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I am not able to deal with the following problem. Giving function $$G(x-\xi,y-\eta,k)=\frac{1}{4\pi^2i}\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{i\mu(x-\xi)-i\nu(y-\eta)}}{\nu-i\mu(\mu+2k)}d\nu d\mu$$ and to make clear, let $k=k_1+ik_2$, how to prove $$\frac{\partial G(x-\xi,y-\eta,k)}{\partial \bar k}=\frac{1}{2\pi}\text{sgn}(-k_1)e^{-2ik_1[(x-\xi)-2k_2(y-\eta)]}?$$ It seems to me that $\bar \partial \frac{1}{\pi(z-\zeta)}=\delta(z-\zeta)$ should be made use of, but I am not sure how to carry out that.

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Is there a typo somewhere? The integrand doesn't seem to depend on $\bar k$, so the derivative of the integral w/r/t $\bar k$ will be zero. –  Gunnar Magnusson Jul 2 '11 at 17:57
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1 Answer

up vote 2 down vote accepted

Sketch: Just take the differentiation inside.

$$ \bar{\partial}_k \frac{1}{\nu - i \mu^2 - 2i\mu k} = -2\pi i \mu \delta(\nu - i \mu^2 - 2i\mu k) $$

the delta function can be decomposed

$$ \delta(\nu - i \mu^2 - 2i\mu k) = \delta(\nu + 2\mu k_2) \delta(\mu(\mu + 2k_1)) $$

the rest follows from the usual change of variable formula involving the $\delta$ distribution. (Integrate first in $\mu$. Because of the $\mu$ outside already, the only root of $\mu(\mu+2k_1)$ that contributes is the one $\mu = -2k_1$.)

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It may be worth noting that $\overline{\partial}(1/z)$ isn't $0$, because $1/z$ is not holomorphic at $z=0$. Yes, the support of whatever $\overline{\partial}(1/z)$ is is $\{0\}$, etc. Also, given the delicate convergence of the integral at infinity, already that integral probably needs interpretation as Fourier transform of a tempered distribution, entailing some regularization. Then the "formal" riff as in the answer completely crushes it. –  paul garrett Jul 2 '11 at 20:34
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Many thanks to all! I've got the answer following Willie's approach. As to the chain rule of Wirtinger derivative, $\frac{\partial}{\partial\bar k}(f\circ g)=(\frac{\partial f}{\partial k}\circ g)\frac{\partial g}{\partial\bar k}+(\frac{\partial f}{\partial\bar k}\circ g)\frac{\partial\bar g}{\partial\bar k}$, the negative sign of the $\bar\partial_k$ formula should be removed, that is, $$\bar{\partial}_k\frac{1}{\nu-i\mu^2-2i\mu k}=0+\pi\delta(\nu-i\mu^2-2i\mu k)\cdot\frac{\partial \overline{-2i\mu k}}{\partial \bar k}= 2\pi i \mu\delta(\nu-i\mu^2-2i\mu k)$$ whereupon the result holds. –  Ansel Jul 3 '11 at 16:34
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