$\int_0^{\pi/2}sin^p\,\theta\;cos^q\,\theta\;d\theta = \frac{\Gamma{\frac{p+1}{2}}\Gamma{\frac{q+1}{2}}}{2\Gamma{\frac{p+q+2}{2}}},\; p,q > -1$

Question

Show that: $$\int_0^{\pi/2}sin^p\,\theta\;cos^q\,\theta\;d\theta = \frac{\Gamma{\frac{p+1}{2}}\Gamma{\frac{q+1}{2}}}{2\Gamma{\frac{p+q+2}{2}}},\; p,q > -1$$

Here's the question.

Please, have a look at this: https://en.wikipedia.org/wiki/Beta_function#Properties — Olivier Oloa, Apr 10 '17 at 14:31
I have never seen Gamma function written like that in print! — Paramanand Singh, Apr 10 '17 at 16:10

Stefan4024 · Accepted Answer · 2017-04-10T15:12:47.283

7

I think that the OP meant the Gamma function instead of square root. If so the relation is true, as from the Beta function we have:

$$\int_0^1 v^{y-1}(1-v)^{x-1} dv = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

Then substituting $v = \sin^2 \theta$ will provide you with the result.

edited Apr 10 '17 at 15:12

answered Apr 10 '17 at 14:33

Stefan4024

35,843

score 5 · Answer 2 · answered Apr 10 '17 at 14:25

5

The announced result is not right, as may be seen by taking $q=1$ giving $$ \int_0^{\pi/2}\sin^p\,\theta\;\cos^q\,\theta\;d\theta =\int_0^1u^p\,du=\frac1{p+1}\ne \frac{\sqrt{p+1}}{2\sqrt{p+3}},\quad p>-1, $$ in general.

answered Apr 10 '17 at 14:25

Olivier Oloa

120,989

score 3 · Answer 3 · answered Apr 10 '17 at 14:28

3

if $p=q=1$, we get

$$\int_0^\frac {\pi}{2} \frac {\sin (2\theta)d\theta}{2} $$

$$=\frac {1}{4}[-\cos (2\theta)]_0^\frac {\pi}{2} $$

$$=\frac {1}{2} \neq \frac {1}{2\sqrt {2}} $$

Your formula doesn't seem to be correct.

answered Apr 10 '17 at 14:28

hamam_Abdallah

62,951

$\int_0^{\pi/2}sin^p\,\theta\;cos^q\,\theta\;d\theta = \frac{\Gamma{\frac{p+1}{2}}\Gamma{\frac{q+1}{2}}}{2\Gamma{\frac{p+q+2}{2}}},\; p,q > -1$

3 Answers3