Evaluate Gaussian Integral $$\int_0^\infty e^{-x^2}dx$$ I'm wondering if there are any interesting method to solve gaussian integral.
I'll write a few things I know below. Welcome everyone to correct the mistakes (if there are any) and offer greater solutions.
Evaluate Gaussian Integral $$\int_0^\infty e^{-x^2}dx$$
let $I=\int_0^\infty e^{-x^2}dx$, we have $$I^2=(\int_0^\infty e^{-x^2}dx)^2=\int_0^\infty e^{-x^2}dx\cdot\int_0^\infty e^{-y^2}dy$$ Using Fubini's theorem, $$I^2=\iint_{[0,\infty]\times[0,\infty]} e^{-(x^2+y^2)}d(x,y)$$ To convert it into polar system, we let $${x=rcos\theta}\quad{y=rsin\theta}$$ Then, calculate the determinant of the Jacobian matrix $$|J|=\bigg|\matrix{{\frac{dx}{dr}}{\frac{dx}{d\theta}}\\{\frac{dy}{dr}}{\frac{dy}{d\theta}}}\bigg|=\bigg|\matrix{{cos\theta}\quad{-rsin\theta}\\{sin\theta}\quad{rcos\theta}}\bigg|=r$$ Thus, $$I^2=\int_0^{\frac\pi2}\int_0^\infty re^{-r^2}drd\theta=-\frac12\int_0^{\frac\pi2}[e^{-r^2}]_0^\infty d\theta=\frac12[\theta]_0^{\frac\pi2}=\frac\pi4$$ $$I=\frac{\sqrt{\pi}}2$$
Let $$f(x)=(\int_0^xe^{-t^2}dt)^2\quad g(x)=\int_0^1\frac{e^{-x^2(t^2+1)}}{1+t^2}dt$$ $$f'(x)=2e^{-x^2}\int_0^xe^{-t^2}dt$$ $$g'(x)=-2xe^{-x^2}\int_0^1e^{-x^2t^2}dt$$ Using substitution, let $s=xt$ $$g'(x)=-2e^{-x^2}\int_0^xe^{-s^2}ds=-f'(x)$$ Consider the derivative of the sum of the two functions: $$[f(x)+g(x)]'=f'(x)+g'(x)=0$$ $$\Rightarrow f(x)+g(x)=f(0)+g(0)=0+\int_0^1\frac1{t^2+1}dt=[arctant]_0^1=\frac\pi4$$ When $x\to\infty$ $$\frac{e^{-x^2(t^2+1)}}{1+t^2}\to0$$ for $t\in[0,1]$ $$\Rightarrow g(x)\to0$$ $$f(\infty)=\frac\pi4-g(\infty)=\frac\pi4$$ $$\Rightarrow(\int_0^\infty e^{-x^2}dx)^2=\frac\pi4$$ $$\Rightarrow\int_0^\infty e^{-x^2}dx=\frac{\sqrt\pi}2$$