advanced-calculus-prove-that-0-1-e-x-2-x-2-dx-pi-

Question Number 136750 by mnjuly1970 last updated on 25/Mar/21

\dots . . a d v a n c e d c a l c u l u s \dots . .

p r o v e t h a t ::

\dots ϕ = \int_{0}^{\infty} \frac{1 - e^{- x^{2}}}{x^{2}} d x = \sqrt{π}

Answered by Ñï= last updated on 25/Mar/21

\int_{0}^{\infty} \frac{1 - e^{- x^{2}}}{x^{2}} d x = \int_{0}^{\infty} \frac{e^{- 0} - e^{- x^{2}}}{x^{2}} d x = \int_{0}^{\infty} \int_{0}^{1} e^{- {a x}^{2}} d a d x

= \int_{0}^{1} \frac{\sqrt{π}}{2 \sqrt{a}} d a = \sqrt{π}

Commented by mnjuly1970 last updated on 25/Mar/21

t h a n k s a l o t \dots

Answered by Dwaipayan Shikari last updated on 25/Mar/21

η (ϕ) = \int_{0}^{\infty} \frac{e^{- ϕ x^{2}} - e^{- x^{2}}}{x^{2}} d x \Rightarrow η^{'} (ϕ) = - \int_{0}^{\infty} e^{- ϕ x^{2}} d x = - \frac{\sqrt{π}}{2 \sqrt{ϕ}}

η (ϕ) = - \sqrt{π ϕ} + C \Rightarrow η (1) = 0 \Rightarrow C = \sqrt{π}

η (ϕ) = \sqrt{π} (1 - \sqrt{ϕ}) \Rightarrow η (0) = \sqrt{π}

Answered by mathmax by abdo last updated on 25/Mar/21

let f (a) = \int_{0}^{\infty} \frac{e^{- {ax}^{2}} - e^{- x^{2}}}{x^{2}} dx with a > 0 \Rightarrow f^{^{'}} (a) = - \int_{0}^{\infty} e^{- {ax}^{2}} dx

= - \int_{0}^{\infty} e^{- {(\sqrt{a} x)}^{2}} dx =_{\sqrt{a} x = t} - \int_{0}^{\infty} e^{- t^{2}} \frac{dt}{\sqrt{a}} = - \frac{1}{\sqrt{a}} . \frac{\sqrt{π}}{2} \Rightarrow f (a) = \sqrt{π} \int \frac{da}{2 \sqrt{a}} + C

= - \sqrt{π a} + C

f (1) = 0 = - \sqrt{π} + C \Rightarrow C = \sqrt{π} \Rightarrow f (a) = \sqrt{π} - \sqrt{π a}

Φ = f (0) = \sqrt{π}