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gerryliyana
 2 years ago
Find the solution of gamma and beta function!
gerryliyana
 2 years ago
Find the solution of gamma and beta function!

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gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1Find the solution of gamma and beta function for \[\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} }\]

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1@ajprincess @amistre64 @BAdhi

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0i dont have enough experience with beta and gamma

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0I am sorry @gerryliyana I havnt learnt this.:(

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0I don't understand your original question, but this is an elliptic integral.

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1If the root wasn't there I would have an idea but it's making it a problem with it there.

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0@gerryliyana can you state more clearly what you wish to do?

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1Isn't it \[\frac{ \Gamma (m) }{ 2 \times \Gamma (m) }\]

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1ok ok...,for eample i have \[\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} } \] based on Beta function; \[B(p,q) = 2 \int\limits_{0}^{\pi/2} (\sin \theta)^{2p1} (\cos \theta)^{2q1} d \theta \] 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 \[\frac{ 1 }{ 2 } \frac{ \Gamma (1/4) \Gamma (1/2)}{ \Gamma (\frac{ 1 }{ 4 }+\frac{ 1 }{ 2 }) }\]

oldrin.bataku
 2 years ago
Best ResponseYou've already chosen the best response.0Oh... okay. You didn't clarify that's what you wanted.

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1Ok guys actually, i wanna check my work.., hei how about these "find the solution of gamma and beta function for \[\int\limits_{0}^{\pi/2} (\tan^{3}\theta + \tan^{5} \theta) e^{\tan^{2}\theta} d \theta\]

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2let \(u=\tan^2 \theta\) see what happens, have u tried it yet?

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1yes., i've tried for this one, but it didn't work, oh my bad :(

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2it becomes\[\frac{1}{2}\int_{0}^{\infty} u e^{u} du\]right?

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1then, \[\frac{ 1 }{ 2 } \int\limits_{0}^{\infty} ue^{u} du = \frac{ 1 }{ 2 } \int\limits_{0}^{\infty} u^{21} e^{u} du\] then p=2

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1\[= \frac{ 1 }{ 2 } \Gamma(2)\]

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2and u know that\[\Gamma(n+1)=n!\]we are done :)

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.1ah yeah.., thank you so much @mukushla and other woow we are done :)
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