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experimentX
 3 years ago
Prove:
\[ \Gamma(z)\Gamma(1z) = \frac{\pi}{\sin{(\pi z)}} \]
experimentX
 3 years ago
Prove: \[ \Gamma(z)\Gamma(1z) = \frac{\pi}{\sin{(\pi z)}} \]

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experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1349012805707:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1349012926410:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1this is Euler's reflection formula ... http://en.wikipedia.org/wiki/Reflection_formula I never tried this. Looks worth!!

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0\[\displaystyle \sin(\pi z) = \pi z \prod_{n \neq 0} \left(1 \frac zn\right) \exp\left( \frac z n \right)\]

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1beta function\[B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\]so\[\Gamma(z)\Gamma(1z)=B(z,1z)=\int_{0}^{\infty} \frac{u^{z1}}{1+u} \text{d}u\]

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1later integral can be done complex integration on this dw:1349016144043:dwcontour

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1thats a circle i believe...lol

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1this contour will work i know...but there are some more

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0\[\displaystyle \frac 1 {\Gamma \left({z}\right)} = z e^{\gamma z} \prod_{n=1}^\infty \left({ 1 + \frac z n}\right) \exp\left({\frac z n}\right)\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yeah i can do that contour!!

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1that must be equal to ... directly from the formula \[ \pi \over \sin (\pi z)\] \[ \int_0^{\infty}{x^{a1} \over {1 +x^b}}dx = {\pi \over b \sin \left ( {a \pi \over b} \right )} \]
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