0-Cos-ax-x-2-b-2-dx-

Question Number 101970 by Rohit@Thakur last updated on 05/Jul/20

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

Answered by mathmax by abdo last updated on 06/Jul/20

let I = \int_{0}^{\infty} \frac{\cos (ax)}{x^{2} + b^{2}} dx for b > 0 we do the changement x = bt \Rightarrow

I = \int_{0}^{\infty} \frac{\cos (abt)}{b^{2} (1 + t^{2})} bdt = \frac{1}{b} \int_{0}^{\infty} \frac{\cos (abt)}{t^{2} + 1} dt = \frac{1}{2 b} \int_{- \infty}^{+ \infty} \frac{\cos (abt)}{t^{2} + 1} dt

= \frac{1}{2 b} Re (\int_{- \infty}^{+ \infty} \frac{e^{iabt}}{t^{2} + 1} dt) let φ (z) = \frac{e^{iabz}}{z^{2} + 1} \Rightarrow φ (z) = \frac{e^{iabz}}{(z - i) (z + i)}

residus tbeorem give \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, i)

= 2 i π \times \frac{e^{iab (i)}}{2 i} = π e^{- ab} \Rightarrow I = \frac{π}{2 b} e^{- ab}