Inverse Laplace Transform of s over s^3 + a^3

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Theorem

$\map {\laptrans {\dfrac {e^{a t / 2} } {3 a} \paren {\cos \dfrac {\sqrt 3} 2 a t + \sqrt 3 \sin \dfrac {\sqrt 3} 2 a t - e^{-3 a t / 2} } } } s = \dfrac s {s^3 + a^3}$

where:

$s$ is a complex number with $\map \Re s > a$
$\laptrans f$ denotes the Laplace transform of $f$ evaluated at $s$.


Proof

\(\ds \map {\laptrans {\dfrac {e^{a t / 2} } {3 a} \paren {\cos \dfrac {\sqrt 3} 2 a t + \sqrt 3 \sin \dfrac {\sqrt 3} 2 a t - e^{-3 a t / 2} } } } s\) \(=\) \(\ds \dfrac 1 {3 a} \map {\laptrans {e^{a t / 2} \cos \dfrac {\sqrt 3} 2 a t} } s + \dfrac {\sqrt 3} {3 a} \map {\laptrans {e^{a t / 2} \sin \dfrac {\sqrt 3} 2 a t} } s - \dfrac 1 {3 a} \map {\laptrans {e^{a t / 2} e^{-3 a t / 2} } } s\) Linear Combination of Laplace Transforms
\(\ds \) \(=\) \(\ds \dfrac 1 {3 a} \paren {\dfrac {s - a / 2} {\paren {s - a / 2}^2 + \paren {\sqrt 3 a / 2}^2} } + \dfrac {\sqrt 3} {3 a} \map {\laptrans {e^{a t / 2} \sin \dfrac {\sqrt 3} 2 a t} } s - \dfrac 1 {3 a} \map {\laptrans {e^{-a t / 2} } } s\) Laplace Transform of Exponential times Cosine and simplification
\(\ds \) \(=\) \(\ds \dfrac 1 {3 a} \paren {\dfrac {s - a / 2} {\paren {s - a / 2}^2 + 3 a^2 / 4} } + \dfrac {\sqrt 3} {3 a} \paren {\dfrac {\sqrt 3 a / 2} {\paren {s - a / 2}^2 + \paren {\sqrt 3 a / 2}^2} } - \dfrac 1 {3 a} \map {\laptrans {e^{-a t / 2} } } s\) Laplace Transform of Exponential times Sine
\(\ds \) \(=\) \(\ds \dfrac 1 {3 a} \paren {\dfrac {s - a / 2} {\paren {s - a / 2}^2 + 3 a^2 / 4} } + \dfrac {\sqrt 3} {3 a} \paren {\dfrac {\sqrt 3 a / 2} {\paren {s - a / 2}^2 + 3 a^2 / 4} } - \dfrac 1 {3 a} \paren {\dfrac 1 {s + a} }\) Laplace Transform of Exponential
\(\ds \) \(=\) \(\ds \dfrac 1 {3 a} \paren {\dfrac {s - a / 2 + 3 a / 2} {\paren {s - a / 2}^2 + 3 a^2 / 4} - \dfrac 1 {s + a} }\) simplification
\(\ds \) \(=\) \(\ds \dfrac 1 {3 a} \paren {\dfrac {s + a} {\paren {s^2 - a s + a^2 / 4} + 3 a^2 / 4} - \dfrac 1 {s + a} }\) multiplying out square
\(\ds \) \(=\) \(\ds \dfrac 1 {3 a} \paren {\dfrac {s + a} {s^2 - a s + a^2} - \dfrac 1 {s + a} }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 {3 a} \paren {\dfrac {\paren {s + a}^2 - \paren {s^2 - a s + a^2} } {\paren {s^2 - a s + a^2} \paren {s + a} } }\) common denominator
\(\ds \) \(=\) \(\ds \dfrac 1 {3 a} \paren {\dfrac {s^2 + 2 a s + a^2 - s^2 + a s - a^2} {s^3 + a^3} }\) multiplying out, and Sum of Two Cubes
\(\ds \) \(=\) \(\ds \dfrac s {s^3 + a^3}\) simplifying

$\blacksquare$


Sources

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