gamma function question relating to normal distribution

Question

I'm trying to show that $\Gamma(1/2)=\sqrt\pi$. A hint I've been given is to use a change of variable and then relate it to normal distribution density. However, I'm really confused as to how I would go about it this way.

Thanks for the help.

Answer 1

$$ \Gamma(1/2) = \int_0^ \infty e^{-t} \frac{ dt}{\sqrt{t}} $$

Now with $du = \frac{ dt}{\sqrt{t}}$, you find that $$ u = 2\sqrt{t} \implies t = u^ 2/4 $$ Let $\sigma^2 = 2$. $$ \Gamma(1/2) = \int_0^ \infty e^{-u^2/2\sigma^2} du = \sqrt{2\pi \sigma^ 2}/2 = \sqrt{\pi} $$

Answer 2

Proof 1: using the Reflection Formula

$$\Gamma\left(\frac12\right)^2=\Gamma\left(\frac12\right)\Gamma\left(1-\frac12\right)=\frac\pi{\sin\frac\pi2}=\pi\implies\Gamma\left(\frac12\right)=\sqrt\pi$$

Proof 2: With Normal Dist. and stuff:

$$\Gamma(t):=\int\limits_0^\infty x^{1-t}e^{-x}dx\stackrel{x=u^2}=2\int\limits_0^\infty u^{2t-1}e^{-u^2}du\implies$$

$$\Gamma\left(\frac12\right)=2\int\limits_0^\infty e^{-u^2}du=\sqrt\pi$$