I was reading about the Gamma function, however, I have some trouble to figure out why the domain of

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

How can I get $ \alpha > -1$?

  • 1
    $\begingroup$ Actually, $\alpha>0$. $\endgroup$ – Mark Viola Jun 5 '18 at 2:41
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    $\begingroup$ maybe because if you allow $\alpha<0$ then the integral does not converge due to singularity at $x=0$ $\endgroup$ – Frank Moses Jun 5 '18 at 2:48
  • $\begingroup$ en.wikipedia.org/wiki/Gamma_function $\endgroup$ – Dzoooks Jun 5 '18 at 3:11

$$ \int_0^\infty x^{\alpha-1} e^{-x} \,dx \le \int_0^1 x^{\alpha-1} \,dx + \int_1^\infty x^{\alpha-1} e^{-x}\,dx. $$ The first integral on the right converges if $\alpha>0.$ $$ \int_0^\infty x^{\alpha-1} e^{-x} \, dx \ge \frac 1 e \int_0^1 x^{\alpha-1} \,dx + \int_1^\infty x^{\alpha-1} e^{-x} \, dx. $$ The first integral on the right diverges to $+\infty$ if $\alpha<0.$

The second integral on the right can be shown to converge if $\alpha>0$ by going through the usual integration by parts.

By using the functional equation $\Gamma(\alpha+1) = \alpha\Gamma(\alpha),$ you can extend the domain to negative non-integers, but then it's not equal to the integral. By analytic continuation, you can extend it to complex numbers.


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