Q. 5.21 - A First Course in Probability

Question

Show that $\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$

Hint: $\Gamma \left(\frac{1}{2}\right)={\int }_0^{\infty }{e}^{-x}{x}^{-1/2}dx$ Make the change of variables $y=\sqrt{2x}$ and then relate the resulting expression to the normal distribution.

Step-by-Step Solution

Answer

Using the proper substitution, find the density of Standard Normal under the integral by expressing $\Gamma \left(\frac{1}{2}\right)$the condition.

Substitutions of variables.

We have that

$\Gamma \left(\frac{1}{2}\right)={\int }_0^{\infty }{x}^{\frac{1}{2}-1}{e}^{-\mathrm{mtight}>\mathrm{x}}dx={\int }_0^{\infty }{x}^{-\frac{1}{2}}{e}^{-x}dx$

Now, make the substitution $x=\frac{y^2}{2}$which implies that $dx=ydy$and $y=\sqrt{2x}$

We have that

${\int }_0^{\infty }{x}^{-\frac{1}{2}}{e}^{-x}dx={\int }_0^{\infty }\frac{\sqrt{2}}{y}{e}^{-\frac{y^2}{2}}ydy=\sqrt{2}{\int }_0^{\infty }{e}^{-\frac{y^2}{2}}dy$

Evaluate the integral.

In order to evaluate the remaining integral, observe that ${\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{y^2}{2}}dy=\frac{1}{2}$since the function under the integral is the density of Standard Normal and we integrate it over the positive half of the real line.

So, we have that

$\sqrt{2}{\int }_0^{\infty }{e}^{-\frac{y^2}{2}}dy=\sqrt{2}·\sqrt{2\pi }{\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{y^2}{2}}dy=\sqrt{2}·\sqrt{2\pi }·\frac{1}{2}=\sqrt{\pi }$

So, we have proved that

$\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }$