Question

Find the solution of gamma and beta function!

Answer 1

Find the solution of gamma and beta function for
\[\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} }\]

Answer 2

@ajprincess @amistre64 @BAdhi

Answer 3

i dont have enough experience with beta and gamma

Answer 4

me neither :(

Answer 5

I am sorry @gerryliyana I havnt learnt this.:(

Answer 6

I don't understand your original question, but this is an elliptic integral.

Answer 7

If the root wasn't there I would have an idea but it's making it a problem with it there.

Answer 8

@gerryliyana can you state more clearly what you wish to do?

Answer 9

Isn't it \[\frac{ \Gamma (m) }{ 2 \times \Gamma (m) }\]

Answer 10

For your beta

Answer 11

ok ok...,for eample i have
\[\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} } \]
based on Beta function;
\[B(p,q) = 2 \int\limits_{0}^{\pi/2} (\sin \theta)^{2p-1} (\cos \theta)^{2q-1} d \theta \]
then, i have
\[\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} } \]
\[\int\limits_{0}^{\pi/2} (\sin \theta)^{-1/2} (\cos \theta)^{0} d \theta\]
\[\int\limits\limits_{0}^{\pi/2} (\sin \theta)^{2 (1/4)-1} (\cos \theta)^{2 (1/2)-1} d \theta \]
then i have
\[B(p,q)=2 \int\limits\limits\limits_{0}^{\pi/2} (\sin \theta)^{2 (1/4)-1} (\cos \theta)^{2 (1/2)-1} d \theta \]
\[\int\limits\limits\limits_{0}^{\pi/2} (\sin \theta)^{2 (1/4)-1} (\cos \theta)^{2 (1/2)-1} d \theta = \frac{ 1 }{ 2 } B(p,q)\]
with p = 1/4 and q = 1/2; then (relation Beta and gamma)
\[B(p,q) = \frac{ \Gamma (p) \Gamma (q) }{ \Gamma (p+q) }\]
\[B(p,q) = \frac{ \Gamma (1/4) \Gamma (1/2) }{ \Gamma (1/4+1/2) }\]
then i have
\[\int\limits\limits\limits_{0}^{\pi/2} (\sin \theta)^{2 (1/4)-1} (\cos \theta)^{2 (1/2)-1} d \theta = \frac{ 1 }{ 2 } B(p,q)\]
\[\int\limits\limits\limits\limits_{0}^{\pi/2} (\sin \theta)^{2 (1/4)-1} (\cos \theta)^{2 (1/2)-1} d \theta = \frac{ 1 }{ 2 } \frac{ \Gamma(1/4)\Gamma (1/2) }{ \Gamma (1/4+1/2 )}\]
finally
The solution of gamma and Beta function for
\(\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} } \) is \[\frac{ 1 }{ 2 } \frac{ \Gamma (1/4) \Gamma (1/2)}{ \Gamma (\frac{ 1 }{ 4 }+\frac{ 1 }{ 2 }) }\]

Answer 12

Oh... okay. You didn't clarify that's what you wanted.

Answer 13

Ok guys actually, i wanna check my work.., hei how about these
"find the solution of gamma and beta function for
\[\int\limits_{0}^{\pi/2} (\tan^{3}\theta + \tan^{5} \theta) e^{-\tan^{2}\theta} d \theta\]

Answer 14

let \(u=\tan^2 \theta\) see what happens, have u tried it yet?

Answer 15

yes., i've tried for this one, but it didn't work, oh my bad :(

Answer 16

it becomes\[\frac{1}{2}\int_{0}^{\infty} u e^{-u} du\]right?

Answer 17

wait a sec..,

Answer 18

then,
\[\frac{ 1 }{ 2 } \int\limits_{0}^{\infty} ue^{-u} du = \frac{ 1 }{ 2 } \int\limits_{0}^{\infty} u^{2-1} e^{-u} du\]
then p=2

Answer 19

\[= \frac{ 1 }{ 2 } \Gamma(2)\]

Answer 20

and u know that\[\Gamma(n+1)=n!\]we are done :)

Answer 21

ah yeah.., thank you so much @mukushla and other woow we are done :)

Answer 22

very welcome :)