# Laplace Transform of Sine of t over t

## Theorem

Let $\sin$ denote the real sine function.

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

Then:

$\laptrans {\dfrac {\sin t} t} = \arctan \dfrac 1 s$

### Corollary

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

## Proof

$\ds \lim_{x \mathop \to 0} \frac {\sin x} x = 1$
$(1): \quad \laptrans {\sin t} = \dfrac 1 {s^2 + 1}$
$(2): \quad \ds \laptrans {\dfrac {\map f t} t} = \int_s^{\to \infty} \map F u \rd u$

Hence:

 $\ds \laptrans {\dfrac {\sin t} t}$ $=$ $\ds \int_s^{\to \infty} \dfrac 1 {u^2 + 1} \rd u$ $(1)$ and $(2)$ $\ds$ $=$ $\ds \lim_{L \mathop \to \infty} \int_s^L \dfrac 1 {u^2 + 1} \rd u$ Definition of Improper Integral $\ds$ $=$ $\ds \lim_{L \mathop \to \infty} \bigintlimits {\arctan u} s L$ Primitive of $\dfrac 1 {x^2 + a^2}$ $\ds$ $=$ $\ds \lim_{L \mathop \to \infty} \paren {\arctan L - \arctan s}$ $\ds$ $=$ $\ds \dfrac \pi 2 - \arctan s$ $\ds$ $=$ $\ds \arccot s$ Sum of Arctangent and Arccotangent $\ds$ $=$ $\ds \arctan \dfrac 1 s$ Arctangent of Reciprocal equals Arccotangent

$\blacksquare$