R-e-its-s-3-ds-

Question Number 143190 by Ar Brandon last updated on 11/Jun/21

\int_{R} \frac{e^{its}}{s + 3} ds

Commented by mohammad17 last updated on 11/Jun/21

w = \frac{1}{s + 3} \Rightarrow s = \frac{1}{w} - 3 \Rightarrow d s = - \frac{1}{w^{2}} d w

\int {w e}^{i t (\frac{1}{w} - 3)} (- \frac{1}{w^{2}}) d w = \frac{1}{e^{3 i t}} \int {w e}^{(\frac{i t}{w})} (- \frac{1}{w^{2}}) d w

b y b a r t :: u = w \to d u = d w, d v = e^{(\frac{i t}{w})} (- \frac{1}{w^{2}}) \to v = \frac{1}{i t} e^{(\frac{i t}{w})}

= \frac{1}{{i t e}^{3 i t}} {w e}^{(\frac{i t}{w})} - \frac{1}{{i t e}^{3 i t}} \int e^{(\frac{i t}{w})} d w

= \frac{1}{{i t e}^{3 i t}} {w e}^{(\frac{i t}{w})} - \frac{1}{{i t e}^{3 i t}} {w e}^{(\frac{i t}{w})} + \frac{1}{e^{3 i t}} E i (\frac{i t}{w}) + C

\int_{R} \frac{e^{i t s}}{s + 3} d s = \frac{1}{e^{3 i t}} E i (i t (s + 3)) + C

⟨ m o h a m m a d a l d o l a i m y ⟩

Commented by Ar Brandon last updated on 11/Jun/21

Thanks Sir . Can you insert

the limits and evaluate ?

Answered by mathmax by abdo last updated on 11/Jun/21

Φ = \int_{- \infty}^{+ \infty} \frac{e^{itx}}{x + 3} dx = \int_{- \infty}^{0} \frac{e^{itx}}{x + 3} dx + \int_{0}^{\infty} \frac{e^{itx}}{x + 3} dx but

\int_{- \infty}^{0} \frac{e^{itx}}{x + 3} dx =_{x = - u} \int_{0}^{\infty} \frac{e^{- itu}}{3 - u} du = \int_{0}^{\infty} \frac{e^{- itx}}{3 - x} dx \Rightarrow

Φ = \int_{0}^{\infty} \frac{e^{itx}}{x + 3} dx + \int_{0}^{\infty} \frac{e^{- itx}}{3 - x} dx [we have

\int_{0}^{\infty} \frac{e^{itx}}{x + 3} dx = \int_{0}^{\infty} (\int_{0}^{\infty} e^{- (x + 3) z} dz) e^{itx} dx

= \int_{0}^{\infty} (\int_{0}^{\infty} e^{(- z + it) x} dx) e^{- 3 z} dz

= \int_{0}^{\infty} ({[\frac{1}{- z + it} e^{(- z + it) x}]}_{0}^{\infty}) e^{- 3 z} dz = \int_{0}^{\infty} (\frac{1}{- z + it}) e^{- 3 z} dz

= - \int_{0}^{\infty} (\frac{1}{z - it}) e^{- 3 z} dz = - \int_{0}^{\infty} \frac{(z + it) e^{- 3 z}}{z^{2} + t^{2}} dz

\int_{0}^{\infty} \frac{e^{- itx}}{3 - x} dx = \int_{0}^{\infty} (\int_{0}^{\infty} e^{- (3 - x) z} dz) e^{- itx} dx

= \int_{0}^{\infty} (\int_{0}^{\infty} e^{(z - it) x} dx) e^{- 3 z} dz

= \int_{0}^{\infty} ({[\frac{1}{z - it} e^{(z - it) x}]}_{0}^{\infty}) e^{- 3 z} dz

= \int_{0}^{\infty} (- \frac{1}{z - it}) e^{- 3 z} dz = - \int_{0}^{\infty} \frac{(z + it) e^{- 3 z}}{z^{2} + t^{2}} dz \Rightarrow

Φ = - \int_{0}^{\infty} \frac{(z + it) e^{- 3 z}}{z^{2} + t^{2}} dz - \int_{0}^{\infty} \frac{(z + it) e^{- 3 z}}{z^{2} + t^{2}} dz

= - 2 \int_{0}^{\infty} \frac{(z + it) e^{- 3 z}}{z^{2} + t^{2}} dz \dots . be continued \dots .