Laplace Transform of Sine Integral Function/Proof 4

From ProofWiki
Jump to navigation Jump to search

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
\(\ds \) \(=\) \(\ds \int_0^1 \dfrac {\sin t v} v \rd v\) 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 $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \dfrac 1 s \arctan \dfrac 1 s\)

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Laplace_Transform_of_Sine_Integral_Function/Proof_4&oldid=511753"