Calculus operators in Laplace space?

Question

I was playing with Laplace transforms and found something curious.

Suppose we have an expression $4x+3$ that we want to take the derivative of or the integral of. If we instead took its Laplace transform, we have $$\mathcal{L} [4x+3](p)= \frac{3p+4}{p^2}$$ Now, let's multiply this and divide this by $p$, and then invert the transform. We can see that $$\frac{3p+4}{p^2}\cdot p = \frac{3p+4}{p}$$ $$\mathcal{L}^{-1}\left[\frac{3p+4}{p}\right](x) = 3\delta(x)+4$$ $$\frac{3p+4}{p^2}\cdot \frac1p = \frac{3p+4}{p^3}$$ $$\mathcal{L}^{-1}\left[\frac{3p+4}{p^3}\right](x) = 2x^2+3x$$ I noticed an odd thing. The results are quite similar to taking the integral in the divide case, and taking a derivative in the multiplying case. There's an extra dirac delta function which gives me doubts however, so I tried with something different.

Using $\sin(2x)+3$ gives me a similar result. If we take its Laplace transform, multiply/divide by $p$, and then invert the transform, we get $3\delta(x)+2\cos(2x)$ from multiplying $p$, and $3x-\frac12\cos(2x)+\frac12$ from dividing by $p$, which in the vein of integration can be said to be the same as the indefinite integral since we just collect constants.

Clearly, there seems to be some relationship between dividing in Laplace space and integrating in real space, and vice versa, but it doesn't seem it's exactly the same either, since there's extra constants and Dirac delta functions. I'm guessing I'm supposed to account for some type of initial condition? but i'm not sure.

Any ideas on what's going on?

Answer 1

As it turns out, it's simply just a property of the Laplace transform.

We know that \begin{align} \mathcal{L}\{f'(x)\}(s) = sF(s) - f(0)\tag{1} \end{align}

and $$\mathcal{L}\left\{\int_0^xf(t)\text{ d}t\right\}(s) = \frac{F(s)}{s}\tag{2}$$

The derivative and integration properties come from $(1)$ and $(2)$ respectively.

The constants in the integration come from the fact that $g(x)=\int_0^xf(t)\text{ d}t$ must equal $0$ when $x=0$ from the bounds.

Meanwhile, the Dirac deltas seem to come from the fact that I didn't subtract off the initial condition, which meant that the Laplace transform "preserved" information that an actual derivative wouldn't preserve. This means that within the scope of the Laplace transform ($0$ to infinity), a derivative taken via the "Laplace transform method" preserves constants with a Dirac delta. This can be seen by integrating $3\delta(x)+4$ which gives us the constant $3$ for the positive reals, showing how this method preserves this constant which a normal derivative removes, if we don't account for the initial condition.