# Laplace Transform of Sine Integral Function

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## 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 1

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

Then:

$\map f 0 = 0$

and:

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

which leads to:

$c = \dfrac \pi 2$

Thus:

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

$\blacksquare$

## Proof 2

 $\ds \laptrans {\dfrac {\sin t} t}$ $=$ $\ds \arctan \dfrac 1 s$ Laplace Transform of $\dfrac {\sin t} t$ $\ds \leadsto \ \$ $\ds \laptrans {\int_{u \mathop \to 0}^{u \mathop = t} \frac {\sin u} u \rd u}$ $=$ $\ds \dfrac 1 s \arctan \dfrac 1 s$ Laplace Transform of Integral $\ds \leadsto \ \$ $\ds \laptrans {\map \Si t}$ $=$ $\ds \dfrac 1 s \arctan \dfrac 1 s$ Definition of Sine Integral

$\blacksquare$

## Proof 3

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

Then:

$\map f 0 = 0$

and:

 $\ds \map \Si t$ $=$ $\ds \int_0^t \dfrac {\sin u} u \rd u$ Definition of Sine Integral Function $\ds$ $=$ $\ds \int_0^t \dfrac 1 u \paren {u - \dfrac {u^3} {3!} + \dfrac {u^5} {5!} - \dfrac {u^7} {7!} + \dotsb} \rd u$ Definition of Real Sine Function $\ds$ $=$ $\ds t - \dfrac {t^3} {3 \times 3!} + \dfrac {t^5} {5 \times 5!} - \dfrac {t^7} {7 \times 7!} + \dotsb$ Primitive of Power $\ds \leadsto \ \$ $\ds \laptrans {\map \Si t}$ $=$ $\ds \laptrans {t - \dfrac {t^3} {3 \times 3!} + \dfrac {t^5} {5 \times 5!} - \dfrac {t^7} {7 \times 7!} + \dotsb}$ $\ds$ $=$ $\ds \dfrac 1 {s^2} - \dfrac 1 {3 \times 3!} \dfrac {3!} {s^4} + \dfrac 1 {5 \times 5!} \dfrac {5!} {s^6} - \dfrac 1 {7 \times 7!} \dfrac {7!} {s^8} + \dotsb$ Laplace Transform of Positive Integer Power $\ds$ $=$ $\ds \dfrac 1 {s^2} - \dfrac 1 {3 s^4} + \dfrac 1 {5 s^6} - \dfrac 1 {7 s^8} + \dotsb$ simplifying $\ds$ $=$ $\ds \dfrac 1 s \paren {\dfrac {\paren {1 / s} } 1 - \dfrac {\paren {1 / s}^3} 3 + \dfrac {\paren {1 / s}^5} 5 - \dfrac {\paren {1 / s}^7} 7 + \dotsb}$ rearranging $\ds$ $=$ $\ds \dfrac 1 s \arctan \dfrac 1 s$ Power Series Expansion for Real Arctangent Function

$\blacksquare$

## Proof 4

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

Then:

$\map f 0 = 0$

and:

 $\ds \map \Si t$ $=$ $\ds \int_0^t \dfrac {\sin u} u \rd u$ Definition of Sine Integral Function $\ds$ $=$ $\ds \int_0^1 \dfrac {\sin t v} v \rd v$ Integration by Substitution $u = t v$ $\ds \leadsto \ \$ $\ds \laptrans {\map \Si t}$ $=$ $\ds \laptrans {\int_0^1 \dfrac {\sin t v} v \rd v}$ $\ds$ $=$ $\ds \int_0^\infty e^{-s t} \paren {\int_0^1 \dfrac {\sin t v} v \rd v} \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \int_0^1 \dfrac 1 v \paren {\int_0^\infty e^{-s t} \sin t v \rd t} \rd v$ exchanging order of integration $\ds$ $=$ $\ds \int_0^1 \dfrac {\laptrans {\sin t v} } v \rd v$ Definition of Laplace Transform $\ds$ $=$ $\ds \int_0^1 \dfrac {\d v} {s^2 + v^2}$ Laplace Transform of Sine $\ds$ $=$ $\ds \intlimits {\dfrac 1 s \arctan \dfrac v s} 0 1$ Primitive of $\dfrac 1 {x^2 + a^2}$ $\ds$ $=$ $\ds \dfrac 1 s \arctan \dfrac 1 s$

$\blacksquare$