# Laplace Transform of t by Sine a t

## Theorem

Let $\sin$ denote the real sine function.

Let $\laptrans f$ denote the Laplace transform of a real function $f$.

Then:

$\laptrans {t \sin a t} = \dfrac {2 a s} {\paren {s^2 + a^2}^2}$

## Proof 1

 $\ds \laptrans {t \sin a t}$ $=$ $\ds -\map {\dfrac \d {\d s} } {\laptrans {\sin a t} }$ Derivative of Laplace Transform $\ds$ $=$ $\ds -\map {\dfrac \d {\d s} } {\dfrac a {s^2 + a^2} }$ Laplace Transform of Sine $\ds$ $=$ $\ds \dfrac {2 a s} {\paren {s^2 + a^2}^2}$ Quotient Rule for Derivatives

$\blacksquare$

## Proof 2

We have:

 $\ds \laptrans {\cos a t}$ $=$ $\ds \int_0^\infty e^{-s t} \cos a t \rd t$ Definition of Laplace Transform $\text {(1)}: \quad$ $\ds$ $=$ $\ds \dfrac s {s^2 + a^2}$ Laplace Transform of Cosine

Hence:

 $\ds \dfrac \d {\d a} \int_0^\infty e^{-s t} \cos a t \rd t$ $=$ $\ds \int_0^\infty e^{-s t} \paren {\map {\dfrac \partial {\partial a} } {\cos a t} } \rd t$ Derivative of Integral $\ds$ $=$ $\ds \int_0^\infty e^{-s t} \paren {-t \sin a t} \rd t$ Derivative of Cosine Function $\ds$ $=$ $\ds -\laptrans {t \sin a t}$ Definition of Laplace Transform $\ds$ $=$ $\ds \map {\dfrac \d {\d a} } {\dfrac s {s^2 + a^2} }$ from $(1)$ $\ds$ $=$ $\ds -\dfrac {2 a s} {\paren {s^2 + a^2}^2}$ Quotient Rule for Derivatives

and the result follows.

$\blacksquare$