## experimentX 3 years ago Prove: $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{(\pi z)}}$

1. experimentX

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2. experimentX

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3. experimentX

this is Euler's reflection formula ... http://en.wikipedia.org/wiki/Reflection_formula I never tried this. Looks worth!!

4. estudier

$\displaystyle \sin(\pi z) = \pi z \prod_{n \neq 0} \left(1- \frac zn\right) \exp\left( \frac z n \right)$

5. mukushla

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

later integral can be done complex integration on this |dw:1349016144043:dw|contour

7. mukushla

thats a circle i believe...lol

8. mukushla

this contour will work i know...but there are some more

9. mukushla

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10. estudier

$\displaystyle \frac 1 {\Gamma \left({z}\right)} = z e^{\gamma z} \prod_{n=1}^\infty \left({ 1 + \frac z n}\right) \exp\left({-\frac z n}\right)$

11. experimentX

yeah i can do that contour!!

12. experimentX

that must be equal to ... directly from the formula $\pi \over \sin (\pi z)$ $\int_0^{\infty}{x^{a-1} \over {1 +x^b}}dx = {\pi \over b \sin \left ( {a \pi \over b} \right )}$