Poisson's Equation : $$\nabla^2 \psi = -\frac{\rho}{\epsilon_0}.........(i)$$ Green function satisfy : $$ \nabla^2G(\mathbf r_1,\mathbf r_2) = -\delta^3(\mathbf r_1-\mathbf r_2) ........(ii)$$ Delta function : $$\delta^3(\mathbb r_1-r_2) = \frac{1}{(2\pi)^3}\iiint e^{i\mathbf q\bullet(\mathbf r_1-\mathbf r_2)} {d^3q}$$ Let $$ G(\mathbf r_1, \mathbf r_2) =\frac{1}{(2\pi)^3}\iiint G^\sim(\mathbf q)e^{\mathbf q\bullet(\mathbf r_1- \mathbf r_2)} $$ Now from (ii) by substituting above values in (ii) $$G\sim(\mathbf q) =\frac{1}{q^2}$$ Therefore$$G(\mathbf r_1,\mathbf r2)=\frac{1}{2\pi}\iiint d^3q \frac {e^{i\mathbf q \bullet \mathbf(r_1-r_2 )}}{q^2}$$
$$\Rightarrow G(\mathbf r_1,\mathbf r2)= \frac{1}{4\pi^2}\int_{q=-\infty}^\infty dq\int_{Cos\theta=-1}^1d(Cos\theta) e^{iq|\mathbf r_1-\mathbf r_2|Cos\theta}$$
$$\Rightarrow G(\mathbf r_1,\mathbf r2)= \frac{1}{4\pi^2}\int_{q=-\infty}^\infty dq\int_{Cos\theta=-1}^1d(Cos\theta) e^{iq|\mathbf r_1-\mathbf r_2|Cos\theta}$$
$$\Rightarrow G(\mathbf r_1,\mathbf r2)= \frac{1}{4\pi^2}\int_{q=-\infty}^\infty dq \frac{e^{iq|\mathbf r_1-\mathbf r_2|}}{iq|\mathbf r_1 - \mathbf r_2|} - \frac{1}{4\pi^2}\int_{q=-\infty}^\infty dq \frac{e^{-iq|\mathbf r_1-\mathbf r_2|}}{iq|\mathbf r_1 - \mathbf r_2|}$$
$$\Rightarrow G(\mathbf r_1,\mathbf r2)= I_1 - I_2$$ Now by using residue method $$I_1 = \frac{1}{4\pi^2i|\mathbf r_1 - \mathbf r_2|}(-\pi i)$$ and $$I_2 = \frac{1}{4\pi^2i|\mathbf r_1 - \mathbf r_2|}(-\pi i)$$ Which give me $G(\mathbf r_1, r_2) = 0;$ If I_2 =0 then It give correct result :$ G(\mathbf r_1,r_2) = -\frac{1}{4\pi|\mathbf r_1-r_2|}$.
I don't understand how $I_2$ get zero.I appreciate Any suggestions or comment.Thank You
I am not quite sure but I think ,I should shift the pole into the upper half semi circle and so lower half semicircle do not having any pole and doing integration second integral in the contour of lower half semi circle which gives zero .