# How to compute Integral

I wonder how to compute the following integral (for any natural number $n\geq 0$): $$\int_a^b (b-x)^{\frac{n-1}{2}}(x-a)^{-1/2}dx.$$ Does anyone know the final answer and how to get there? Is there a trick? Thank you very much!

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Hint:

$$\int_0^1 dt\: t^{m-1} (1-t)^{n-1} = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}$$

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The second factor in the numerator should be an $m$ right? Thank you very much! :) –  Dan Mar 13 '13 at 17:25
@Dan: typo - sorry about that. –  Ron Gordon Mar 13 '13 at 17:27
$\dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$ is also known as the Beta function $\mathrm{B}(m,n)$. –  robjohn Mar 13 '13 at 18:01
@robjohn: of course, but I felt it sufficient to provide the raw functions for the purpose of the hint. Thanks for pointing it out, though. –  Ron Gordon Mar 13 '13 at 18:10

$$I=\int_a^b\frac{\sqrt{b-x}^{\,n}}{\sqrt{(b-x)(x-a)}}\,dx$$

$$x=a\cos^2 t+b\sin^2 t:$$

\begin{align*}I&=\int_0^{\pi/2}\frac{2(b-a)\cos t\sin t\cdot\sqrt{(b-a)\cos^{2}t}^{\,n}}{\sqrt{(b-a)^2\cos^2 t\sin^2 t}}\,dt\\[7pt]&=2\sqrt{b-a}^{\,n}\int_0^{\pi/2}\cos^{n}t\,dt\\[7pt]&= \left\{ \begin{array}{l l} \pi\sqrt{b-a}^{\,n}\frac{1\cdot 3\cdots (n-1)}{2\cdot 4\cdots n}\quad(n\;\text{even})\\[7pt] 2\sqrt{b-a}^{\,n}\frac{2\cdot 4\cdots (n-1)}{1\cdot 3\cdots n}\quad(n\;\text{odd}) \end{array} \right.\end{align*}

$$\star$$ The latter integral is well known, and can easily be evaluated by parts:

\begin{align*}I_n=\int_0^{\pi/2}\cos^nt\,dt\Rightarrow I_n&=(n-1)\int_0^{\pi/2}\sin^2 t\cos^{n-2}t\,dt\\[7pt]&=(n-1)I_{n-2}-(n-1)I_{n}\\[7pt]&\qquad\Rightarrow I_n=\frac{n-1}{n}I_{n-2}\\[7pt]&\qquad\qquad\Rightarrow\left\{ \begin{array}{l l} I_n=\frac{1\cdot 3\cdots (n-1)}{2\cdot 4\cdots n}I_0\;(n\;\text{even})\\[7pt] I_n=\frac{2\cdot 4\cdots (n-1)}{1\cdot 3\cdots n}I_1\;(n\;\text{odd})\end{array}\right. \\[7pt]&\qquad\qquad\qquad\big(I_0=\dfrac{\pi}{2}\,\;\;I_1=1\big)\end{align*}

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Although my answer was arrived at independently, if you feel it is too close in appearance, I will delete it. –  robjohn Mar 13 '13 at 19:35
@robjohn Nah, the more the better! –  L. F. Mar 13 '13 at 21:22

Let $x=(b-a)t+a$, then $b-x=(b-a)(1-t)$, $x-a=(b-a)t$, and $\mathrm{d}x=(b-a)\mathrm{d}t$.

Then let $t=\sin^2(\theta)$. \begin{align} \int_a^b(b-x)^{(n-1)/2}\,(x-a)^{-1/2}\,\mathrm{d}x &=(b-a)^{n/2}\int_0^1(1-t)^{(n-1)/2}\,t^{-1/2}\,\mathrm{d}t\\ &=2(b-a)^{n/2}\int_0^{\pi/2}\cos^{n}(\theta)\,\mathrm{d}\theta\tag{1} \end{align} Integration by parts yields $$\int_0^{\pi/2}\cos^n(\theta)\,\mathrm{d}\theta=\frac{n-1}{n}\int_0^{\pi/2}\cos^{n-2}(\theta)\,\mathrm{d}\theta\tag{2}$$ For even $n$, we get \begin{align} \int_0^{\pi/2}\cos^n(\theta)\,\mathrm{d}\theta&=\frac{\pi}{2^{n+1}}\binom{n}{n/2}\\ \int_a^b(b-x)^{(n-1)/2}\,(x-a)^{-1/2}\,\mathrm{d}x&=(b-a)^{n/2}\frac{\pi}{2^n}\binom{n}{n/2}\tag{3} \end{align} For odd $n$, we get \begin{align} \int_0^{\pi/2}\cos^n(\theta)\,\mathrm{d}\theta&=\frac{2^{n-1}}{n\binom{n-1}{(n-1)/2}}\\ \int_a^b(b-x)^{(n-1)/2}\,(x-a)^{-1/2}\,\mathrm{d}x&=(b-a)^{n/2}\frac{2^n}{n\binom{n-1}{(n-1)/2}}\tag{4} \end{align}

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