= Gamma integral
{c}
{title2=$\int_0^\infty t^{s-1}e^{-at}\,dt=a^{-s}\Gamma(s)$}
For real $a>0$ and $\operatorname{Re}s>0$, the <change of variables> $u=at$ in the defining <improper integral> for the <Gamma function> gives
$$
\int_0^\infty t^{s-1}e^{-at}\,dt=a^{-s}\Gamma(s).
$$
The <improper integral> converges absolutely: near zero its absolute <integrand> behaves like $t^{\operatorname{Re}s-1}$, and at infinity the <exponential function> overcomes every fixed power.https://dlmf.nist.gov/5.2#E1[NIST DLMF, equation 5.2.1: Euler's integral and its domain]{ref}
For real $s=1,2,4$ and $a=1$, the <integrands> are $e^{-t}$, $te^{-t}$, and $t^3e^{-t}$, with total areas $\Gamma(1)=1$, $\Gamma(2)=1$, and $\Gamma(4)=6$. The third <integrand> is taller and has six times the area of the first, although the <exponential function> makes each tail decay.
\Image[/gamma-integral-shape-comparison.png]
{title=Gamma integral kernels for shapes 1, 2, and 4 have total areas 1, 1, and 6 before probability normalization}
For real $s>0$, the kernel $t^{s-1}e^{-t}$ becomes the <probability density function> of a <Gamma distribution> with shape $s$ and unit rate only after division by its <normalizing constant> $\Gamma(s)$. In particular, the shape-four <probability density function> is $t^3e^{-t}/6$. More generally, the <Gamma integral> verifies that $a^st^{s-1}e^{-at}/\Gamma(s)$ integrates to one for shape $s>0$ and rate $a>0$.
For the half-integer evaluation, the <change of variables> $t=u^2$ reduces the <Gamma integral> to the <Gaussian integral>:
$$
\Gamma\left(\frac12\right)=2\int_0^\infty e^{-u^2}\,du=\sqrt\pi.
$$
Consequently, for real $u>0$,
$$
u^{-1/2}=\frac1{\sqrt\pi}\int_0^\infty t^{-1/2}e^{-ut}\,dt.
$$
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