find-0-e-x-2-3-x-2-2-dx-

Question Number 142990 by mathmax by abdo last updated on 08/Jun/21

find \int_{0}^{\infty} \frac{e^{- x^{2}}}{{(3 + x^{2})}^{2}} dx

Answered by qaz last updated on 08/Jun/21

\int_{0}^{\infty} \frac{e^{- x^{2}}}{{(3 + x^{2})}^{2}} dx

= \int_{0}^{\infty} e^{- x^{2}} \int_{0}^{\infty} {ue}^{- u (3 + x^{2})} dudx

= \int_{0}^{\infty} {ue}^{- 3 u} du \int_{0}^{\infty} e^{- (u + 1) x^{2}} dx

= \int_{0}^{\infty} {ue}^{- 3 u} \frac{Γ (\frac{1}{2})}{2 {(u + 1)}^{\frac{1}{2}}} du

= \frac{\sqrt{π}}{2} \int_{0}^{\infty} \frac{{ue}^{- 3 u}}{\sqrt{u + 1}} du

= \frac{e^{3} \sqrt{π}}{2} \int_{1}^{\infty} \frac{(u - 1) e^{- 3 u}}{\sqrt{u}} du

= \frac{e^{3} \sqrt{π}}{2} (\int_{1}^{\infty} \sqrt{u} e^{- 3 u} du - \int_{1}^{\infty} \frac{e^{- 3 u}}{\sqrt{u}} du)

= \frac{e^{3} \sqrt{π}}{2} (- \frac{1}{3} \sqrt{u} e^{- 3 u} ∣_{1}^{\infty} + \frac{1}{3} \int_{1}^{\infty} \frac{e^{- 3 u}}{2 \sqrt{u}} du - \int_{1}^{\infty} \frac{e^{- 3 u}}{\sqrt{u}} du)

= \frac{e^{3} \sqrt{π}}{2} (\frac{1}{3} e^{- 3} - \frac{5}{6} \int_{1}^{\infty} \frac{e^{- 3 u}}{\sqrt{u}} du)

= \frac{e^{3} \sqrt{π}}{2} (\frac{1}{3} e^{- 3} - \frac{5}{6 \sqrt{3}} \int_{3}^{\infty} \frac{e^{- u}}{\sqrt{u}} du)

== \frac{e^{3} \sqrt{π}}{2} (\frac{1}{3} e^{- 3} - \frac{5}{3 \sqrt{3}} \int_{\sqrt{3}}^{\infty} e^{- v^{2}} dv)

= \frac{\sqrt{π}}{6} - \frac{5 \sqrt{3} π e^{3}}{36} erfc (\sqrt{3})

Commented by qaz last updated on 08/Jun/21

\erf (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} dt

erfc (x) = 1 - \erf (x) = \frac{2}{\sqrt{π}} \int_{x}^{\infty} e^{- t^{2}} dt

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

thank you sir .

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

parametric method let f (a) = \int_{0}^{\infty} \frac{e^{- x^{2}}}{x^{2} + a^{2}} \Rightarrow f^{^{'}} (a) = - 2 a \int_{0}^{\infty} \frac{e^{- x^{2}}}{{(x^{2} + a^{2})}^{2}}

\Rightarrow f^{^{'}} (\sqrt{3}) = - 2 \sqrt{3} \int_{0}^{\infty} \frac{e^{- x^{2}}}{{(x^{2} + 3)}^{2}} \Rightarrow \int_{0}^{\infty} \frac{e^{- x^{2}}}{{(x^{2} + 3)}^{2}} dx = - \frac{1}{2 \sqrt{3}} f^{^{'}} (\sqrt{3})

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

= \int_{0}^{\infty} (\int_{0}^{\infty} e^{- (t + 1) x^{2}} dx) e^{- {ta}^{2}} dt but

\int_{0}^{\infty} e^{- (t + 1) x^{2}} dx =_{\sqrt{t + 1} x = u} \int_{0}^{\infty} e^{- u^{2}} \frac{du}{\sqrt{t + 1}} = \frac{\sqrt{π}}{2 \sqrt{t + 1}} \Rightarrow

f (a) = \frac{\sqrt{π}}{2} \int_{0}^{\infty} \frac{e^{- a^{2} t}}{\sqrt{t + 1}} dt but

\int_{0}^{\infty} \frac{e^{- a^{2} t}}{\sqrt{t + 1}} dt =_{\sqrt{t + 1} = z} \int_{1}^{\infty} e^{- a^{2} (z^{2} - 1)} \times \frac{1}{z} (2 z) dz

= {2 e}^{a^{2}} \int_{1}^{\infty} e^{- a^{2} z^{2}} dz =_{az = α} {2 e}^{a^{2}} \int_{1}^{\infty} e^{- α^{2}} \frac{d α}{a}

= \frac{2}{a} e^{a^{2}} \int_{1}^{\infty} e^{- α^{2}} d α let k_{0} = \int_{1}^{\infty} e^{- α^{2}} d α \Rightarrow

f (a) = \frac{\sqrt{π}}{2} \frac{{2 k}_{0}}{a} e^{a^{2}} = \frac{k_{0} \sqrt{π}}{a} e^{a^{2}} \Rightarrow

f^{^{'}} (a) = k_{0} \sqrt{π} {- \frac{1}{a^{2}} e^{a^{2}} + \frac{2 a}{a} e^{a^{2}}} = k_{0} \sqrt{π} {{2 e}^{a^{2}} - \frac{1}{a^{2}} e^{a^{2}}}

= k_{0} \sqrt{π} (2 - \frac{1}{a^{2}}) e^{a^{2}} \Rightarrow f^{^{'}} (\sqrt{3}) = k_{0} \sqrt{π} (2 - \frac{1}{3}) e^{3} \Rightarrow

\int_{0}^{\infty} \frac{e^{- x^{2}}}{{(x^{2} + 3)}^{2}} dx = - \frac{1}{2 \sqrt{3}} . k_{0} \sqrt{π} . \frac{5}{3} e^{3} = - k_{0} \sqrt{π} . \frac{5}{6 \sqrt{3}} e^{3}