How to close the contour of integration when there is a branch cut from -infinity to +infinity at a negative imaginary value?

Question

From some loop calculation in an EFT, I have found the Green's function containing this square root: $$G(\omega, k)\sim \sqrt{\frac{D k^2 \tau - (i+\tau \omega)^2}{(i+\tau \omega)^2(-D k^2 \tau +\omega \tau(2 i +\tau \omega))}}$$ I need to find $G(t,k)$. However, "InverseFourierTransform" cannot do this. Therefore, I restrict myself to the special case $k=0$: $$G(\omega, 0)\sim \sqrt{\frac{-1}{\omega \tau (2 i +\tau \omega)}}$$ Even in this case, "InverseFourierTransform" doesn't work. In particular, the "ComplexPlot" shows that the analytical structure of $G(t,0)$ is a bit strange: there are two branch cuts, one starting from 0 to $-2i /\tau$, the other from $-i/\tau-\infty$ to $-i/\tau+\infty$. Now I want to understand how to close the integration contour in $$G(t,0)=\int_{-\infty}^{+\infty}G(\omega,0)e^{-i\omega t} d\omega$$ on the lower half-plane of complex $\omega$! The goal is to find some asymptotic expansion of $G(t,0)$ within the limit of $t>>\tau$ or $t<<\tau$.

If anyone could comment on this issue I would be grateful.

Answer 1

This is a partial answer, but before we can go further we need to clarify some points about your question. As you are using Fourier analysis am I correct that you are aiming to integrate along the imaginary axis and thus go from the frequency domain to the time domain? I also assume that omega is a real value along the imaginary axis -please confirm.

First we make a function for your expression. You have both omega and tor together so we can absorb them into a single parameter which I will call omega.

ClearAll[G, \[Omega]];
G[\[Omega]_] := Sqrt[-1/(\[Omega] (2 I  + \[Omega]))]

Now we can plot the absolute and argument values of the function

Plot3D[Evaluate[Abs[G[x + I y]]], {x, -3, 3}, {y, -5, 5}, 
 PlotRange -> {All, All, {0, 2}}, PlotPoints -> {50, 50}, 
 BoxRatios -> {3, 5, 2}]

The plot shows that there are singularities at 0 and -2 I as expected. The argument gives us a clue about the location of the branch cuts.

Plot3D[Evaluate[Arg[G[x + I y]]], {x, -3, 3}, {y, -5, 5}, 
 PlotRange -> {All, All, {-\[Pi], \[Pi]}}, PlotPoints -> {50, 50}, 
 BoxRatios -> {3, 5, 2}]

If you rotate the graphics you will see that Mathematica has put one branch cut from 0 to -2 I and another along -inf -2 I to + inf -2 I. This is also clear if you plot the contour plot.

ContourPlot[Evaluate[Arg[G[x + I y]]], {x, -3, 3}, {y, -5, 5}, 
 PlotRange -> {All, All, {-\[Pi], \[Pi]}}, PlotPoints -> {50, 50}, 
 BoxRatios -> {3, 5, 2}, ContourShading -> False, 
 Contours -> Table[h, {h, -\[Pi]/2, \[Pi]/2, \[Pi]/100}], 
 AspectRatio -> Automatic]

With these branch cuts we will need to do work with the four quadrants of the complex plane. The values obtained from the left half plane will correspond to time greater than zero and the values obtained from the right half plane for times less then zero. Were you expecting a response for time less than zero? Negative times? This suggests that your problem is non -causal events start before time is equal to zero.

I am stopping here hoping that is helpful and that the points raised can be clarified.

