determine-using-laplce-transformation-this-integrale-0-tsin-tx-a-2-t-2-dt-

Question Number 106691 by Laplace last updated on 06/Aug/20

d e t e r m i n e u s i n g l a p l c e t r a n s f o r m a t i o n t h i s

i n t e g r a l e

\int_{0}^{+ \infty} \frac{t s i n (t x)}{a^{2} + t^{2}} d t

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

f_{a} (x) = \int_{0}^{\infty} \frac{tsin (tx)}{a^{2} + t^{2}} dt \Rightarrow f_{a} (x) = \int_{0}^{\infty} tsin (tx) \int_{0}^{\infty} e^{- (a^{2} + t^{2}) x} dx dt

= \int_{0}^{\infty} (\int_{0}^{\infty} \sin (tx) e^{- (a^{2} + t^{2}) x} dx) tdt (by use of fubini)

\int_{0}^{\infty} \sin (tx) e^{- (a^{2} + t^{2}) x} dx = Im (\int_{0}^{\infty} e^{itx - (a^{2} + t^{2}) x} dx) and

\int_{0}^{\infty} e^{(it - (a^{2} + t^{2})) x} dx = {[\frac{1}{it - (a^{2} + t^{2})} e^{(it - (a^{2} + t^{2})) x}]}_{0}^{\infty}

= \frac{- 1}{it - (a^{2} + t^{2})} = \frac{1}{(a^{2} + t^{2}) - it} = \frac{a^{2} + t^{2} + it}{{(a^{2} + t^{2})}^{2} + t^{2}} \Rightarrow

Im (\dots) = \frac{t}{{(a^{2} + t^{2})}^{2} + t^{2}} \Rightarrow

f_{a} (x) = \int_{0}^{\infty} \frac{t^{2}}{{(a^{2} + t^{2})}^{2} + t^{2}} dt = \int_{0}^{\infty} \frac{t^{2}}{(a^{4} + {2 a}^{2} t^{2} + t^{4} + t^{2}} dt

= \int_{0}^{\infty} \frac{t^{2}}{t^{4} + ({2 a}^{2} + 1) t^{2} + a^{4}} dt = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{t^{2}}{t^{4} + ({2 a}^{2} + 1) t^{2} + a^{4}} dt

let φ (z) = \frac{z^{2}}{z^{4} + ({2 a}^{2} + 1) z^{2} + a^{4}} poles of φ ?

Δ = {({2 a}^{2} + 1)}^{2} - {4 a}^{4} = {4 a}^{4} + {4 a}^{2} + 1 - {4 a}^{4} = {4 a}^{2} + 1 (z^{2} = u)

u_{1} = \frac{- ({2 a}^{2} + 1) + \sqrt{{4 a}^{2} + 1}}{2} and u_{2} = \frac{- ({2 a}^{2} + 1) - \sqrt{{4 a}^{2} + 1}}{2}

\Rightarrow φ (z) = \frac{z^{2}}{(z^{2} - u_{1}) (z^{2} - u_{2})} = \frac{z^{2}}{(z - \sqrt{u_{1}}) (z + \sqrt{u_{1}}) (z - \sqrt{u_{2}}) (z + \sqrt{u_{2}})}

\sqrt{u_{1}} = i \sqrt{\frac{{2 a}^{2} + 1 - \sqrt{{4 a}^{2} + 1}}{2}} and \sqrt{u_{2}} = i \sqrt{\frac{{2 a}^{2} + 1 + \sqrt{{4 a}^{2} + 1}}{2}}

\dots be continued \dots

Commented by Ar Brandon last updated on 06/Aug/20

��

Answered by abdomathmax last updated on 07/Aug/20

residus method let I = \int_{0}^{\infty} \frac{tsin (tx)}{t^{2} + a^{2}} dt

\Rightarrow I =_{t =∣ a ∣ u} \int_{0}^{\infty} \frac{∣ a ∣ usin (∣ a ∣ ux)}{a^{2} (u^{2} + 1)} ∣ a ∣ du

= \int_{0}^{\infty} \frac{u \sin (x ∣ a ∣ u)}{u^{2} + 1} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{u \sin (x ∣ a ∣ u)}{u^{2} + 1} du

= \frac{1}{2} Im (\int_{- \infty}^{+ \infty} \frac{u e^{ix ∣ a ∣ u}}{u^{2} + 1} du) let φ (z) = \frac{z e^{ix ∣ a ∣ z}}{z^{2} + 1}

\int_{- \infty}^{+ \infty} φ (z) dx = 2 i π Res (φ, i) = 2 i π \times \frac{{ie}^{- x ∣ a ∣}}{2 i}

= i π e^{- x ∣ a ∣} \Rightarrow I = \frac{π}{2} e^{- x ∣ a ∣}