similar to gamma function

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July 9th 2007, 11:03 AM
emily28

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similar to gamma function

ive got this question in for tomorrow and i have no idea whatsoever how to attempt it :(

evaluate the following intergral. your final answer should not contain the gamma function.

maths 6.JPG

thank you
July 9th 2007, 03:58 PM
galactus

Since you can't use gamma, you could try rewriting and making a substitution.

$(1-x^{3})^{\frac{5}{2}}\cdot{x^{\frac{7}{2}}}$

$=(1-x^{3})^{2}(1-x^{3})^{\frac{1}{2}}x^{\frac{7}{2}}$

Let $u=\sqrt{1-x^{3}}, \;\ du=\frac{-3x^{2}}{2\sqrt{1-x^{3}}}$

Making the subs gives:

$\frac{-2}{3}\int{u^{6}\sqrt{1-u^{2}}}du$

Now, let $u=sin{\theta}, \;\ du=cos{\theta}d{\theta}$

Make the subs and you get:

$\frac{-2}{3}\int{sin^{6}{\theta}cos^{2}{\theta}}d{\theta}$

$=\frac{-2}{3}\int[sin^{6}{\theta}-sin^{8}{\theta}]d{\theta}$

Now, you can use the reduction formulas:

$\int{sin^{n}{\theta}}d{\theta}=\frac{sin^{n-1}{\theta}cos{\theta}}{n}+\frac{n-1}{n}\int{sin^{n-2}}d{\theta}$

Just a thought. It ain't pretty no matter how you look at it.
July 9th 2007, 07:05 PM
ThePerfectHacker

This can be achieved with the Beta Integral.

$\int_0^1 (1-x^3)^{5/2} x^{7/2} dx$
We can write this as,
$\frac{1}{3} \int_0^1 (1-x^3)^{5/2} \cdot x^{3/2} \cdot (3x^2) \ dx$

Let $t=x^3 \implies t'=3x^2$ thus,

$\frac{1}{3}\int_0^1 (1-t)^{5/2} \cdot t^{1/2} dt$

Use the Beta Integral and get,
$\bold{B}\left(\frac{7}{2} , \frac{3}{2} \right)$

Converte to Gamma Function,

$\frac{\Gamma \left( \frac{7}{2} \right) \cdot \Gamma \left( \frac{3}{2} \right) }{\Gamma (5)}$

But,
$\Gamma \left( \frac{1}{2} \right) = \sqrt{\pi}$

Thus, using reduction identity for Gamma functions,
$\Gamma \left( \frac{7}{2} \right) = \frac{15\sqrt{\pi}}{8}$

And,
$\Gamma \left( \frac{3}{2} \right) = \frac{\sqrt{\pi}}{2}$

And,
$\Gamma (5) = 4!=24$

Thus,
$\mbox{ answer } = \frac{15\pi}{384}$
July 9th 2007, 07:19 PM
ThePerfectHacker

Galactus' approach is not completely way off. He arrives at the integral,
$\int_0^{\pi/2} \sin ^{\alpha} x \cos^{\beta} x \ dx$

Which is discussed here.