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Question Number 116806 by Backer last updated on 07/Oct/20

Hi

Prove that :

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

Thanks beforehand

Answered by Bird last updated on 07/Oct/20

l e t A_{ξ} = \int \int_{] - ξ, ξ [^{2}} e^{- x^{2} - y^{2}} d x d y

w e h s v e {l i m}_{ξ \to + \infty} A_{ξ} = {(\int_{- \infty}^{\infty} e^{- x^{2}} d x)}^{2}

l e t u s e t h e d i f f e o m o r p h i s m

{\begin{cases} x = r c o s θ \\ y = r s i n θ w e h a v e - ξ ⩽ x ⩽ ξ \end{cases}

a n d - ξ ⩽ y ⩽ ξ \Rightarrow 0 ⩽ x^{2} + y^{2} ⩽ 2 ξ^{2}

\Rightarrow 0 ⩽ r ⩽ ξ \sqrt{2} \Rightarrow

A_{ξ} = \int_{0}^{ξ \sqrt{2}} r e^{- r^{2}} d r \int_{- π}^{π} d θ

= 2 π {[- \frac{1}{2} e^{- r^{2}}]}_{0}^{ξ \sqrt{2}}

= π (1 - e^{- 2 ξ^{2}}) \Rightarrow {l i m}_{ξ \to + \infty} A_{ξ} = π

= {(\int_{- \infty}^{+ \infty} e^{- x^{2}} d x)}^{2} b u t \int_{- \infty}^{\infty} e^{- x^{2}} d x > 0

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

Answered by bobhans last updated on 07/Oct/20

I = \underset{- \infty}{\int^{\infty}} e^{- x^{2}} dx \Rightarrow I^{2} = \underset{- \infty}{\int^{\infty}} \underset{- \infty}{\int^{\infty}} e^{- x^{2} - y^{2}} dx dy

I^{2} = \underset{0}{\int^{\infty}} \underset{0}{\int^{2 π}} r e^{- r^{2}} d θ dr = \underset{0}{\int^{\infty}} ∣ ({re}^{- r^{2}} (θ)) ∣_{0}^{2 π}) dr

I^{2} = 2 π \underset{0}{\int^{\infty}} {re}^{- r^{2}} dr = π \underset{0}{\int^{\infty}} e^{- r^{2}} d (r^{2})

I^{2} = - π ∣ (e^{- r^{2}}) ∣_{0}^{\infty} = - π (0 - 1) = π

Hence I = \sqrt{π}

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