Question

Prove the following!!!

$\displaystyle \int\limits_{- \infty}^{\infty} e^{-x^2} \, dx =\sqrt{\pi}$
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Answer 1

This is an old favorite of mine.

Define

I=∫

+∞

−∞

e−x2dx

Then

I2=(∫

+∞

−∞

e−x2dx)(∫

+∞

−∞

e−y2dy)

I2=∫

+∞

−∞

∫

+∞

−∞

e−(x2+y2)dxdy

Now change to polar coordinates

I2=∫

+2π

0

∫

+∞

0

e−r2rdrdθ

The θ integral just gives 2π, while the r integral succumbs to the substitution u=r2

I2=2π∫

+∞

0

e−udu/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.

Answer 2

Answer:

hi

Step-by-step explanation:

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