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The Gaussian integral


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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