How to move forward with Laplace transform of $e^{i(5t)}$

Question

I'm trying to solve the Laplace transform of $e^{i(5t)}$. This is my work so far:

$$\begin{align} \mathcal{L}\{ e^{i(5t)}\} &= \int_0^\infty e^{-st} e^{i(5t)} \ dt \\ &= \int_0^\infty e^{-t(s - 5i)} \ dt \end{align}$$

Doing the change of variables $u = -t(s - 5i) = t(5i - s)$, we get

$$\begin{align} \int_0^\infty e^{-t(s - 5i)} \ dt &= -\dfrac{1}{s - 5i} \int_0^{\infty (5i - s)} e^u \ du \\ &= \dfrac{1}{5i - s} \left[ e^u \right]^{\infty (5i - s)}_0 \end{align}$$

But then we need to consider the value of $\infty(5i - s)$. For Laplace transforms, I think we require that $\Re(s - 5i) = \Re(s) > 0$ for convergence, right? I'm a bit confused with how to move forward here. Can someone please explain this? Thank you.

Answer 1

We have $$\int_0^\infty e^{-(s-5i)t}\,dt=\left[-\frac{e^{-(s-5i)t}}{s-5i}\right]_0^\infty=-\frac1{s-5i}\left[\frac{e^{5it}}{e^{st}}\right]_0^\infty=-\frac1{s-5i}[0-1]=\frac1{s-5i}$$ since $\left|e^{5it}/e^{st}\right|\le1/e^{st}\to0$ as $t\to\infty$.

Answer 2

Hint: write $e^{i5t}$ as $\cos (5t)+i\sin (5t)$ and use integration by parts for each of the two terms.