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## The Gaussian integral

### Solution

Consider the double integrals: $$\int_{0}^{\infty} \int_{0}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d x d y$$.
Looking at the limits of integration, we see that the first integration is along a vertical strip extending from $$y=0$$ to $$y=\infty$$. The second integration is along an horizontal strip from $$x=0$$ to $$x=\infty$$. The next figure shows the region of integration
By changing to polar co-ordinates, this region can be covered by radial strips going from $$r=0$$ to $$r=\infty$$. The strip begins from $$\theta=0$$ and goes upto $$\theta=\frac{\pi}{2}$$
Therefore, \begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d x d y &=\int_{0}^{ \frac{\pi}{2}} \int_{0}^{\infty} e^{-r^{2}} r d r d \theta \\\\ &=-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty}(-2 r) e^{-r^{2}} d r d \theta\\\\&=-\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left[e^{-r^{2}}\right]_{0}^{\infty} d \theta \\\\ &=-\frac{1}{2} \int_{0}^{\frac{\pi}{2}}(0-1) d \theta\\\\&=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 d \theta\\\\&=\frac{\pi}{4} \end{aligned}
Let, $$I=\int_{0}^{\infty} e^{-x^{2}} d x \;\;\;\;\;\;\; (1)$$
Also, $$I=\int_{0}^{\infty} e^{-y^{2}} d y \;\;\;\;\;\;\; (2)$$
By multiplying $$(1)$$ and $$(2)$$, \begin{aligned} &I^{2}=\int_{0}^{\infty} \int_{0}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d x d y=\frac{\pi}{4} \\\\ &I=\sqrt{\frac{\pi}{4}}=\frac{\sqrt{\pi}}{2} \end{aligned}
$\large \Rightarrow\int_{0}^{\infty} e^{-x^{2}} d x=\frac{\sqrt{\pi}}{2}$
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