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Question Number 143083 by Mathspace last updated on 09/Jun/21

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

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

Answered by Olaf_Thorendsen last updated on 10/Jun/21

f_{b} (a) = Ψ (a, b) = \int_{0}^{\infty} \frac{e^{- {a x}^{2}}}{{(x^{2} + b^{2})}^{2}} d x

f_{b}^{'} (a) = \frac{\partial Ψ}{\partial a} (a, b) = - \int_{0}^{\infty} \frac{x^{2} e^{- {a x}^{2}}}{{(x^{2} + b^{2})}^{2}} d x

f_{b}^{″} (a) = \frac{\partial^{2} Ψ}{\partial a^{2}} (a, b) = + \int_{0}^{\infty} \frac{x^{4} e^{- {a x}^{2}}}{{(x^{2} + b^{2})}^{2}} d x

f_{b}^{″} (a) - 2 b^{2} f_{b}^{'} (a) + b^{4} f_{b} (a)

= \int_{0}^{\infty} \frac{(x^{4} + 2 b^{2} x^{2} + b^{4}) e^{- {a x}^{2}}}{{(x^{2} + b^{2})}^{2}} d x

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

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

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

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

= \frac{1}{2} \sqrt{\frac{π}{a}}

f_{b}^{″} (a) - 2 b^{2} f_{b}^{'} (a) + b^{4} f_{b} (a) = \frac{1}{2} \sqrt{\frac{π}{a}}

f_{b} (a) = Ψ (a, b) =

(c_{1} + c_{2} a) e^{{a b}^{2}} + \frac{\sqrt{π}}{2} e^{{a b}^{2}} [a Γ (\frac{1}{2}, b^{2} a) + \frac{Γ (\frac{3}{2}, {a b}^{2})}{b^{3}}]

\dots to be continued \dots