How to derive Eq. (2.1.24) in Polchinski's string theory book Excuse me, I got one more stupid question in Polchinski's string theory book :(
$$\partial \bar{\partial} \ln |z|^2 = 2 \pi \delta^2 (z,\bar{z}) (1) $$
I shall check this equation by integrating both sides over $\int \int d^2z $
The right hand side is obviously $2\pi$.
The left hand side is evaluated as following
$$ \partial \bar{\partial} \ln |z|^2  = \partial \bar{\partial} \left( \ln z + \ln \bar{z} \right)  = \partial \left( \bar{\partial} \ln \bar{z} \right) + \bar{\partial} \left( \partial   \ln z \right) (2) $$
with the help of Eq. (2.1.9) in that book, we have
$$ \int \int_R d^2 z \left[  \partial \left( \bar{\partial} \ln \bar{z} \right) + \bar{\partial} \left( \partial   \ln z \right)  \right] = i \oint_{\partial R} 
\bar{\partial} \ln \bar{z} d \bar z - \partial \ln z d z (3) $$
$$= i \oint_{\partial R} \frac{1}{\bar{z}} d \bar z - \frac{1}{z} d z = 2 \pi +  \oint_{\partial R} \frac{1}{\bar{z}} d \bar z $$
Here I have used the contour integral. But is the a remaing term $\oint_{\partial R} \frac{1}{\bar{z}} d \bar z$ zero? Why?
 A: Obviously you are trying to solve problem 2.1 from Polchinski´s book which states:
"Verfify that $\partial\overline{\partial}\ln\left|z\right|^{2}=\partial\frac{1}{\overline{z}}=\overline{\partial}\frac{1}{z}=2\pi\delta^{2}\left(z,\:\overline{z}\right)$
(a) by use of the divergence theorem (2.1.9) 
($\int_{R}d^{2}z\left(\partial_{z}v^{z}+\partial_{\overline{z}}v^{\overline{z}}\right)=i\oint_{\partial R}\left(v^{z}d\overline{z}-v^{z}dz\right)$);
(b) by regulating the singularity and then taking the limit."
(a) You have 


*

*For holomorphic test functions $f(z)$:


$\int_{R}d^{2}z\partial\overline{\partial}\ln\left|z\right|^{2}f\left(z\right)=\int_{R}d^{2}z\overline{\partial}\frac{1}{z}f\left(z\right)=-i\oint_{\partial R}dz\frac{1}{z}f\left(z\right)=2\pi f\left(0\right)$.


*

*For antiholomorphic Test functions $f\left(\overline{z}\right)$:


$\int_{R}d^{2}z\partial\overline{\partial}\ln\left|z\right|^{2}f\left(\overline{z}\right)=\int_{R}d^{2}z\partial\frac{1}{\overline{z}}f\left(\overline{z}\right)=-i\oint_{\partial R}d\overline{z}\frac{1}{\overline{z}}f\left(\overline{z}\right)=2\pi f\left(0\right)$.
(b) Now comes the second part of the problem:
To regulate $\ln\left|z\right|^{2}$ use the good old $\epsilon$-environment trick and rewrite it as $\ln\left(\left|z\right|^{2}+\epsilon\right)$. This regularizes you also $\frac{1}{z}$ and $\frac{1}{\overline{z}}$:
$\partial\overline{\partial}\ln\left(\left|z\right|^{2}+\epsilon\right)=\partial\frac{z}{\left|z\right|^{2}+\epsilon}=\overline{\partial}\frac{\overline{z}}{\left|z\right|^{2}+\epsilon}=\frac{\epsilon}{\left(\left|z\right|^2+\epsilon\right)^{2}}$.
From this point the symmetry of the problem makes the use of polar coordinates more convenient. There consider a general test function $f\left(r,\:\theta\right)$, and define $g\left(r^{2}\right)\equiv\int d\theta f\left(r,\:\theta\right)$ which is assumed to be sufficiently well behaved in the asymptotic cases $0$ and $\infty$, then
$\int d^{2}z\frac{\epsilon}{\left(\left|z\right|^{2}+\epsilon\right)^{2}}f\left(z,\:\overline{z}\right)=\int^{\infty}_{0} du\frac{\epsilon}{\left(u+\epsilon\right)^{2}}g\left(u\right)=\left.\left(-\frac{\epsilon}{u+\epsilon}g\left(u\right)+\epsilon\ln\left(u+\epsilon\right)g^{\prime}\left(u\right)\right)\right|_{0}^{\infty}-\int^{\infty}_{0} du \epsilon\ln\left(u+\epsilon\right)g^{\prime\prime}\left(u\right)=g\left(0\right)=2\pi f\left(0\right).$
