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## Homework Statement

Page 35 of Jackson's Electrodynamics (3rd ed), it gives the following equation (basically trying to prove your standard 1/r potential is a solution to Poisson equation):

[tex] \nabla^2 \Phi_a = \frac{ -1 }{ \epsilon_0 } \int \frac{ a^2 }{( r^2 + a^2)^{5/2} } \rho( \boldsymbol{x'} ) d^3 x' [/tex]

Where r is distance from center of a sphere with radius R, centered at [itex] \boldsymbol{x} [/itex] and [itex] a [/itex] is a parameter much smaller than R whose limit approaches to 0. And note that [itex] r = \sqrt{ (x' - x)^2 + (y' - y)^2 + (z' - z)^2} [/itex]. R is chosen such that [itex] \rho( \boldsymbol{x'} ) [/itex] changes little inside the sphere. The integral is zero outside the sphere as [itex] a [/itex] approaches zero, so only the inside of the sphere is considered.

Then the book says, "with a Taylor series expansion of the well behaved [itex] \rho( \boldsymbol{x'} ) [/itex] around [itex] \boldsymbol{x'} = \boldsymbol{x} [/itex], one finds":

[tex] \nabla^2 \Phi_a = \frac{ -1 }{ \epsilon_0 } \int_0^R \frac{ a^2 }{( r^2 + a^2)^{5/2} } [ \rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \nabla^2 \rho + \ldots ] r^2 dr + O( a^2 ) [/tex]

Essentially I tried to fill in the gaps between those two steps.

## Homework Equations

Taylor expansion for scalar field:

[tex] f(\boldsymbol{x'}) = f(\boldsymbol{x}) + \boldsymbol{x} \cdot \nabla f + \frac{ 1 }{ 2 } (\boldsymbol{x} \cdot \nabla)^2 f [/tex]

## The Attempt at a Solution

Well, essentially I attempted to solve the problem by using the relevant equations. First of all, I changed [itex] d^3 x' [/itex] into spherical polar and the angular parts integrate into [itex] 4 \pi [/itex].

This is where I get lost because when I did the Taylor expansion, the first order terms did not go away. Secondly the second order term is a problem because I never managed to get a six at the bottom. In fact, what I get is:

[tex] \rho + r \frac{ \partial \rho }{ \partial r } + \frac{1}{2} r^2 \frac{ \partial \rho }{ \partial r } + \frac{1}{2} r^2 \frac{ \partial^2 \rho }{ \partial r^2 } [/tex]

Which is not the same as

[tex] \rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \nabla^2 \rho = \rho( \boldsymbol{x} ) + \frac{ r^2 }{ 6 } \frac{\partial^2 \rho}{\partial r^2} [/tex]