experimentX
  • experimentX
Prove: \[ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{(\pi z)}} \]
Mathematics
jamiebookeater
  • jamiebookeater
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experimentX
  • experimentX
|dw:1349012805707:dw|
experimentX
  • experimentX
|dw:1349012926410:dw|
experimentX
  • experimentX
this is Euler's reflection formula ... http://en.wikipedia.org/wiki/Reflection_formula I never tried this. Looks worth!!

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anonymous
  • anonymous
\[\displaystyle \sin(\pi z) = \pi z \prod_{n \neq 0} \left(1- \frac zn\right) \exp\left( \frac z n \right)\]
anonymous
  • anonymous
beta function\[B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\]so\[\Gamma(z)\Gamma(1-z)=B(z,1-z)=\int_{0}^{\infty} \frac{u^{z-1}}{1+u} \text{d}u\]
anonymous
  • anonymous
later integral can be done complex integration on this |dw:1349016144043:dw|contour
anonymous
  • anonymous
thats a circle i believe...lol
anonymous
  • anonymous
this contour will work i know...but there are some more
anonymous
  • anonymous
|dw:1349016381131:dw|
anonymous
  • anonymous
\[\displaystyle \frac 1 {\Gamma \left({z}\right)} = z e^{\gamma z} \prod_{n=1}^\infty \left({ 1 + \frac z n}\right) \exp\left({-\frac z n}\right)\]
experimentX
  • experimentX
yeah i can do that contour!!
experimentX
  • experimentX
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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