calculate-0-x-sin-2x-x-2-4-dx-

Question Number 35217 by abdo.msup.com last updated on 16/May/18

c a l c u l a t e \int_{0}^{\infty} \frac{x s i n (2 x)}{x^{2} + 4} d x

Commented by prof Abdo imad last updated on 20/May/18

l e t p u t I = \int_{0}^{\infty} \frac{x s i n (2 x)}{x^{2} + 4} d x

2 I = \int_{- \infty}^{+ \infty} \frac{x s i n (2 x)}{x^{2} + 4} d x = I m (\int_{- \infty}^{+ \infty} \frac{{x e}^{i 2 x}}{x^{2} + 4}) d x

l e t c o n s i d e r t h e c o m p o l e x f i n c t i o n

φ (z) = \frac{z e^{i 2 z}}{z^{2} + 4}

φ (z) = \frac{z e^{i 2 z}}{(z - 2 i) (z + 2 i)} s o t h e p o l e s o f φ a r e

2 i a n d - 2 i

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, 2 i)

R e s (φ, 2 i) = {l i m}_{z \to 2 i} (z - 2 i) φ (z)

= \frac{(2 i) e^{2 i (2 i)}}{4 i} = \frac{1}{2} e^{- 4} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{e^{- 4}}{2} = i π e^{- 4} \Rightarrow

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

★ I = \frac{π}{2} e^{- 4} ★