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Let $x,y\in\mathbb{R}$. Then the Green's function for the Helmholtz equation is given by

$$\left(\Delta+\frac{\omega^{2}}{c^{2}}\right)G(x,y,\omega)=\delta(x-y).$$

Now what is the idea for deriving the Green's function here? Intuitively, I would take Fourier transforms on both sides, which would give me a convolution on the LHS and an exponential function on the RHS, but this would be very messy and I think I am missing something here. I believe the answer should be

$$G(x,y,\omega)=\frac{ic}{2\omega}e^{i\omega|x-y|/c}.$$

Apologies if this is a duplicate, but I could not find what I was looking for on the search bar.

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    If I remember correctly, in this case the Green's function should look like $\frac{\exp(\pm i |x-y|)}{|x-y|}$ up to some multiplicative normalisation factors. – 2017-01-18

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I don't see where you get the convolutions on the LHS. The FT would give you (up to some multiplicative constants from FT normalisation, who cares)

$$(\omega^2/c^2-|\xi|^2) G(\xi,y,\omega ) = \exp(i y \xi),$$ which should be easy to manipulate - I think a residue theorem would do the trick.

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    Because if $f$ and $g$ are functions, then $\mathcal{F}\{f(\cdot) g(\cdot)\}(\xi)=\hat{f}(\xi)\ast\hat{g}(\xi)$. – 2017-01-18
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    @user3482534 in the LHS I see a differential operator $\Delta$ (it gives $-|\xi|^2$) and a constant (with respect to $x$) factor $\omega^2/c^2$, which remains as it is - a constant factor. – 2017-01-18
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    Also, I want to determine $G(x,y,\omega)$, not $G(\xi,y,\omega)$. – 2017-01-28
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    @user3482534 when you apply the inverse FT to your $G(\xi, y,\omega)$, what did you obtain? Does it correspond to your initial hypothesis? – 2017-01-29
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    Well, multiplying both sides by $(\omega^2/c^2-|\xi|^2)$, from where you left off, I get $$\hat{g}(\omega,\xi,y)=\exp(iy\xi)(c^2/\omega^2-|\xi|^{-2})$$ So applying the IFT gives $$\begin{aligned}g(\omega,x,y)&=\frac{1}{2\pi}\int_{\mathbb{R}}\exp(iy\xi)(c^2/\omega^2-|\xi|^{-2})\cdot\exp(-ix\xi)\,dx\\&=\frac{c^2}{2\pi\omega^2}\int_{\mathbb{R}}\exp(iy\xi-ix\xi)\,dx-\frac{1}{2\pi|\xi|^{2}}\int_{\mathbb{R}}\exp(iy\xi-ix\xi)\,dx\end{aligned}$$ But this seems to be heading in the wrong direction as those integrals are equivalent to Dirac distributions (which is what I began with anyway(!) – 2017-02-07
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    All normalisation constants aside, you need to be able to find the inverse FT of a distribution $\xi\to PV\frac{\exp(i\xi y)}{\xi^2 -1}$. After a couple of algrbraic transformations this reduces to finding inverse FT of distributions $\xi\to PV\frac{\exp(i\xi y)}{\xi \pm1}$, which should give you - the residue theorem helps - what you want. – 2017-02-07
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    I have to admit that I get $$ \hat{G}(\omega,\xi,y)=\frac{1}{2\xi}\left(\frac{1}{\frac{\omega}{c}-\xi}-\frac{1}{\frac{\omega}{c}+\xi}\right)e^{iy\cdot\xi},$$ but I just can't seem to reach the point whereby we have a principle value, as you suggest. – 2017-03-09