First of all, I am a physicist, so please excuse me if I make basic mistakes in the following, I will try to be as rigorous as possible. In my research, I recently came across the following integral for $\Omega>0$:
$\psi(t,\Omega)=\int_{0}^{+\infty}\mathrm{d}\omega\,\Phi(\omega)\frac{\mathrm{e}^{-\mathrm{i}\omega t}-\mathrm{e}^{-\mathrm{i}\Omega t}}{\omega-\Omega}$
$\Phi$ being a smooth and integrable function. Typically, I have to deal with functions of the kind:
$\Phi(\omega)\underset{\omega\to0^+}{\longrightarrow}0$ and $\Phi(\omega)\underset{\omega\to+\infty}{\simeq}K\omega^\alpha\mathrm{e}^{-\omega}$
I would like to estimate $\psi(t,\Omega)$ for $t\to+\infty$.
I tried to rewrite the integrand as follows:
$\frac{\mathrm{e}^{-\mathrm{i}\omega t}-\mathrm{e}^{-\mathrm{i}\Omega t}}{\omega-\Omega}=-\frac{2\mathrm{i}\mathrm{e}^{-\mathrm{i}\frac{\omega+\Omega}{2}t}}{\omega-\Omega}\sin(\frac{\omega-\Omega}{2}t)$
I then used the approximation (fairly widely used in my field, see https://en.wikipedia.org/wiki/Dirac_delta_function#Oscillatory_integrals):
$\frac{\sin(\omega t)}{\omega}\underset{t\to+\infty}{\simeq}\pi\delta(\omega)$, $\delta$ being the Dirac distribution
That would lead me to (warning, here comes the lack of mathematical rigor):
$\frac{\mathrm{e}^{-\mathrm{i}\omega t}-\mathrm{e}^{-\mathrm{i}\Omega t}}{\omega-\Omega}\underset{t\to+\infty}{\simeq}-2\mathrm{i}\pi\mathrm{e}^{-\mathrm{i}\frac{\omega+\Omega}{2}t}\delta(\omega-\Omega)=-2\mathrm{i}\pi\mathrm{e}^{-\mathrm{i}\Omega t}\delta(\omega-\Omega)$
This yields:
$\psi(t,\Omega)\underset{t\to+\infty}{\simeq}-2\mathrm{i}\pi\Phi(\Omega)\mathrm{e}^{-\mathrm{i}\Omega t}$
I tried to perform some numerical calculations with Wolfram Mathematica to check this result, and, if the oscillation induced by the factor $\mathrm{e}^{-\mathrm{i}\Omega t}$ seems present, the modulus of $\psi(t,\Omega)$ does not seem to follow the above approximation.
Since I am no expert in this kind of questions, I don't know how to find information that could help me with this question. If someone has an idea or a good reference to advise it would be of great help to me. Thank you in advance.
I followed Ian's suggestion of splitting the integral and, although I am still unable to justify rigorously every point, I think I eventually came up with the correct answer. First of all, the asymptotics of my integral seems to be:
$\psi(t,\Omega)\underset{t\to+\infty}{\simeq}\left(\mathrm{pv}\int_{0}^{\infty}\mathrm{d}\omega\,\frac{\Phi(\omega)}{\Omega-\omega}-\mathrm{i}\pi\Phi(\Omega)\right)\mathrm{e}^{-\mathrm{i}\Omega t}$
To get this result, you split the integral:
$\psi(t,\Omega)=\lim_{\delta\to0^+}\left(\mathrm{e}^{-\mathrm{i}\Omega t}\int_{|\omega-\Omega|>\delta}\mathrm{d}\omega\,\frac{\Phi(\omega)}{\Omega-\omega}+\int_{|\omega-\Omega|>\delta}\mathrm{d}\omega\,\frac{\Phi(\omega)\mathrm{e}^{-\mathrm{i}\omega t}}{\Omega-\omega}+\int_{|\omega-\Omega|<\delta}\mathrm{d}\omega\,\Phi(\omega)\frac{\mathrm{e}^{-\mathrm{i}\omega t}-\mathrm{e}^{-\mathrm{i}\Omega t}}{\omega-\Omega}\right)$
The real part of the asymptotics obviously comes from the first term in this development. The imaginary part comes from the second one, I still need to work a bit to prove this rigorously though. Finally, the third term should vanish, I did not prove it properly but this shouldn't be too difficult to do for someone more literate than I am in this field.