Laplace Transform of Sine Integral Function/Proof 1

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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

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 $\dfrac 1 {x^2 + a^2}$


By the Initial Value Theorem of Laplace Transform:

$\ds \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$

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
\(\ds \leadsto \ \ \) \(\ds \laptrans {\map f t}\) \(=\) \(\ds \dfrac 1 s \arctan \dfrac 1 s\)

$\blacksquare$


Sources

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