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I am trying to solve a PDE via a Fourier transform and I am stuck with a final integral which is of the following form

$$I = \int_{-\infty}^{\infty}\frac{k}{\sqrt{k^2 - \dot{a}^2}}\sin{(kr)}\sin{\left(\sqrt{k^2-\dot{a}^2}~\tau\right)}~dk~,$$

where $r > 0$ and $\tau < 0$. Now, I have a separate computation of the case when $\dot{a} = 0$ obtained independently and that gives a result proportional to a delta function (by expanding the $\sin$ functions into exponentials). From dimensional arguments about the full integral, I expect the result for the $\dot{a} \neq 0$ case to be ~ $r^2$ along with a delta function factor. However, I'm not sure how to proceed with this integral.

I've tried to do a Taylor expansion of the integrand with respect to $\dot{a}$ and perform the integral order by order, but that doesn't seem to help. For details about the full fourier transform please see here.

newtothis
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  • For $|k|<|\dot{a}|$ there is a negative number under the square root, which makes me think there's a mistake somewhere. Looking at the document you've linked, I see that's indeed the case - the computation of the integral over $\omega$ doesn't handle the case $|k|<|\dot{a}|$ correctly. – metamorphy May 16 '23 at 04:15
  • Set $\varphi(k,\tau) = \frac{k}{\sqrt{k^2-a^2}} \sin(kr)\sin(\sqrt{k^2 -a^2} \tau)$. Differentiating $\int_{-\infty}^{\infty} \varphi(k,\tau) dk$ with respect to $\tau$ gives a simpler integral – Victor May 16 '23 at 05:14
  • @metamorphy when I am computing the $\omega$ integral, isn't $k$ kept as a constant? Or would I have to take two cases - $k > a$ (which is hopefully this calculation) and $k < a$ as a separate case? – newtothis May 16 '23 at 05:39
  • Yes, the distinction is needed here. Moreover, if you consider $0<\dot{a}<|k|$ and $\tau<0$ (as in the document), then the chosen contour doesn't have anything to do with $\int_{-\infty}^\infty$; instead, use a lower-semicircular one (which yields $0$ as the result). In contrast, the case $\tau>0$ needs an upper semicircle, which captures the poles (and residues). – metamorphy May 17 '23 at 03:33
  • @metamorphy I am a little confused why you say that the chosen contour doesnt have anything to do with $\int_{-\infty}^{\infty}$. Retaining the lower semicircular arc without including the poles would give me zero as you suggest, but that doesn't make sense given the context. I am actually calculating a retarded Green's function in a physical context, and an almost identical contour is also chosen when $\dot{a} = 0$, except then both the poles lie on the X Axis. – newtothis May 17 '23 at 04:46
  • The result is zero (again, under $\color{blue}{0<\dot{a}<|k|}$ and $\tau<0$), whether that makes sense or not. And yes, $\dot{a}=0$ is a corner case, with the poles on the real axis (which needs something like that contour or another form of the "half-residue" trick). – metamorphy May 17 '23 at 05:56

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