find-0-e-t-1-x-2-1-x-2-dx-with-t-0-

Question Number 140977 by Mathspace last updated on 14/May/21

f i n d \int_{0}^{\infty} \frac{e^{- t (1 + x^{2})}}{1 + x^{2}} d x w i t h t ⩾ 0

Answered by mathmax by abdo last updated on 14/May/21

let f (t) = \int_{0}^{\infty} \frac{e^{- (1 + x^{2}) t}}{1 + x^{2}} dx \Rightarrow f^{^{'}} (t) = - \int_{0}^{\infty} e^{- (1 + x^{2}) t} dx

= - e^{- t} \int_{0}^{\infty} e^{- {tx}^{2}} dx =_{u = \sqrt{t} x} - e^{- t} \int_{0}^{\infty} e^{- u^{2}} \frac{du}{\sqrt{t}}

= - \frac{e^{- t}}{\sqrt{t}} . \frac{\sqrt{π}}{2} = - \frac{\sqrt{π}}{2} \frac{e^{- t}}{\sqrt{t}} \Rightarrow f (t) = - \frac{\sqrt{π}}{2} \int_{0}^{t} \frac{e^{- u}}{\sqrt{u}} du + k

f (0) = \frac{π}{2} = k \Rightarrow f (t) = \frac{π}{2} - \frac{\sqrt{π}}{2} \int_{0}^{t} \frac{e^{- u}}{\sqrt{u}} du (\sqrt{u} = x)

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

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

Answered by qaz last updated on 14/May/21

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

= - \int_{0}^{t} d t \int_{0}^{\infty} e^{- t (1 + x^{2})} d x + \frac{π}{2}

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

= - \int_{0}^{t} e^{- t} \frac{Γ (\frac{1}{2})}{2 t^{\frac{1}{2}}} d t + \frac{π}{2}

= - \frac{\sqrt{π}}{2} \int_{0}^{t} t^{- \frac{1}{2}} e^{- t} d t + \frac{π}{2}

= \frac{π}{2} [1 - e r f (t^{2})]

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