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How can I evaluate the following integral

$$ \int_{0}^{\infty} x^{-1/2} \exp({-x/2})\ dx $$

I know the answer is $\sqrt{2\pi}$.

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    $\begingroup$ Hint: Let $u=\sqrt{x}$ and write the exponent as $-(\sqrt{x})^2/2$ to translate the integral to an integral involving the standard normal. $\endgroup$ – Mark McClure Feb 3 '15 at 14:55
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    $\begingroup$ @MarkMcClure Thanks, I was able to do the rest. $\endgroup$ – Vesnog Feb 3 '15 at 15:11
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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.

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  • $\begingroup$ Thanks for pointing out the most general solution. $\endgroup$ – Vesnog Feb 3 '15 at 15:22
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    $\begingroup$ You're welcome! I thought of the indefinite integral first. $\endgroup$ – Aaron Maroja Feb 3 '15 at 15:24
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$$ \Large{\color{#66f}{\int_{0}^{\infty} x^{-\frac{1}{2}} e^{-\frac{x}{2}} dx}}$$

$$\bbox[8pt,border:2pt solid #66f]{\begin{matrix}{\color{crimson}{\text{Substitute:}}\\ u=x/2 \implies du= \frac{1}{2} dx }\end{matrix}}$$

$$\color{crimson}{-------------------}$$ $$\text{The Integral Becomes:}$$

$$2 \int_0^{\infty} {x}^{-\frac{1}{2}} e^{-u} du$$

$$2\int_0^{\infty} (2u)^{-\frac{1}{2}} e^{-u} du = \sqrt{2} {\int_0^{\infty} u^{-\frac{1}{2}} e^{-u} du}=\sqrt{2} \cdot \color{#ff6600}{\Gamma\left(\frac{1}{2}\right)}$$

Where $\Gamma(n)$ is the Gamma Function

$$\bbox[8pt,border:2pt solid crimson]{\color{#ff6600}{\Gamma\left(\frac{1}{2}\right)}=\sqrt{\pi}}$$ $$(\spadesuit)$$ $$ \Large{\int_{0}^{\infty} x^{-\frac{1}{2}} e^{-\frac{x}{2}} dx=\sqrt{2\pi}}$$

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    $\begingroup$ See $(\spadesuit)$ $\endgroup$ – The Artist Feb 3 '15 at 16:00
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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:

$$ \int_0^\infty x^{-1/2} e^{-x/2} dx = 2\int_0^\infty e^{-u^2/2} du = \int_{-\infty}^\infty e^{-u^2/2} du\\ = \sqrt{ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(u^2+v^2)/2} dudv} = \sqrt{ 2\pi \int_0^{\infty} e^{-r^2/2} rdr} = \sqrt{2\pi} $$

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