Sumudu transform

Remarks

The integral converges absolutely, since $\abs{f(ut)} e^{-t} \le M e^{-(1 - cu) t}$ and $cu < 1$. At $u = 0$ one gets $G(0) = f(0)$. For $u > 0$ the substitution $s = ut$ gives $G(u) = \frac{1}{u} \int_0^\infty f(s) e^{-s/u}\, ds = \frac{1}{u} F\big(\frac{1}{u}\big)$, where $F$ is the Laplace transform of $f$. So every rule for the Laplace transform has a Sumudu counterpart; for example, the transform of $f'$ is $\frac{G(u) - f(0)}{u}$ when $f'$ is continuous and of exponential order.

Example: $f(t) = e^{at}$ with $a \in \R$. Then $G(u) = \int_0^\infty e^{-(1 - au) t}\, dt = \frac{1}{1 - au}$ for $0 \le u$ and $au < 1$. Also $f(t) = t^n$ gives $G(u) = u^n \int_0^\infty t^n e^{-t}\, dt = n!\, u^n$.

Non-example: $f(t) = e^{t^2}$ is not of exponential order, and $\int_0^\infty e^{u^2 t^2 - t}\, dt$ diverges for every $u > 0$.