show-thats-true-e-x-2-pi-

Question Number 8490 by fernandodantas1996 last updated on 14/Oct/16

s h o w t h a t s t r u e :

\int_{- \infty}^{+ \infty} e^{- x^{2}} = \sqrt{π}

Commented by prakash jain last updated on 14/Oct/16

Y o u n e e d t o d e f i n e l i m i t s .

Commented by fernandodantas1996 last updated on 23/Oct/16

Answered by prakash jain last updated on 23/Oct/16

{(\int_{- \infty}^{\infty} e^{- x^{2}})}^{2} = \int_{- \infty}^{\infty} e^{- x^{2}} d x \int_{- \infty}^{\infty} e^{- y^{2}} d y

= \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- (x^{2} + y^{2})} d x d y

Convert to polar coordinates

x = r \cos θ

y = r \sin θ

d x d y = r d r d θ

\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- (x^{2} + y^{2})} d x d y

= \int_{0}^{2 π} \int_{0}^{\infty} e^{- r^{2}} r d r d θ

= \int_{0}^{2 π} d θ \int_{0}^{\infty} e^{- r^{2}} r d r

subtitute s = - r^{2} \Rightarrow d s = - 2 r d r

= \int_{0}^{2 π} d θ \int_{0}^{- \infty} - \frac{1}{2} e^{s} d s

= 2 π (\frac{1}{2}) (- [e^{- \infty} - e^{0}])

= π

I = \int_{- \infty}^{\infty} e^{- x^{2}} d x

I^{2} = π \Rightarrow I = \int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}