## gerryliyana Group Title Find the solution of gamma and beta function! one year ago one year ago

1. gerryliyana Group Title

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

2. gerryliyana Group Title

3. amistre64 Group Title

i dont have enough experience with beta and gamma

me neither :(

5. ajprincess Group Title

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

6. oldrin.bataku Group Title

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

7. abb0t Group Title

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

8. oldrin.bataku Group Title

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

9. abb0t Group Title

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

10. abb0t Group Title

11. gerryliyana Group Title

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 $\frac{ 1 }{ 2 } \frac{ \Gamma (1/4) \Gamma (1/2)}{ \Gamma (\frac{ 1 }{ 4 }+\frac{ 1 }{ 2 }) }$

12. oldrin.bataku Group Title

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

13. gerryliyana Group Title

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

14. mukushla Group Title

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

15. gerryliyana Group Title

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

16. mukushla Group Title

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

17. gerryliyana Group Title

wait a sec..,

18. gerryliyana Group Title

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

19. gerryliyana Group Title

$= \frac{ 1 }{ 2 } \Gamma(2)$

20. mukushla Group Title

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

21. gerryliyana Group Title

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

22. mukushla Group Title

very welcome :)