Laplace Transform of t by Sine a t

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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.

$\blacksquare$


Sources

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