En-utilisant-la-transforme-de-laplace-Calculer-0-tsin-xt-a-2-t-2-dt-a-x-R-

Question Number 144057 by lapache last updated on 21/Jun/21

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

C a l c u l e r

\int_{0}^{+ \infty} \frac{t s i n (x t)}{a^{2} + t^{2}} d t \forall a, x \in R^{*}

Answered by qaz last updated on 21/Jun/21

I (x) = \int_{0}^{\infty} \frac{tsin (xt)}{a^{2} + t^{2}} dt

L (I (x)) (s) = \int_{0}^{\infty} \int_{0}^{\infty} \frac{tsin (xt)}{a^{2} + t^{2}} e^{- sx} dtdx

= \int_{0}^{\infty} \frac{t}{a^{2} + t^{2}} \cdot L (\sin (xt)) (s) dt

= \int_{0}^{\infty} \frac{t}{a^{2} + t^{2}} \cdot \frac{t}{s^{2} + t^{2}} dt

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

= \frac{1}{s^{2} - a^{2}} ({stan}^{- 1} \frac{t}{s} - {atan}^{- 1} \frac{t}{a}) ∣_{0}^{\infty}

= \frac{1}{s^{2} - a^{2}} \cdot \frac{π}{2} (s - a)

= \frac{π}{2 (s + a)}

I (x) = L^{- 1} (\frac{π}{2 (s + a)}) = \frac{π}{2} e^{- ax}

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

residus method Φ = \int_{0}^{\infty} \frac{tsin (xt)}{a^{2} + t^{2}} dt \Rightarrow

Φ =_{t = au} \int_{0}^{\infty} \frac{ausin (xau)}{a^{2} (1 + u^{2})} (adu) = \int_{0}^{\infty} \frac{usin (xau)}{u^{2} + 1} du (we suppose a > 0)

Φ = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{{ue}^{ixau}}{u^{2} + 1} du let φ (z) = \frac{{ze}^{ixaz}}{z^{2} + 1} \Rightarrow φ (z) = \frac{{ze}^{ixaz}}{(z - i) (z + i)}

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

Φ = \frac{π}{2} e^{- xa} if a < 0 we do the changement t = - au

generally we get Φ = \frac{π}{2} e^{- x ∣ a ∣}