Laplace Transform of Sine of t over t/Corollary

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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:

$\laptrans {\dfrac {\sin a t} t} = \arctan \dfrac a s$


Proof

\(\ds \laptrans {\dfrac {\sin t} t}\) \(=\) \(\ds \arctan \dfrac 1 s\) Laplace Transform of Sine of t over t
\(\ds \leadsto \ \ \) \(\ds \laptrans {\dfrac {\sin a t} {a t} }\) \(=\) \(\ds \dfrac 1 a \arctan \dfrac 1 {s / a}\) Laplace Transform of Constant Multiple
\(\ds \) \(=\) \(\ds \dfrac 1 a \arctan \dfrac a s\)

But:

\(\ds \laptrans {\dfrac {\sin a t} {a t} }\) \(=\) \(\ds \int_0^\infty e^{-s t} \dfrac {\sin a t} {a t} \rd t\) Definition of Laplace Transform
\(\ds \) \(=\) \(\ds \dfrac 1 a \int_0^\infty e^{-s t} \dfrac {\sin a t} t \rd t\)
\(\ds \) \(=\) \(\ds \dfrac 1 a \laptrans {\dfrac {\sin a t} t}\)

The result follows.

$\blacksquare$


Sources

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