calculate-0-tsin-2t-t-2-4-dt-

Question Number 106899 by abdomathmax last updated on 07/Aug/20

calculate \int_{0}^{\infty} \frac{tsin (2 t)}{t^{2} + 4} dt

Answered by mathmax by abdo last updated on 08/Aug/20

A = \int_{0}^{\infty} \frac{t \sin (2 t)}{t^{2} + 4} dt \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{tsin (2 t)}{t^{2} + 4} dt

= Im (\int_{- \infty}^{+ \infty} \frac{t e^{2 it}}{t^{2} + 4} dt) let φ (z) = \frac{z e^{2 iz}}{z^{2} + 4} \Rightarrow φ (z) = \frac{z e^{2 iz}}{(z - 2 i) (z + 2 i)}

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, 2 i) = 2 i π \times \frac{{2 ie}^{2 i (2 i)}}{4 i} = i π e^{- 4} \Rightarrow

2 A = π e^{- 4} \Rightarrow A = \frac{π}{2} e^{- 4}