Question-149773

Question Number 149773 by Samimsultani last updated on 07/Aug/21

Answered by puissant last updated on 07/Aug/21

I = \int_{0}^{\infty} \frac{c o s (a x)}{x^{2} + b^{2}} d x

= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{c o s (a x)}{x^{2} + b^{2}} d x c a r c o s (x) e s t p a i r e

= \frac{1}{2} R e (\int_{- \infty}^{+ \infty} \frac{e^{a i x}}{x^{2} + b^{2}} d x)

o r ψ = \int_{- \infty}^{+ \infty} \frac{e^{i a x}}{x^{2} + b^{2}} d x = 2 i π R e s (ψ, i b)

(t h e o r e m e d u r e s i d u .) . .

= 2 i π \frac{e^{i a (i b)}}{2 i b}

= \frac{π}{b} e^{- a b}

d^{'} o u ∵ I = \frac{π}{2 b} e^{- a b} \dots .

\dots . L e p u i s s a n t \dots .

Facebook Tweet Pin