How to prove the value of this integral?

suchith

[itex]\int^{∞}_{-∞} e^{-x^{2}}dx[/itex] = [itex]\frac{\sqrt{\pi}}{2}[/itex]

pwsnafu

Ah, that one. Hint: consider
##\int_{-\infty}^{\infty} \int_{-\infty}^\infty e^{-x^2-y^2}\, dx \, dy##

HallsofIvy

That should be found in pretty much any Calculus text. Look at the integral pwsnafu suggests. Note that, by symmetry, that is [itex]4\int_0^\infty\int_0^\infty e^{-x^2- y^2} dx dy[/itex], over the first quadrant. That can be converted into a "doable" integral by changing to polar coordinates.

suchith

Quote by HallsofIvy

That should be found in pretty much any Calculus text. Look at the integral pwsnafu suggests. Note that, by symmetry, that is [itex]4\int_0^\infty\int_0^\infty e^{-x^2- y^2} dx dy[/itex], over the first quadrant. That can be converted into a "doable" integral by changing to polar coordinates.

I don't know Double integrals. Is it possible to prove the result only using Single variable calculus? At first I tried Integration by parts, but I failed :(

Isn't it possible to integrate using limit of sums and symmetry with suitable manipulations? I tried to sum directly but failed :(

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suchith	How to prove the value of this integral? [itex]\int^{∞}_{-∞} e^{-x^{2}}dx[/itex] = [itex]\frac{\sqrt{\pi}}{2}[/itex]

pwsnafu	Ah, that one. Hint: consider ##\int_{-\infty}^{\infty} \int_{-\infty}^\infty e^{-x^2-y^2}\, dx \, dy##

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	That should be found in pretty much any Calculus text. Look at the integral pwsnafu suggests. Note that, by symmetry, that is [itex]4\int_0^\infty\int_0^\infty e^{-x^2- y^2} dx dy[/itex], over the first quadrant. That can be converted into a "doable" integral by changing to polar coordinates.