Which methods can be used to evaluate the following integral? How can I evaluate the following integral
$$ \int_{0}^{\infty} x^{-1/2} \exp({-x/2})\ dx $$
I know the answer is $\sqrt{2\pi}$.
 A: $$ \Large{\color{#66f}{\int_{0}^{\infty} x^{-\frac{1}{2}} e^{-\frac{x}{2}} dx}}$$
$$\bbox[8pt,border:2pt solid #66f]{\begin{matrix}{\color{crimson}{\text{Substitute:}}\\ u=x/2 \implies du= \frac{1}{2} dx }\end{matrix}}$$
$$\color{crimson}{-------------------}$$
$$\text{The Integral Becomes:}$$
$$2 \int_0^{\infty} {x}^{-\frac{1}{2}} e^{-u} du$$
$$2\int_0^{\infty} (2u)^{-\frac{1}{2}} e^{-u} du = \sqrt{2} {\int_0^{\infty} u^{-\frac{1}{2}} e^{-u} du}=\sqrt{2} \cdot \color{#ff6600}{\Gamma\left(\frac{1}{2}\right)}$$
Where $\Gamma(n)$ is the Gamma Function
$$\bbox[8pt,border:2pt solid crimson]{\color{#ff6600}{\Gamma\left(\frac{1}{2}\right)}=\sqrt{\pi}}$$
$$(\spadesuit)$$
$$ \Large{\int_{0}^{\infty} x^{-\frac{1}{2}} e^{-\frac{x}{2}} dx=\sqrt{2\pi}}$$
A: The standard elementary proof is set $u = \sqrt{x}$, convert it to the square root of a 2-d integral and then evaluate the 2-d integral in polar coordinates:
$$
\int_0^\infty x^{-1/2} e^{-x/2} dx
= 2\int_0^\infty e^{-u^2/2} du = \int_{-\infty}^\infty e^{-u^2/2} du\\
= \sqrt{ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}  e^{-(u^2+v^2)/2} dudv}
= \sqrt{ 2\pi \int_0^{\infty} e^{-r^2/2} rdr} = \sqrt{2\pi}
$$
A: Let $u = \frac{\sqrt{x}}{\sqrt{2}}$
$$\begin{align}\lim_{t \to \infty}\int_{0}^{t} \frac{1}{\sqrt{x}} e^{-x/2} dx &= \lim _{t \to \infty} 2\sqrt{2}\int_{0}^{\sqrt{t}/\sqrt{2}}  e^{-u^2} du =  \lim _{t \to \infty} \frac{2}{\sqrt{\pi}}\sqrt{2\pi}\int_{0}^{\sqrt{t}/\sqrt{2}}  e^{-u^2} du \\&= \lim_{t\to\infty} \sqrt{2\pi}\  erf\Bigg(\frac{\sqrt{t}}{\sqrt{2}}\Bigg) \end{align}$$
where $erf$ is the Error function.
Alternatively, 
Let $u = \sqrt{x}$ then $2u\ du = dx$ and 
$$\lim_{t \to \infty}\int_{0}^{t} \frac{1}{\sqrt{x}} e^{-x/2} dx = \lim _{t \to \infty} 2\int_{0}^{\sqrt{t}}  u\ e^{-u^2/2} du $$
I'm sure you can take from here. 
