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$$\Gamma(z)\Gamma(1-z)=\int_0^{\infty}x^{z-1}e^{-x}dx\int_0^{\infty}x^{-z}e^{-x}dx=\int_0^{\infty}(2x)^{-1}e^{-2x}(2dx)=\Gamma(0)\to\infty??$$

RE60K
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  • $$\Gamma(z)\Gamma(1-z)=\int_0^\infty x^{z-1}e^{-x}dx\int_0^\infty x^{-z}e^{-x}dx=\int_0^\infty\int_0^\infty x^{z-1}y^{-z}e^{-(x+y)}~dx~dy.$$ – Lucian Jan 09 '15 at 12:03

1 Answers1

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I don't think your operation is valid. Compare:

$$ \int_a^b x \; dx \int_a^b 1/x \; dx = \frac12(b^2-a^2)(\ln b - \ln a) \neq \int_a^b 1 dx =b-a $$

draks ...
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