Laplace Transform of t^2 cosine a t

Theorem

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

where:

$s$ is a complex number with $\map \Re s > a$

$\laptrans f$ denotes the Laplace transform of $f$ evaluated at $s$.

Proof

$\ds \map {\laptrans {t^2 \cos a t} } s$

$=$

$\ds \map {\dfrac \d {\d s} } {-\map {\laptrans {t \cos a t} } s}$

Derivative of Laplace Transform

$\ds $

$=$

$\ds \map {\dfrac \d {\d s} } {-\dfrac {s^2 - a^2} {\paren {s^2 + a^2}^2} }$

Laplace Transform of $t \cos a t$

$\ds $

$=$

$\ds \map {\dfrac \d {\d s} } {\dfrac {a^2 - s^2} {\paren {s^2 + a^2}^2} }$

simplifying

$\ds $

$=$

$\ds \dfrac {\paren {s^2 + a^2}^2 \map {\dfrac \d {\d s} } {a^2 - s^2} - \paren {a^2 - s^2} \map {\dfrac \d {\d s} } {\paren {s^2 + a^2}^2} } {\paren {s^2 + a^2}^4}$

Quotient Rule for Derivatives

$\ds $

$=$

$\ds \dfrac {\paren {s^2 + a^2}^2 \paren {-2 s} - \paren {a^2 - s^2} \paren {2 \paren {s^2 + a^2} \paren {2 s} } } {\paren {s^2 + a^2}^4}$

Power Rule for Derivatives, Chain Rule for Derivatives

$\ds $

$=$

$\ds \dfrac {-2 s \paren {s^2 + a^2} - 4 s \paren {a^2 - s^2} } {\paren {s^2 + a^2}^3}$

simplifying

$\ds $

$=$

$\ds \dfrac {-2 s^3 - 2 s a^2 - 4 s a^2 + 4 s^3} {\paren {s^2 + a^2}^3}$

multiplying out

$\ds $

$=$

$\ds \dfrac {2 s^3 - 6 a^2 s} {\paren {s^2 + a^2}^3}$

simplifying

$\blacksquare$

Sources

1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Multiplication by Powers of $t$: $20 \ \text{(b)}$