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Question Number 142989 by mathmax by abdo last updated on 08/Jun/21

calculate \int_{0}^{\infty} \frac{e^{- {3 x}^{2}}}{1 + x^{2}} dx

Answered by Dwaipayan Shikari last updated on 08/Jun/21

\int_{0}^{\infty} e^{- 3 x^{2}} \int_{0}^{\infty} e^{- u (1 + x^{2})} d x d u

= \frac{\sqrt{π}}{2} \int_{0}^{\infty} \frac{e^{- u}}{\sqrt{u + 3}} d u u + 3 = t^{2}

= e^{3} \sqrt{π} \int_{\sqrt{3}}^{\infty} e^{- t^{2}} d t

= e^{3} \sqrt{π} (\int_{0}^{\infty} e^{- t^{2}} d t - \int_{0}^{\sqrt{3}} e^{- t^{2}} d t)

N o w \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} d x = e f r (x)

= e^{3} \frac{π}{2} - e^{3} \frac{e r f (\sqrt{3})}{2} π = \frac{e^{3} π}{2} (1 - e r f (\sqrt{3}))

Answered by mathmax by abdo last updated on 09/Jun/21

Φ = \int_{0}^{\infty} \frac{e^{- {3 x}^{2}}}{1 + x^{2}} dx \Rightarrow Φ = \int_{0}^{\infty} (\int_{0}^{\infty} e^{- t (1 + x^{2})} dt) e^{- {3 x}^{2}} dx

= \int_{0}^{\infty} (\int_{0}^{\infty} e^{- (t + 3) x^{2}} dx) e^{- t} dt but

\int_{0}^{\infty} e^{- (t + 3) x^{2}} dx =_{\sqrt{t + 3} x = y} \int_{\sqrt{3}}^{\infty} e^{- y^{2}} \frac{dy}{\sqrt{t + 3}} = \frac{1}{\sqrt{t + 3}} \int_{\sqrt{3}}^{\infty} e^{- y^{2}} dy \Rightarrow

Φ = \int_{\sqrt{3}}^{\infty} e^{- y^{2}} dy \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t + 3}} dt and

\int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t + 3}} dt =_{\sqrt{t + 3} = z} \int_{\sqrt{3}}^{\infty} \frac{e^{- (z^{2} - 3)}}{z} (2 z) dz = {2 e}^{3} \int_{\sqrt{3}}^{\infty} e^{- z^{2}} dz \Rightarrow

Φ = {2 e}^{3} {(\int_{\sqrt{3}}^{\infty} e^{- y^{2}} dy)}^{2} let λ_{0} = \int_{\sqrt{3}}^{\infty} e^{- y^{2}} dy \Rightarrow

Φ = {2 e}^{3} λ_{0}^{2}