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The integral form definition of the Gamma function is as follows. It is valid for all complex numbers with $\mathrm{Re}(z)>0$:

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

See this Wikipedia page for justification.

It is well-known that the Gamma function also has this infinite product expression that is valid for all complex numbers $z$ except for the negative integers.

$$\Gamma(z)=\frac{1}{z}\prod_{n=1}^\infty\left[\frac{1}{1+\frac{z}{n}} \left(1+\frac{1}{n}\right)^z\right] $$

See this Wikipedia page for the justification.

How do we rigorously prove that these two definitions of $\Gamma(z)$ are equivalent on their common domain $\mathrm{Re}(z)>0$?

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    $\begingroup$ Take a look at this. $\endgroup$ – mrtaurho Aug 12 '18 at 9:31
  • $\begingroup$ Through the dominated convergence theorem and integration by parts. Have a look at my notes around page 72. $\endgroup$ – Jack D'Aurizio Aug 13 '18 at 5:55

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