$\int_0^{\infty} \exp(-0.5x^2dx)$ I tried to compute $\int_0^{\infty} \exp(-0.5x^2)dx$ using the polar coordinate method in https://en.wikipedia.org/wiki/Gaussian_integral as follows
$$\begin{align} \int_0^{\infty} \exp(-0.5 x^2)dx \int_0^{\infty} \exp(-0.5 y^2)dy
    &= \int_0^{2\pi} \int_0^{\infty} \exp(-0.5 r^2)rdrd\theta \\     
    &= \int_0^{2\pi} -\exp(-0.5 r^2)\big|_0^{\infty} d\theta \\     
    &= \int_0^{2\pi} 1 d\theta \\     
    &= 2\pi
\end{align}$$
and therefore
$$\int_0^{\infty} \exp(-0.5 x^2)dx = \sqrt{2\pi}$$
But this answer is not correct. I think I applied the first line incorrectly because the lower limits are only at 0. Is there an easy way to adapt this method to this lower limit?
 A: I'd approach this a different way, using the (hopefully) known fact that
$$\int_0^\infty e^{-x^2}dx = \frac{\sqrt \pi}{2}$$
Now notice that
$$-0.5x^2 = - \left( \frac{1}{\sqrt 2} x \right)^2$$
This suggests a certain substitution: $u = x / \sqrt 2$ which gives $dx = \sqrt 2 \; du$. The bounds remain unchanged. Then
$$\int_0^\infty e^{-0.5x^2}dx=\sqrt 2\int_0^\infty e^{-u^2}du=\sqrt 2 \cdot \frac{\sqrt \pi}{2} = \sqrt{ \frac{\pi}{2} }$$

But let's address your method. Firstly, bear in mind the integral on Wikipedia is given over $(-\infty,\infty)$ where yours is over $(0,\infty)$. This changes up the polar substitution a fair bit. Namely, your integral, in effect, should only be over $(0,\pi/2)$ for $\theta$: the Wikipedia double integral is over all of $\Bbb R^2$, whereas yours should be only in the top-right quadrant.
This is somewhat easy to rectify, however. $e^{-x^2}$ is an even function, and whenever $f$ is even over a symmetric interval $(a,a)$,
$$\int_{-a}^a f(x)dx = 2 \int_0^a f(x)dx$$
You could use this property to start at the $(-\infty,\infty)$ integral and convert to yours, if desired; I won't though. So going forward I'll readjust the bounds on $\theta$ accordingly.

Another error is introduced at this step:
$$\int_0^{2\pi} \int_0^{\infty} \exp(-0.5 r^2)rdrd\theta = \int_0^{2\pi} -\exp(-0.5 r^2)\int_0^{\infty} d\theta$$
You need to remember that the $r$ integral is on $(0,\infty)$; the $\theta$ integral is on $(0,\pi/4)$ (once the aforementioned error is fixed). Also, you should have an extra factor of $r$ and $dr$ for the $r$ integral on the right-hand-side. So, if anything, I would start by saying
$$\int_0^{\pi/2} \int_0^{\infty} e^{-0.5 r^2}rdrd\theta = \int_0^\infty re^{-0.5r^2} dr \int_0^{\pi/2} d\theta$$
The $\theta$ integral here becomes $\pi/2$, obviously. The $r$ integral can be solved with the substitution $u=-0.5r^2$ which gives $du = -r\;dr$. Then
$$\int_0^\infty re^{-0.5r^2} dr = - \int_0^{-\infty} e^udu$$
Use the minus sign to swap the integral bounds, and then evaluate it:
$$- \int_0^{-\infty} e^udu = \int_{-\infty}^0 e^u du = e^0 - \lim_{u \to -\infty} e^u = 1 - 0 = 1$$

Thus, the product of the two integrals is $\pi/2$. Thus, you have shown (aside from some rigor justifying the separation and combination of integrals as you've done them),
$$\left( \int_0^\infty e^{-0.5x^2}dx \right)^2 = \frac{\pi}{2} \implies \int_0^\infty e^{-0.5x^2}dx = \sqrt{ \frac \pi 2 }$$
For what it's worth, you can also see a touch on how to make these properties' use rigorous, and a diagram exemplifying why we use $\theta \in (0, \pi/2)$ instead of $\theta \in (0, 2\pi)$ here on ProofWiki.
A: If
$$\int{e^{- \frac{x^{2}}{2}} d x}=\frac{\sqrt{2\pi} \operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)}}{2}$$
then, being $\text{Gauss error function}$ called $\text{erf}$ we have:
$$\int_{0}^{\infty}\left( e^{- \frac{x^{2}}{2}} \right)dx=\lim_{x \to \infty}\left[\left(\frac{\sqrt{2\pi} \operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)}}{2}\right)-\left(\frac{\sqrt{2\pi} \operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)}}{2}\right)\right]_{\left(x=0\right)}=\frac{\sqrt{2\pi}}{2}$$
