Asymptotic expansion of an inverse Fourier integral

Question

To find the $t\rightarrow +\infty$ limit of the following integral: $$ G(t)=\int_{-\infty}^{+\infty} \left(\frac{\sqrt{\frac{1}{1- 2 i \omega}}(3 i \omega -1)\omega^2}{(i \omega+1)(1-i \omega)^2}-\frac{\sqrt{\frac{1}{1+ 2 i \omega}}(3 i \omega +1)\omega^2}{(-i \omega+1)(1+i \omega)^2}\right) e^{- i \omega t} $$ I have tried several methods:

1. trying to explicitly calculate the "inverse Fourier transform",
2. trying to convert it into a "inverse Laplace transform" problem,
3. trying to perform the integration by closing the integral contour in the lower half of the complex $\omega$ plane.

However, these methods are of no use here! I then turned to using steepest descent method to find the asymptotics of the integral. Exponentiating the expression within the brackets, I'm trying to find the stationary phase point. But finding the extremum value of the exponent is analytically impossible. Now I'm not sure whether to stick with this approach or try yet another one.

If anyone can provide technical help I would be grateful.

Answer 1

Thanks a lot to @Luca Armstrong for the reference paper; I can now solve this problem using Watson's Lemma given in Olver's textbook.

I proceed as follows. Let us first focus on the first part of the integral, which is non-zero for $t>0$: $$G^{>}(t)=\int_{-\infty}^{+\infty}\frac{\sqrt{\frac{1}{1-2i \omega}}(3i \omega - 1) \omega^2}{(i \omega+1)(1-i \omega)^2}e^{- i \omega t}$$ Note that the integrand has a branch cut on the Im axis between $-i \infty$ and $-i/2$. The integral can then be calculated by pushing the integral contour downwards (as shown in the figure below); now we only need to calculate the integral along the branch cut: $$G^{>}(t)= \lim_{\epsilon\rightarrow 0} 2 \int_{-i\infty-\epsilon}^{-i/2-\epsilon}\frac{\sqrt{\frac {1}{1-2i \omega}}(3i \omega - 1)\omega^2}{(i \omega+1)(1-i \omega)^2}e^{- i \omega t} $$ Changing the variables to $\omega = - i s$ and then $ 1- 2 s = -2 z$ we get $$G^{>}(t)= \frac{1}{\sqrt{2}} e^{-t/2} \int_{0}^{+\infty}\frac{1}{\sqrt {z}}\frac{(1+2z)^2(1+6z)}{(1-2z)^2(3+2z)}e^{- z t}$$ Now it is the same as Watson's lemma: $I(t)=\int_{0}^{\infty}q(z)e^{- z t }dz$. According to the lemma, we need to expand $q(z)$ as $$q(z)\sim \sum_{s=0}a_s z^{(s+\lambda-\mu)/\mu}$$ where $\mu$ is a positive constant and $Re(\lambda$$)>0$. In our current case, it is found: $$\mu=1,\,\,\,\lambda=\frac{1}{2},\,\,\,\,a_0=\frac{1}{3}$$ This information is enough to find the main contribution of the integral using Watson's lemma: $$G^{>}(t)\sim \frac{1}{\sqrt{2}} e^{-t/2}\sum_{s=0}\Gamma\left(\frac{s+\lambda }{\mu}\right)\frac{a_s}{t^{(s+\lambda)/\mu}}\sim \frac{1}{\sqrt{2}} e^{-t/2}\Gamma\left(\frac{1}{2}\right)\frac{a_0}{t^{1/2}}$$ Finally we find: $$G^{>}(t)\sim \frac{1}{3}\sqrt{\frac{\pi}{2}}\frac{e^{-t/2}}{\sqrt{t }}$$ For the second part of $G(t)$, we can use the same method.