Question-130326

Question Number 130326 by rs4089 last updated on 24/Jan/21

Answered by Lordose last updated on 24/Jan/21

Ω (p) = \int_{0}^{\infty} \frac{\sin (px)}{x (x^{2} + 1)} dx

Ω^{'} (p) = \int_{0}^{\infty} \frac{\cos (px)}{x^{2} + 1} dx = \frac{π}{2} e^{- p}

Ω (p) = \frac{π}{2} e^{- p} + C

Ω (0) = \frac{π}{2} = C

Ω (p) = \frac{π}{2} e^{- p} + \frac{π}{2}

Answered by mathmax by abdo last updated on 24/Jan/21

let f (p) = \int_{0}^{\infty} \frac{\sin (px)}{x (x^{2} + 1)} dx \Rightarrow f^{^{'}} (p) = \int_{0}^{\infty} \frac{\cos (px)}{x^{2} + 1} dx

= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{\cos (px)}{x^{2} + 1} dx = \frac{1}{2} Re (\int_{R} \frac{e^{ipx}}{x^{2} + 1} dx)

= \frac{1}{2} Re (2 i π \times \frac{e^{- p}}{2 i}) = \frac{1}{2} Re (π e^{- p}) = \frac{π}{2} e^{- p} \Rightarrow

f (p) = \frac{π}{2} e^{- p} + C

f (0) = 0 = \frac{π}{2} + C \Rightarrow C = - \frac{π}{2} \Rightarrow f (p) = \frac{π}{2} e^{- p} - \frac{π}{2}