Laplace inverse as a double integral

Question

Show that $$\mathcal{L}^{-1}\left[\frac{f(s)}{s^2}\right]=\int_{0}^{t}\int_{0}^{x}F(x)dxdy.$$ I tried using the formula $$\mathcal{L}^{-1}\left[\frac{f(s)}{s}\right]=\int_{0}^{t}F(x)dx.$$

Answer 1

Let $G(s) = \frac{F(s)}{s}$

Then, as you stated,

(Selecting variables y for g and x for f) $$\mathcal{L}^{-1}[G(s)]= g(y) = \mathcal{L}^{-1}\left[\frac{F(s)}{s}\right] = \int^x_0{f(x)dx}$$

$$\mathcal{L}^{-1}\left[\frac{G(s)}{s}\right] = \mathcal{L}^{-1}\left[\frac{F(s)}{s^2}\right] = \int^y_0{g(y)dy} = \int^y_0{\int^x_0{f(x)dx}dy}$$

Thus the property is proved!