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experimentX

  • 2 years ago

Prove: \[ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{(\pi z)}} \]

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  1. experimentX
    • 2 years ago
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    |dw:1349012805707:dw|

  2. experimentX
    • 2 years ago
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    |dw:1349012926410:dw|

  3. experimentX
    • 2 years ago
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    this is Euler's reflection formula ... http://en.wikipedia.org/wiki/Reflection_formula I never tried this. Looks worth!!

  4. estudier
    • 2 years ago
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    \[\displaystyle \sin(\pi z) = \pi z \prod_{n \neq 0} \left(1- \frac zn\right) \exp\left( \frac z n \right)\]

  5. mukushla
    • 2 years ago
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    beta function\[B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\]so\[\Gamma(z)\Gamma(1-z)=B(z,1-z)=\int_{0}^{\infty} \frac{u^{z-1}}{1+u} \text{d}u\]

  6. mukushla
    • 2 years ago
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    later integral can be done complex integration on this |dw:1349016144043:dw|contour

  7. mukushla
    • 2 years ago
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    thats a circle i believe...lol

  8. mukushla
    • 2 years ago
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    this contour will work i know...but there are some more

  9. mukushla
    • 2 years ago
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    |dw:1349016381131:dw|

  10. estudier
    • 2 years ago
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    \[\displaystyle \frac 1 {\Gamma \left({z}\right)} = z e^{\gamma z} \prod_{n=1}^\infty \left({ 1 + \frac z n}\right) \exp\left({-\frac z n}\right)\]

  11. experimentX
    • 2 years ago
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    yeah i can do that contour!!

  12. experimentX
    • 2 years ago
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    that must be equal to ... directly from the formula \[ \pi \over \sin (\pi z)\] \[ \int_0^{\infty}{x^{a-1} \over {1 +x^b}}dx = {\pi \over b \sin \left ( {a \pi \over b} \right )} \]

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