I was solving an integral and i stepped in a Laplace transormation of the form $\mathcal{L}_t\left\{sin\left(at^n\right)\right\}\left(s\right)$ and I was curious on a generalized solution. After some work I wrote the following identity:
$$\mathcal{L}_t\left\{sin\left(at^n\right)\right\}\left(s\right)=\frac{a}{s^{n+1}}\sum_{k\ \in\ \mathbb{N}_0}{\left(\frac{(-1)^k\left(a^2s^{-2n}\right)^k}{\left(2k+1\right)!}\Gamma\left(2kn+n+1\right)\right)}$$
where
$\sum_{k\ \in\ \mathbb{N}_0}{\left(\frac{(-1)^k\left(a^2s^{-2n}\right)^k}{\left(2k+1\right)!}\Gamma\left(2kn+n+1\right)\right)} = \frac{sin\left(a\right)}{a}\ \ \ \ \ \ \ $ for $n=0$ (trivial)
$\sum_{k\ \in\ \mathbb{N}_0}{\left(\frac{(-1)^k\left(a^2s^{-2n}\right)^k}{\left(2k+1\right)!}\Gamma\left(2kn+n+1\right)\right)} = \frac{s^2}{a^2+s^2}\ \ \ \ \ \ \ $ for $n=1$ (geometric series)
and
$\sum_{k\ \in\ \mathbb{N}_0}{\left(\frac{(-1)^k\left(a^2s^{-2n}\right)^k}{\left(2k+1\right)!}\Gamma\left(2kn+n+1\right)\right)} = \frac{\sqrt{\pi}}{2}e^{-\frac{a^2}{4s}}\ \ \ \ \ \ \ $ for $n=\frac{1}{2}$ $\left(\Gamma\left(x+\frac{3}{2}\right)=\frac{\sqrt{\pi}\left(2x+1\right)!}{2^{2x+1}x!}\right)$
Do you know how to get a general closed formula of this series?