Laplace Transform of Cosine Integral Function
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Theorem
- $\laptrans {\map \Ci t} = \dfrac {\map \ln {s^2 + 1} } {2 s}$
where:
- $\laptrans f$ denotes the Laplace transform of the function $f$
- $\Ci$ denotes the cosine integral function.
Proof 1
Let $\map f t := \map \Ci t = \ds \int_t^\infty \dfrac {\cos u} u \rd u$.
Then:
\(\ds \map {f'} t\) | \(=\) | \(\ds -\dfrac {\cos t} t\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds t \map {f'} t\) | \(=\) | \(\ds -\cos t\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \laptrans {t \map {f'} t}\) | \(=\) | \(\ds -\laptrans {\cos t}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac s {s^2 + 1}\) | Laplace Transform of Cosine | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -\dfrac \d {\d s} \laptrans {\map {f'} t}\) | \(=\) | \(\ds -\dfrac s {s^2 + 1}\) | Derivative of Laplace Transform | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}\) | \(=\) | \(\ds \dfrac s {s^2 + 1}\) | Laplace Transform of Derivative | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds s \laptrans {\map f t}\) | \(=\) | \(\ds \int \dfrac s {s^2 + 1} \rd s\) | $\map f 0 = 0$, and integrating both sides with respect to $s$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds s \laptrans {\map f t}\) | \(=\) | \(\ds \dfrac 1 2 \map \ln {s^2 + 1} + C\) | Primitive of $\dfrac x {x^2 + a^2}$ |
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:
- $c = 0$
Thus:
\(\ds s \laptrans {\map f t}\) | \(=\) | \(\ds \dfrac 1 2 \map \ln {s^2 + 1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \laptrans {\map f t}\) | \(=\) | \(\ds \dfrac {\map \ln {s^2 + 1} } {2 s}\) |
$\blacksquare$
Sources
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $9$