I was trying to find the Laplace transform of $e^{3t}$:
$$\int^\infty_0 e^{3t}e^{-st} \ dt = \int_0^\infty e^{3t – st} \ dt = \lim_{x \to \infty}\int_0^x e^{3t – st} \ dt$$
So if we then attempt to solve this using change of variables instead of integration by parts, and we change the bounds of integration by making the substitution $t = 0$ and $t = x$, then we get $u = 0$ and $u = 3x – sx$ as the new bounds. But we then get
$$\lim_{x \to \infty} \dfrac{1}{3 – s} \int_0^{3x – sx} e^u \ du = \dfrac{1}{3 – s} [(\infty – \infty) – 1],$$
which is an indeterminate form.
The solution should be $\dfrac{1}{s – 3}$.
What's going on here? Have I made an error?
I would appreciate it if people could please take the time to clarify this.
Best Answer
Let $S=s-3$.
Change of variables $u=(s-3)t=St$ which implies
$$du=S dt \iff dt=\frac{1}{S}du$$
permits to write the integral under the form :
$$\int_{u=0}^{u=+\infty} e^{-u}\frac{1}{S}du$$
whence the result : $\frac{1}{S}$