Laplace transformation with imaginary unit $i$

Question

Does there exist a real-valued function $f(t)$ such that its Laplace transform $\mathcal{L}{f(t)}$ is an expression that contains the imaginary unit $i$?

For example $$ \mathcal{L}(f(t))=\frac{1}{(-ip)^3+1}. $$

I have not yet been able to find such an example, but on the other hand, in some technical applications, images of the Laplace transform arise that contain the imaginary unit $i$. And I don't know how to interpret this, as a mistake?

Answer 1

Here is one way to interpret this: consider the Laplace transform $$ \mathcal L\{f(t)\}(s) = \frac{1}{sp^3 + 1} $$ Evaluating this expression at $s = -i$ gives the expression in your question. In other words, the Laplace transform is evaluated only on the imaginary axis. This corresponds to the Fourier transform.

Answer 2

If $f(t)$ is real then $\overline{f(t)} = f(t)$. Then $$ \overline{F(s)} = \int_0^\infty \overline{f(t)\mathrm e^{-st}} \,\mathrm dt = \int_0^\infty f(t)\mathrm e^{-\overline{s}t} \,\mathrm dt = F(\overline{s}) $$ So $F$ is real for real $s$.