Actually, it's not so hard to find a solution of $\frac{\partial}{\partial\overline{z}}G =\delta$. You have the right thing there already, modulo normalisation. We have, in the sense of distributions, for any $\varphi\in \mathscr{D}(\mathbb{C})$,
$$\begin{align}
\left(\frac{\partial}{\partial\overline{z}}\frac{1}{z-w}\right)[\varphi]
&= - \int_{\mathbb{C}} \frac{\partial\varphi}{\partial\overline{z}}(z)\frac{1}{z-w}\,d\lambda\\
&= -\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert z-w\rvert < R} \frac{\partial\varphi/\partial\overline{z}(z)}{z-w}\,d\lambda
\end{align}$$
where $\lambda$ is the Lebesgue measure on $\mathbb{C}$ and $R$ is sufficiently large, so that the support of $\varphi$ is contained in $D_R(w)$. Now write the integral with differential forms, that is, replace $d\lambda$ with $dx\wedge dy$, and write the latter as $\frac{1}{2i}d\overline{z}\wedge dz$. We get
$$\begin{align}
\left(\frac{\partial}{\partial\overline{z}}\frac{1}{z-w}\right)[\varphi] &= -\lim_{\varepsilon\to 0} \frac{1}{2i}\int_{\varepsilon < \lvert z-w\rvert < R} \frac{\partial\varphi/\partial\overline{z}(z)}{z-w}\,d\overline{z}\wedge dz\\
&= -\frac{1}{2i}\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert z-w\rvert < R} d\left(\frac{\varphi(z)}{z-w}\, dz\right) \tag{Stokes}\\
&= \frac{1}{2i}\lim_{\varepsilon\to 0} \int_{\lvert z-w\rvert = \varepsilon} \frac{\varphi(z)}{z-w}\,dz\\
&= \frac{1}{2i}\lim_{\varepsilon\to 0} \left(\int_{\lvert z-w\rvert = \varepsilon} \frac{\varphi(w)}{z-w}\,dz + \int_{\lvert z-w\rvert =\varepsilon} \frac{\varphi(z)-\varphi(w)}{z-w}\,dz \right)\\
&= \pi \varphi(w),
\end{align}$$
since by the differentiability of $\varphi$ in $w$, the integrand of the second integral in the penultimate line remains bounded as $\varepsilon\to 0$, and so the second integral is $\leqslant C\cdot 2\pi\varepsilon$ in absolute value by the ML-inequality. That means
$$\frac{1}{\pi}\frac{\partial}{\partial\overline{z}} \frac{1}{z-w} = \delta_w.$$
Recall Cauchy Integral Formula for Higher Derivatives: $f^{(n)}(z) = \frac{n!}{2\pi i} \displaystyle\int_\gamma \frac{f(\zeta)}{(\zeta-z)^{n+1}}\ d\zeta$.
Let $g(z)=\dfrac{e^{iz}}{(z+1)^2}$. Since $g$ is analytic on $\gamma: |z-1|=\frac{1}{2}$, $g'(1)=\frac{1}{2\pi i}\displaystyle\int_\gamma \frac{g(z)}{(z-1)^2}\ dz=\frac{1}{2\pi i}\displaystyle\int_\gamma \frac{e^{iz}}{(z^2-1)^2}\ dz$. Thus, the value of the integral is $2\pi i\cdot g'(1)$.
Best Answer
The series for $\sin z$ conveges uniformly on compact sets and this implies that the series for $z^{2}\sin (\overline z)$ does the same. So you are justified in interchanging the sum and the integral. Your computation is correct and you have got the right answer.