I was helping somebody solve the problem. It goes like this:
What is the Laplace transform of derivative of $\sin (at)$
My work:
Well…to get the derivative of $\sin (at)$, let $u = at$ and $du = a.$ Then recall that $\frac{d}{dx}(\sin u) = \cos(u) du$. So….
$$\frac{d}{dx}(\sin u ) = \cos (u) du$$
$$\frac{d}{dx}(\sin (at) ) = \cos (at) (a)$$
$$\frac{d}{dx}(\sin (at) ) = a \cos (at) $$
We will get the Laplace transform of $a \cos (at)$.
Recall that the Laplace transform of $\cos (\omega_o t)$ is $\frac{s}{s^2 + \omega_o ^2}$. Then the Laplace transform of $a \cos (at)$ would be:
$$\mathcal L \{ \cos (\omega_o t) \} = \frac{s}{s^2 + \omega_o ^2}$$
$$\mathcal L \{ a \cos (at) \} = a \left( \frac{s}{s^2 + (a)^2} \right)$$
$$\mathcal L \{ a \cos (at) \} = a \left( \frac{s}{s^2 + a^2} \right)$$
Initially, we conclude that Laplace transform of $a \cos (at)$ is $ \frac{a s}{s^2 + a^2} $.
But I have doubts regading the solution, because the equation above can be re-interpreted as getting the Laplace transform of derivative of $\sin (a (\omega_o)t)$, where $\omega_o = 1$. If that's the case, I have to get another derivative of $\sin (a (\omega_o)t)$ where $\omega_o = 1$, which is $a \omega_o \cos (a \omega_o t)$, where $\omega_o = 1$.
The Laplace transform of $\sin (a (\omega_o)t)$ can be solve by remembering a Laplace transform property called "scale change", where in the time-domain, $f(at) $ (with the restriction a > 0 or a = 0), while in the frequency domain, $\frac{1}{a} F \left( \frac{s}{a}\right)$. With that in mind, the Laplace transform of $a \omega_o \cos (a \omega_o t)$ would be:
$$\mathcal L \{ a \omega_o \cos (a \omega_o t) \} = a \left( \frac{1}{a} \left( \frac{\frac{s}{a}}{(\left( \frac{s}{a}) \right) ^2 + (\left( \frac{\omega_o}{a}) \right)^2} \right) \right)$$
$$ = \left( \frac{\frac{s}{a}}{\frac{s^2}{a^2} + \frac{\omega_o ^2}{a^2}} \right) $$
$$ = \left( \frac{\frac{s}{a}}{\frac{s^2 + \omega_o ^2}{a^2}} \right) $$
$$ = \frac{as}{s^2 + \omega_o ^2} $$
If $\omega_o = 1$, ultimately, the Laplace transform of $a \omega_o \cos (a \omega_o t)$ is $\frac{as}{s^2 + 1}$
Put it in another way, the Laplace transform of derivative of $\sin (at)$ is $\frac{as}{s^2 + 1}$.
I got two different Laplace transforms gotten from a single equation, made from two different train of logic. Which one is correct?
Best Answer
${\cal L}(t \mapsto e^{iat})(s) = {\cal L}(t \mapsto \cos{at})(s)+i{\cal L}(t \mapsto \sin{at})(s) = \int_0^\infty e^{-(s-ia)t} dt = {1 \over s-ia}$.
By considering the above with $\pm a$ you can get ${\cal L}(t \mapsto \cos{at})(s) = {s \over s^2+a^2}$, ${\cal L}(t \mapsto \sin{at})(s) = {a \over s^2+a^2}$.
It follows that if $f(t) = \sin at$, then ${\cal L}(f')(s) = a{\cal L}(t \mapsto \cos{at})(s) = {as \over s^2+a^2}$.