calculate-f-a-b-0-e-ax-2-x-2-b-2-dx-with-a-gt-0-and-b-gt-0-

Question Number 143082 by Mathspace last updated on 09/Jun/21

c a l c u l a t e f (a, b) = \int_{0}^{\infty} \frac{e^{- {a x}^{2}}}{x^{2} + b^{2}} d x

w i t h a > 0 a n d b > 0

Answered by Dwaipayan Shikari last updated on 09/Jun/21

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

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

= \sqrt{π} e^{a b} \int_{\sqrt{a}}^{\infty} e^{- u^{2}} d u

= \sqrt{π} e^{a b} (\frac{\sqrt{π}}{2} - \frac{\sqrt{π}}{2} e r f (\sqrt{a})) = \frac{π}{2} e^{a b} (e r f c (\sqrt{a}))

Answered by mathmax by abdo last updated on 10/Jun/21

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

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

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

f (a, b) = \int_{\sqrt{a}}^{\infty} e^{- y^{2}} dy . \int_{0}^{\infty} \frac{e^{- b^{2} t}}{\sqrt{t + a}} dt (let λ_{0} = \int_{\sqrt{a}}^{\infty} e^{- y^{2}} dy)

=_{t + a = z^{2}} λ_{0} \int_{\sqrt{a}}^{\infty} \frac{e^{- b^{2} (z^{2} - a)}}{z} (2 z) dz = 2 λ_{0} e^{{ab}^{2}} \int_{\sqrt{a}}^{\infty} e^{- b^{2} z^{2}} dz

=_{bz = u} 2 λ_{0} e^{{ab}^{2}} \int_{b \sqrt{a}}^{\infty} e^{- u^{2}} \frac{du}{b}

= \frac{2 λ_{0}}{b} e^{{ab}^{2}} \int_{b \sqrt{a}}^{\infty} e^{- u^{2}} du