Consider the gamma function defined by
$$
\Gamma(t)=\int_0^{\infty} x^{t-1} e^{-x} d x, \quad t>0 .
$$
Prove the following:
(a) $\Gamma(t+1)=t \Gamma(t)$ for all $t>0$. (Since $\Gamma(1)=1$, this implies that $\Gamma(n+1)=n !, n=1,2, \cdots)$
(b) Show that $\Gamma(t)$ can be written as
$$
\Gamma(t)=2^{1-t} \int_0^{\infty} z^{2 t-1} e^{-\frac{1}{2} z^2} d z,
$$
for all $t>0$.
(c) Using (b), we can write
$$
\Gamma\left(\frac{1}{2}\right)=\sqrt{2} \int_0^{\infty} e^{-\frac{1}{2} z^2} d z,
$$
and thus,
$$
\begin{aligned}
{\left[\Gamma\left(\frac{1}{2}\right)\right]^2 } & =2\left\{\int_0^{\infty} e^{-\frac{1}{2} x^2} d x\right\}\left\{\int_0^{\infty} e^{-\frac{1}{2} y^2} d y\right\} \\
& =2 \int_0^{\infty} \int_0^{\infty} e^{-\frac{1}{2}\left(x^2+y^2\right)} d x d y .
\end{aligned}
$$
Now use polar coordinates to evaluate the double integral and thereby show that $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$.