarion Thornton

$I={\int }_{a}^{b}\frac{{\sqrt{b-x}}^{\phantom{\rule{thinmathspace}{0ex}}n}}{\sqrt{\left(b-x\right)\left(x-a\right)}}\phantom{\rule{thinmathspace}{0ex}}dx$
$\begin{array}{rl}I& ={\int }_{0}^{\pi /2}\frac{2\left(b-a\right)\mathrm{cos}t\mathrm{sin}t\cdot {\sqrt{\left(b-a\right){\mathrm{cos}}^{2}t}}^{\phantom{\rule{thinmathspace}{0ex}}n}}{\sqrt{\left(b-a{\right)}^{2}{\mathrm{cos}}^{2}t{\mathrm{sin}}^{2}t}}\phantom{\rule{thinmathspace}{0ex}}dt\\ & =2{\sqrt{b-a}}^{\phantom{\rule{thinmathspace}{0ex}}n}{\int }_{0}^{\pi /2}{\mathrm{cos}}^{n}t\phantom{\rule{thinmathspace}{0ex}}dt\\ & =\left\{\begin{array}{l}\pi {\sqrt{b-a}}^{\phantom{\rule{thinmathspace}{0ex}}n}\frac{1\cdot 3\cdots \left(n-1\right)}{2\cdot 4\cdots n}\phantom{\rule{1em}{0ex}}\left(n\phantom{\rule{thickmathspace}{0ex}}\text{even}\right)\\ 2{\sqrt{b-a}}^{\phantom{\rule{thinmathspace}{0ex}}n}\frac{2\cdot 4\cdots \left(n-1\right)}{1\cdot 3\cdots n}\phantom{\rule{1em}{0ex}}\left(n\phantom{\rule{thickmathspace}{0ex}}\text{odd}\right)\end{array}\end{array}$
The latter integral is well known, and can easily be evaluated by parts:
$\begin{array}{rl}{I}_{n}={\int }_{0}^{\pi /2}{\mathrm{cos}}^{n}t\phantom{\rule{thinmathspace}{0ex}}dt⇒{I}_{n}& =\left(n-1\right){\int }_{0}^{\pi /2}{\mathrm{sin}}^{2}t{\mathrm{cos}}^{n-2}t\phantom{\rule{thinmathspace}{0ex}}dt\\ & =\left(n-1\right){I}_{n-2}-\left(n-1\right){I}_{n}\\ & \phantom{\rule{2em}{0ex}}⇒{I}_{n}=\frac{n-1}{n}{I}_{n-2}\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}⇒\left\{\begin{array}{l}{I}_{n}=\frac{1\cdot 3\cdots \left(n-1\right)}{2\cdot 4\cdots n}{I}_{0}\phantom{\rule{thickmathspace}{0ex}}\left(n\phantom{\rule{thickmathspace}{0ex}}\text{even}\right)\\ {I}_{n}=\frac{2\cdot 4\cdots \left(n-1\right)}{1\cdot 3\cdots n}{I}_{1}\phantom{\rule{thickmathspace}{0ex}}\left(n\phantom{\rule{thickmathspace}{0ex}}\text{odd}\right)\end{array}\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left({I}_{0}=\frac{\pi }{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{I}_{1}=1\right)\end{array}$

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