In Olver's Asymptotics and Special Functions, he gives a statement on page 75: For the integral $$ I(x) := \int_a^\infty dt \ e^{i x t } q(t) \ , $$ if $q(t)$ is continuous and $q'(t)$ is absolutely convergent on $[a,\infty)$ then the asymptotics of $I(x)$ as $x \to \infty$ are given by $$ I(x) \approx \frac{i}{x} e^{i a x} q(a) + \ldots $$ to leading order. Taking $\mathrm{Im}[I(x)]$ then gives you information about the sine-transform of $q(t)$.
My question is: What can you say about the sine-transform if $q(t)$ is not absolutely integrable? To be specific, what if I have a convergent integral like $$ J(x) = \int_0^\infty dt \ \sin(x t) \log(t) e^{-t} $$ and I want access to the asymptotics as $x\to\infty$? The function $q(t) = \log(t) e^{-t}$ has a derivative which is not absolutely integrable on $[0,\infty)$ (even though this integral is convergent).
Is there a way to get asymptotics in an example like this? How can you deal with this?
P.S. I know that the integral $J(x)$ can be done explicitly, but I am interested in how to access the asymptotics by looking at the integral.
There isn't one catch-all formula, but there are various generalizations of Laplace's method and the saddle point method. In this example, it can be shown that $$\int_0^\infty e^{i x t} e^{-t} \ln t \, dt \sim e^{-t} \bigg\rvert_{t = 0} \int_0^{i \epsilon} e^{i x t} \ln t \, dt \sim i \int_0^\infty e^{-x u} \ln u \, du.$$ Olver's book has a section on logarithmic singularities.