# Laplace Transform of Sine Integral Function/Proof 1

## Theorem

$\laptrans {\map \Si t} = \dfrac 1 s \, \arctan \dfrac 1 s$

where:

$\laptrans f$ denotes the Laplace transform of the function $f$
$\Si$ denotes the sine integral function

## Proof

Let $\map f t := \map \Si t = \displaystyle \int_0^t \dfrac {\sin u} u \rd u$.

Then:

$\map f 0 = 0$

and:

 $\displaystyle \map {f'} t$ $=$ $\displaystyle \dfrac {\sin t} t$ $\displaystyle \leadsto \ \$ $\displaystyle t \map {f'} t$ $=$ $\displaystyle \sin t$ $\displaystyle \leadsto \ \$ $\displaystyle \laptrans {t \map {f'} t}$ $=$ $\displaystyle \laptrans {\sin t}$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 {s^2 + 1}$ Laplace Transform of Sine $\displaystyle \leadsto \ \$ $\displaystyle -\dfrac \d {\d s} \laptrans {\map {f'} t}$ $=$ $\displaystyle \dfrac 1 {s^2 + 1}$ Derivative of Laplace Transform $\displaystyle \leadsto \ \$ $\displaystyle \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}$ $=$ $\displaystyle -\dfrac 1 {s^2 + 1}$ Laplace Transform of Derivative $\displaystyle \leadsto \ \$ $\displaystyle s \laptrans {\map f t}$ $=$ $\displaystyle -\int \dfrac 1 {s^2 + 1} \rd s$ $\map f 0 = 0$, and integrating both sides with respect to $s$ $\displaystyle \leadsto \ \$ $\displaystyle s \laptrans {\map f t}$ $=$ $\displaystyle -\arctan s + C$ Primitive of $\dfrac 1 {x^2 + a^2}$
$\displaystyle \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$

$c = \dfrac \pi 2$
 $\displaystyle s \laptrans {\map f t}$ $=$ $\displaystyle \dfrac \pi 2 - \arctan s$ $\displaystyle$ $=$ $\displaystyle \arccot s$ Sum of Arctangent and Arccotangent $\displaystyle$ $=$ $\displaystyle \arctan \dfrac 1 s$ Arctangent of Reciprocal equals Arccotangent $\displaystyle \leadsto \ \$ $\displaystyle \laptrans {\map f t}$ $=$ $\displaystyle \dfrac 1 s \arctan \dfrac 1 s$
$\blacksquare$