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!

\[ \begin{aligned} \therefore \Gamma\left(\frac{1}{2}\right) &= 2\sqrt{\pi} \color{teal}{\left(\frac{1}{2}\right)} = \sqrt{\pi} \\ \end{aligned} \]

\(\Box\)

# posted by rot13(Unafba Pune) @ 8:30 AM