Answer 2

This is not a rigorous solution - just some formal manipulations which, however, can bring some value. Formally, we can present the integral in the form $$G(t,0)=i\int_{-\infty}^\infty\frac{e^{-i\omega t}}{\sqrt{\omega \tau(2i+\omega\tau)}}d\omega=\frac i{\sqrt\pi}\int_{-\infty}^\infty d\omega\int_{-\infty}^\infty e^{-i\omega t-s^2\omega\tau(2i+\omega\tau)}ds$$ Changing the order of integration and integrating with respect to $\omega$ $$=\frac i{\sqrt\pi}\int_{-\infty}^\infty ds\int_{-\infty}^\infty e^{-\tau^2s^2\big(\omega^2+2i\omega\frac{s^2\tau+t/2}{s^2\tau^2}-\frac{(s^2\tau+t/2)^2}{s^4\tau^4}\big)}e^{-\frac{(s^2\tau+t/2)^2}{s^2\tau^2}}d\omega$$ $$=\frac i{\sqrt\pi}\int_{-\infty}^\infty ds\int_{-\infty}^\infty e^{-\tau^2s^2(\omega+ib)^2}e^{-\frac{(s^2\tau+t/2)^2}{s^2\tau^2}}d\omega\,\,\Big(\text{where}\,b=\frac{s^2\tau+t/2}{s^2\tau^2}\Big)$$ $$=i\int_{-\infty}^\infty e^{-\frac{(s^2+t/2\tau)^2}{s^2}}\frac{ds}{|s\tau|}\overset{s^2=x}{=}\frac i\tau \int_0^\infty e^{-\frac{(x+t/2\tau)^2}x}\frac{dx}x$$ $$G(t,0)=\frac i\tau e^{-t/\tau}\int_0^\infty e^{-x-t/2\tau x}\frac{dx}x\tag{1}$$ Now we can consider the cases

$1.\quad\displaystyle \frac t\tau\gg1$. $\quad$ Making the substitution $x=\sqrt{\frac t{2\tau}}\,s$ and using the Laplace' method $$G(t,0)=\frac {i}\tau e^{-t/\tau}\int_0^\infty e^{-\sqrt{\frac t{2\tau}}(s+1/s)}\frac{ds}s\sim\frac {i\sqrt\pi}\tau \left(\frac{2\tau}t\right)^{1/4}e^{-t/\tau}e^{-\sqrt{2t/\tau}}\tag{2}$$ $2.\quad\displaystyle \frac t\tau\ll1$. $\quad$ Making the substitution $x=\sqrt{\frac t{2\tau}}e^{-s}$ and denoting $\alpha=\sqrt{\frac t{2\tau}}\,(\,\ll1)$ $$G(t,0)=\frac i\tau e^{-2\alpha^2}\int_{-\infty}^\infty e^{-2\alpha\cosh s}ds\overset{\cosh s=x}{=}\frac {2i}\tau e^{-2\alpha^2}\int_1^\infty e^{-2\alpha x}\frac{dx}{\sqrt{x^2-1}}$$ $$=\frac {2i}\tau e^{-2\alpha^2}e^{-2\alpha}\int_0^\infty e^{-2\alpha x}\frac{dx}{\sqrt x\sqrt{x+2}}=\frac {4i}\tau e^{-2\alpha^2}e^{-2\alpha}\int_0^\infty e^{-2\alpha s^2}\frac{ds}{\sqrt{s^2+1}}$$ Integrating by part $$=\frac {16i\alpha}\tau e^{-2\alpha^2-2\alpha}\int_0^\infty e^{-2\alpha s^2}\ln(s+\sqrt{s^2+1})\,s\,ds=\frac {8i\alpha}\tau e^{-2\alpha^2-2\alpha}\int_0^\infty e^{-2\alpha x}\ln(\sqrt x+\sqrt{x+1})\,dx$$ $$=\frac {4i}\tau e^{-2\alpha^2-2\alpha}\int_0^\infty e^{-s}\ln\Big(\frac1{\sqrt{2\alpha}}\big(\sqrt s+\sqrt{s+2\alpha}\big)\Big)\,dx$$ Dropping the terms $\sim o(\alpha)$ $$G(t,0)\sim\frac {2i}\tau\left(\ln\frac1\alpha+\ln2-\gamma\right)\tag{3}$$