Proving ∫∞0e−x2dx=√π2\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \frac{\sqrt \pi}{2}

How to prove
0ex2dx=π2

Answer

This is an old favorite of mine.
Define I=+ex2dx
Then I2=(+ex2dx)(+ey2dy)
I2=++e(x2+y2)dxdy
Now change to polar coordinates
I2=+2π0+0er2rdrdθ
The θ integral just gives 2π, while the r integral succumbs to the substitution u=r2
I2=2π+0eudu/2=π
So I=π and your integral is half this by symmetry

I have always wondered if somebody found it this way, or did it first using complex variables and noticed this would work.

Attribution
Source : Link , Question Author : Jichao , Answer Author : Ross Millikan

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