Laplace Transform of Frequency Decay: $\sin(2\pi t \cdot f_0 \cdot e^{-t})u(t)$ For a sine wave of exponentially decaying frequency, is the Laplace Transform numerically solvable? If not, why?
One example could be:
$$\sin(2\pi t \cdot f_0 \cdot e^{-t})u(t)$$
where $f_0$ is the frequency starting at $t=0$, and $u(t)$ is the unit-step function.
I cannot find this equation on any Laplace tables, although there are real-world dynamic systems for which this would be helpful.
Other variations might be:

*

*Simple cosine:   $\cos(2\pi t \cdot f_0 \cdot e^{-t})u(t)$

*Decaying to a different frequency:   $\sin(2\pi t \cdot (f_0 \cdot e^{-t} + f_1))u(t)$
 A: Starting from the definition of the Laplace transform:
$$\mathcal L \{\sin(\omega_0te^{-at})\}(s) = \int_0^\infty \sin(\omega_0te^{-at})e^{-st} dt$$
My logic here was that since the frequency goes to zero quickly, using a Maclaurin series on the sine might prove fruitful. (whether this intuition is correct I don't know, but it does seem to have worked) So, expanding the sine we get:
$$\int_0^\infty e^{-st} \left[\sum_{k = 0}^\infty \frac{(-1)^k (\omega_0 te^{-at})^{2k+1}}{(2k+1)!}\right] dt$$

Here if we can interchange the sum and integral then we will get an infinite series in terms of fairly simple integrals of the form $\int_0^\infty t^ae^{-bt} dt.$ Glazing over the details a little, we should be able to perform this interchange if we can show that the original integral is absolutely convergent.
Showing this is relatively straightforward: begin by rewriting as $$\int_0^\infty |\sin(\omega_0te^{-at})e^{-st}| dt = \int_0^\infty |\sin(\omega_0te^{-at})|e^{-st} dt$$
Now consider the fact that for all $x \geq 0,$ we have $|\sin(x)| \leq x$: to prove this, consider that if $x < 1$ then $\sin(x) \geq 0$ so we can drop the absolute value signs and consider that $x - \sin(x) = \int_0^x 1 - \cos(t) dt,$ where the integrand is nonnegative everywhere, and if $x > 1$ then our statement holds because $\sin(x) < 1.$
Because the argument of sine in our integral ($\omega_0 te^{-at}$) is nonnegative everywhere within the bounds of integration, we can use this as a comparison for our integral like so:
$$\int_0^\infty |\sin(\omega_0te^{-at})|e^{-st} dt \leq \int_0^\infty (\omega_0te^{-at})e^{-st} dt = \int_0^\infty \omega_0te^{-(s+a )t} dt$$
where the final integral clearly converges whenever $s > -a.$

So, interchanging our sum and integral we get:
$$\sum_{k = 0}^\infty \frac{(-1)^k \omega_0^{2k + 1}}{(2k + 1)!} \int_0^\infty t^{2k + 1} e^{-a(2k+1)t - st} dt$$
Let $\lambda = a(2k + 1) + s,$ where our assumption $s > -a$ guarantees $\lambda > 0.$ Now let $u = \lambda t,$ $dt = \frac{du}{\lambda}$ in our integral, yielding
$$\int_0^\infty t^{2k + 1} e^{-\lambda t} dt = \int_0^\infty \left(\frac{u}{\lambda}\right)^{2k + 1} e^{-u} \frac{du}{\lambda} = \frac1{\lambda^{2k+2}} \int_0^\infty u^{2k+1}e^{-u} du = \frac{(2k+1)!}{\lambda^{2k+2}}$$
recognizing the final integral as the definition of the gamma function.
Plugging this into our infinite series and cancelling the factorial terms, we get our answer:
$$\boxed{\mathcal L \{\sin(\omega_0te^{-at})\}(s) = \sum_{k = 0}^\infty \frac{(-1)^k \omega_0^{2k + 1}}{(a(2k + 1) + s)^{2k + 2}}}$$
