Originally, the gamma function $\Gamma(x)$ is defined as
$$ \Gamma(x )=\int_0^\infty e^{-t} t^{x-1} dt .$$
This definition works for $Re(x)> 0 $ only.
So, who extended into the whole complex plane? This is nontrivial.
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$\begingroup$ It is not really that hard. Using integration by parts, for ${\rm Re}(x) > 0$ we have $\Gamma(x+1) = x\Gamma(x)$, so $\Gamma(x) = \Gamma(x+1)/x$. The last equation serves to define $\Gamma(x)$ when ${\rm Re}(x) > -1$ except at $x = 0$. Then that same equation serves to define $\Gamma(x)$ when ${\rm Re}(x) > -2$ except at $x = -1$, and so on. We get analytic continuation of $\Gamma(x)$ to $\mathbf C$ except at $x = 0, -1, -2, -3, \ldots$ where there are simple poles. This approach to analytic (really, meromorphic) continuation is in many complex analysis books. $\endgroup$– KCdCommented Mar 16, 2019 at 2:31
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The formula for $\Gamma$ function which is valid for all complex values of the argument was introduced by Euler, $$\Gamma(z)=\frac{1}{z}\prod_{n=1}^\infty\left\{\left(1+\frac{1}{n}\right)^z\left(1+\frac{z}{n}\right)^{-1}\right\}.$$ This is how he originally defined it, and then he showed that it is equal to the "Euler's integral". Of course at the time of Euler there was no rigorously defined notion of analytic continuation, but this did not prevent Euler from working with it.
A modern definition using the product (which also defines $\Gamma$ for all complex $z$) is due to Weierstrass: $$\frac{1}{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-z/n},$$ where $\gamma$ is the Euler-Mascheroni constant.
Source: Whittaker-Watson, Chapter 12.
answered Mar 13, 2019 at 14:09