# Laplace Transform of Sine of t over t/Corollary

## 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.

$\blacksquare$