Gamma Integral

This section studies a special improper integral of the form $\displaystyle\int_0^\infty e^{-x}x^{n-1}\,dx$, where $n$ is a positive integer.

Preliminary limiting facts. $e^\infty=\lim_{x\to\infty}e^x=\infty$ and $e^{-\infty}=\lim_{x\to\infty}e^{-x}=\dfrac{1}{\lim_{x\to\infty}e^x}=\dfrac1\infty=0$. By L'Hôpital's rule, applied repeatedly, for every positive integer $m$,

— a polynomial can never outrun the decay of $e^{-x}$, which is exactly what makes the integral below converge.

Deriving $\Gamma(n)=(n-1)!$. Applying integration by parts to $I_n=\displaystyle\int_0^\infty e^{-x}x^n\,dx$: $I_n=\big[-x^ne^{-x}\big]_0^\infty-\int_0^\infty(-e^{-x})(nx^{n-1})\,dx=n\int_0^\infty e^{-x}x^{n-1}\,dx=nI_{n-1}$ (the boundary term vanishes at both ends by the limiting fact above and because $x^n=0$ at $x=0$). Iterating, $I_n=n(n-1)I_{n-2}=n(n-1)(n-2)\cdots1\cdot I_0$, and $I_0=\int_0^\infty e^{-x}\,dx=\big[-e^{-x}\big]_0^\infty=1$, so

Definition 9.1 (Gamma integral). $\displaystyle\Gamma(n)=\int_0^\infty e^{-x}x^{n-1}\,dx$, read "gamma of $n$". Re-indexing $I_n=n!$ (with $n\to n-1$) gives

with the recurrence $\Gamma(n+1)=n\,\Gamma(n)$ and base value $\Gamma(1)=\int_0^\infty e^{-x}\,dx=1$.

Extending by substitution (Examples 9.44-9.46). Substituting $t=ax$ ($a>0$) in $\int_0^\infty e^{-ax}x^n\,dx$: $dt=a\,dx$, and $x=0\Rightarrow t=0$, $x=\infty\Rightarrow t=\infty$, so

$\int_0^\infty e^{-ax}x^n\,dx=\dfrac{1}{a^{n+1}}\int_0^\infty e^{-t}t^n\,dt=\dfrac{n!}{a^{n+1}}.$ …