Have you ever wondered why
| |
\[
\begin{aligned}
\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}
\end{aligned}
\] |
where
| |
\[
\begin{aligned}
\Gamma(t) = \int_0^\infty x^{t-1} e^{-x} dx
\end{aligned}
\] |
?
Turns out it's rather straight forward. Let's start with
| |
\[
\begin{aligned}
\\
\Gamma\left(\frac{1}{2}\right) &= \int_0^\infty x^{-\frac{1}{2}} e^{-x} dx = \color{blue}{\int_0^\infty \frac{e^{-x}}{\sqrt{x}} dx} \\
\end{aligned}
\] |
Doesn't this look a little like the standard normal ?
| |
\[
\begin{aligned}
\\
\phi(x) = \frac{e^{-x^2/2}}{\sqrt{2\pi}} \\
\end{aligned}
\] |
Let
| |
\[
\begin{aligned}
\\
x &= y^2 / 2 \\
\therefore dx &= y\, dy \\
\end{aligned}
\] |
| |
\[
\begin{aligned}
\Gamma\left(\frac{1}{2}\right) &= \color{blue}{\int_0^\infty \frac{e^{-y^2/2}}{\sqrt{y^2 / 2}} y\, dy} \\
&= \sqrt{2} \int_0^\infty e^{-y^2/2}\, dy \\
&= 2\sqrt{\pi} \color{teal}{\int_0^\infty \frac{e^{-y^2/2}}{\sqrt{2\pi}}\, dy} \\
\end{aligned}
\] |
But \(\displaystyle\int_0^\infty \frac{e^{-y^2/2}}{\sqrt{2\pi}}\, dy = \color{teal}{\int_0^\infty \phi(y)\, dy}\), which is exactly half of the total area under the standard normal!
| |
\[
\begin{aligned}
\therefore \Gamma\left(\frac{1}{2}\right) &= 2\sqrt{\pi} \color{teal}{\left(\frac{1}{2}\right)} = \sqrt{\pi} \\
\end{aligned}
\] |
\(\Box\)
