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gerryliyana

  • 2 years ago

Find the solution of gamma and beta function!

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

  2. gerryliyana
    • 2 years ago
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    @ajprincess @amistre64 @BAdhi

  3. amistre64
    • 2 years ago
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    i dont have enough experience with beta and gamma

  4. BAdhi
    • 2 years ago
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    me neither :(

  5. ajprincess
    • 2 years ago
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    I am sorry @gerryliyana I havnt learnt this.:(

  6. oldrin.bataku
    • 2 years ago
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    I don't understand your original question, but this is an elliptic integral.

  7. abb0t
    • 2 years ago
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    If the root wasn't there I would have an idea but it's making it a problem with it there.

  8. oldrin.bataku
    • 2 years ago
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    @gerryliyana can you state more clearly what you wish to do?

  9. abb0t
    • 2 years ago
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    Isn't it \[\frac{ \Gamma (m) }{ 2 \times \Gamma (m) }\]

  10. abb0t
    • 2 years ago
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    For your beta

  11. gerryliyana
    • 2 years ago
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    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
    • 2 years ago
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    Oh... okay. You didn't clarify that's what you wanted.

  13. gerryliyana
    • 2 years ago
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    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
    • 2 years ago
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    let \(u=\tan^2 \theta\) see what happens, have u tried it yet?

  15. gerryliyana
    • 2 years ago
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    yes., i've tried for this one, but it didn't work, oh my bad :(

  16. mukushla
    • 2 years ago
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    it becomes\[\frac{1}{2}\int_{0}^{\infty} u e^{-u} du\]right?

  17. gerryliyana
    • 2 years ago
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    wait a sec..,

  18. gerryliyana
    • 2 years ago
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    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
    • 2 years ago
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    \[= \frac{ 1 }{ 2 } \Gamma(2)\]

  20. mukushla
    • 2 years ago
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    and u know that\[\Gamma(n+1)=n!\]we are done :)

  21. gerryliyana
    • 2 years ago
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    ah yeah.., thank you so much @mukushla and other woow we are done :)

  22. mukushla
    • 2 years ago
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    very welcome :)

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