Laplace transform of the square of a first derivative.

Question

What is the Laplace transform of the square of a first derivative, i.e $L (dx/dt)^2$?

Answer 1

There's probably not an enormous amount of simplification possible. You can use the multiplication rule: $$L[f(t)\,g(t)]=\frac{1}{2\pi i}\int_{c-iT}^{c+iT}F(\sigma)\,G(s-\sigma)\,d\sigma,$$ where the integration is done along the vertical line $\operatorname{Re}(\sigma)=c,$ and which must be in the region of convergence of $F.$ Here \begin{align*} F(s)&=L[f(t)]\\ G(s)&=L[g(t)]. \end{align*} So, noting that \begin{align*} L[\dot{f}(t)]&=sF(s)-f(0^-), \end{align*} we can write \begin{align*} L\left[\left(\dot{f}(t)\right)^{\!2}\right] &=\frac{1}{2\pi i}\int_{c-iT}^{c+iT}(\sigma F(\sigma)-f(0^-))((s-\sigma)F(s-\sigma)-f(0^-))\,d\sigma. \end{align*} And that's probably about as far as you can go.