## Sunday, February 28, 2016

### A troublesome integral

I calculate an integral that I need when solving problem 5.57 in Griffiths, Introduction to Electrodynamics. I generalized the exercise to $D$ dimensions.
The integral is $$\label{eq:20160227a} \int_S dx\ \frac{\vec n(x)}{|| y - x ||^{D-2} } = \begin{cases}\dfrac{D-2}{D} A y &\text{if }\quad ||y||\le 1 \\[5mm] \dfrac{D-2}{D} A \dfrac{y }{|| y ||^D}&\text{if }\quad ||y||\ge 1 \\ \end{cases}$$ where
• $D \ge 3$
• $S$ is the sphere with radius $1$ in $\mathbb{R}^D$. The sphere itself has $D-1$ dimensions
• $A$ is the area of this sphere
• $\vec n$ is the unit vector perpendicular on the sphere pointing outwards
As an example, if $D = 3$ the integral written out in more detail is \begin{equation*} \int_0^{\pi}\!\! d\theta \int_0^{2 \pi}\!\!\! d \phi\ \sin\theta\ \frac{\vec n(\theta,\phi)}{||\vec y -\vec x(\theta,\phi )||^{D-2} } = \begin{cases}\dfrac{4 \pi}{3} \vec y &\text{if }\quad ||y||\le 1 \\[5mm] \dfrac{4 \pi}{3} \dfrac{\vec y }{|| \vec y ||^3}&\text{if }\quad ||y||\ge 1 \\ \end{cases} \end{equation*} and \begin{equation*} \vec x(\theta,\phi ) = \left( \begin{array}{c} \sin\theta\ \cos\phi\\ \sin\theta\ \sin\phi\\ \cos\theta \end{array} \right) \end{equation*}
Calculation
Write \begin{equation*} F(y) = \int_S dx\ \frac{\vec n(x)}{|| y - x ||^{D-2} } \end{equation*} Because \begin{equation*} \Delta \dfrac{1}{|| x ||^{D-2} } = (2 - D ) A\ \delta(x) \end{equation*}
it follows that \begin{equation*} \Delta F(y) = \int_S dx\ \vec n(x) (2 - D ) A\ \delta(y-x) = \dfrac{y}{||y||} (2 - D ) A \int_S dx\ \delta(y-x) \end{equation*} Using \begin{equation*} \int_S dx\ \delta(y-x) = \delta( || y || - 1 ) \end{equation*} it follows that $F$ satisfies the partial differential equation $$\label{eq:20160227b} \Delta F(y) = y\ (2 - D )\ A\ \delta( || y || - 1 )$$ It is also clear that $F(0) =0$ and $\lim_{y \to \infty} F(y) = 0$. The integral \eqref{eq:20160227a} is thus given by the solution of the partial differential equation \eqref{eq:20160227b} subject to these boundary conditions. The easiest way to solve this partial differential equation is with the ansatz \begin{equation*} F(y) = y\ G(||y||) \end{equation*} The equation \eqref{eq:20160227b} then becomes \begin{equation*} y \Delta G +2 \frac{y}{||y||} G'(||y||) = y\ (2 - D )\ A\ \delta( || y || - 1 ) \end{equation*} Because \begin{equation*} \Delta G = \dfrac{1}{r^{D-1}} \partial_r \left( r^{D-1} \partial_r G \right) \end{equation*} the partial differential equation has been reduced to the ordinary differential equation \begin{equation*} \dfrac{1}{r^{D-1}} \partial_r \left( r^{D-1} \partial_r G \right) +2 \frac{1}{r} G'(r) = (2 - D )\ A\ \delta( r - 1 ) \end{equation*} which is the same as \begin{equation*} \dfrac{1}{r^{D+1}} \partial_r \left( r^{D-1} \partial_r G \right)= (2 - D )\ A\ \delta( r - 1 ) \end{equation*} This equation is easily solved. For example, the solution is given by the potential of a charged sphere in $D + 2$ dimensions. \begin{equation*} G(r) = \begin{cases}\dfrac{D-2}{D} A &\text{if }\quad r\le 1 \\[5mm] \dfrac{D-2}{D} A \dfrac{1 }{r^D}&\text{if }\quad r\ge 1 \\ \end{cases} \end{equation*} I find the calculation of the integral very long. Please leave a comment if you can calculate the integral in a quicker way or with a method that gives more insight.