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following examples of Fourier transform solutions of the heat equation, tried the same thing for

the electrostatics Poisson equation

[tex]

\nabla^2 \phi &= -\rho/\epsilon_0 \\

[/tex]

With fourier transform pairs

[tex]

\begin{align*}

\hat{f}(\mathbf{k}) &= \frac{1}{(\sqrt{2\pi})^3} \iiint f(\mathbf{x}) e^{-i \mathbf{k} \cdot \mathbf{x} } d^3 x \\

{f}(\mathbf{x}) &= \frac{1}{(\sqrt{2\pi})^3} \iiint \hat{f}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 k \\

\end{align*}

[/tex]

one gets

[tex]

\begin{align*}

\phi(\mathbf{x}) &= \frac{1}{\epsilon_0} \int \rho(\mathbf{x}') G(\mathbf{x-x'}) d^3 x' \\

G(\mathbf{x}) &= \frac{1}{(2 \pi)^3} \iiint \frac{1}{\mathbf{k}^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k

\end{align*}

[/tex]

Now it seems to me that this integral [itex]G[/itex] only has to be evaluated around a small neighbourhood of the origin. For example if one evaluates one of

the

integrals

[tex]

\int_{-\infty}^\infty \frac{1}{{k_1}^2 + {k_2}^2 + {k_3}^3 } e^{ i k_1 x_1 } dk_1

[/tex]

using a an upper half plane contour the result is zero unless [itex]k_2 = k_3 = 0[/itex]. So one is left with something loosely like

[tex]

G(\mathbf{x}) &= \lim_{\epsilon \rightarrow 0} \frac{1}{(2 \pi)^3}

\int_{k_1 = -\epsilon}^{\epsilon} dk_1

\int_{k_2 = -\epsilon}^{\epsilon} dk_2

\int_{k_3 = -\epsilon}^{\epsilon} dk_3

\frac{1}{\mathbf{k}^2} e^{ i \mathbf{k} \cdot \mathbf{x} }

[/tex]

However, from electrostatics we also know that the solution to the Poission equation means that [itex]G(\mathbf{x}) = \frac{1}{4\pi\lvert{\mathbf{x}}\rvert}[/itex].

Does anybody know of a technique that would reduce the integral limit expression above for [itex]G[/itex] to the [itex]1/x[/itex] form? Am thinking something residue related, but I'm a bit rusty with my complex variables and how exactly to procede isn't obvious.

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# Fourier transform solution to electrostatics Poisson equation?

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