Laplace Transform of Sine of t over t

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Theorem

Let $\sin$ denote the real sine function.

Let $\laptrans f$ denote the Laplace transform of a real function $f$.


Then:

$\laptrans {\dfrac {\sin t} t} = \arctan \dfrac 1 s$


Corollary

$\laptrans {\dfrac {\sin a t} t} = \arctan \dfrac a s$


Proof

From Limit of $\dfrac {\sin x} x$ at Zero:

$\ds \lim_{x \mathop \to 0} \frac {\sin x} x = 1$


From Laplace Transform of Sine:

$(1): \quad \laptrans {\sin t} = \dfrac 1 {s^2 + 1}$

From Laplace Transform of Integral:

$(2): \quad \ds \laptrans {\dfrac {\map f t} t} = \int_s^{\to \infty} \map F u \rd u$


Hence:

\(\ds \laptrans {\dfrac {\sin t} t}\) \(=\) \(\ds \int_s^{\to \infty} \dfrac 1 {u^2 + 1} \rd u\) $(1)$ and $(2)$
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \int_s^L \dfrac 1 {u^2 + 1} \rd u\) Definition of Improper Integral
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \bigintlimits {\arctan u} s L\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \paren {\arctan L - \arctan s}\)
\(\ds \) \(=\) \(\ds \dfrac \pi 2 - \arctan s\)
\(\ds \) \(=\) \(\ds \arccot s\) Sum of Arctangent and Arccotangent
\(\ds \) \(=\) \(\ds \arctan \dfrac 1 s\) Arctangent of Reciprocal equals Arccotangent

$\blacksquare$


Also see


Sources

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