I'm trying to evaluate an integral, and the final step is to evaluate $e^{-st + 4t}$ at infinity minus $e^{-st + 4t}$ at $10$. (The limits of integration were $\infty$ and $10$.)
To evaluate the upper limit of integration, infinity, I take the limit of the antiderivative $e^{-st + 4t}$ as $t$ goes to infinity. Except...how do I do that? According to my professor the limit is simply $0$.
$e^{-st + 4t}=e^{t(4-s)}$, so how can we know the limit to infinity is $0$? Isn't it only $0$ in the case that $4-s > 0$?
For context, the $s$ and $t$ here are the $s$ and $t$ we deal with during Laplace transforms, not sure if that has anything to do with my question
you are right, because $e^x$ is continuous: $$\lim_{x \to \infty}e^{-st + 4t}=e^{\lim_{x \to \infty}{-st + 4t}}=e^{(4-s)\lim_{x \to \infty}{t}}=e^{(4-s)\ \infty}=\begin{cases} e^{\infty} & \mbox s<4 \\ e^{-\infty} & \mbox s>4 \\ e^0 & \mbox s=4 \end{cases}=\begin{cases} \infty & \mbox s<4 \\ \ 0 & \mbox s>4 \\ \ 1 & \mbox s=4 \end{cases}$$ so the answer depend on the value of s.