Laplace Transform of Sine Integral Function/Proof 3

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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:

\(\ds \map \Si t\) \(=\) \(\ds \int_0^t \dfrac {\sin u} u \rd u\) 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

$\blacksquare$


Sources

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