# Integration of a potential over a cylinder.

It already happened several times. I'm trying to calculate a field, created by something-distributed-over infinite cylinder. And I get an integral like this: $$f(r)=\int_{-\infty}^{\infty}dz\int_0^{2\pi}d\phi\frac{1}{\sqrt{z^2+r^2+R^2-2rR\cos\phi}}$$ And every time I get something like this I have to turn back miserably.

In that particular case I even know the supposed answer -- it must be somethig like $2/r$. The problem is that I have no idea how to approach this integration.

I've tried to integrate first over $d\phi$ or over $dz$. Hoping then to get a remining integral using compex analysis. But everything got too messy.

Does anyone knows/heard of a practical method of dealing with this kind of integrals?

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Do you really want to evaluate the integral, or would an argument using Gauß' law suffice? – joriki Oct 13 '12 at 22:10
@joriki You got my intentions right. I don't want Gauss law. I want to take that integral. – Kostya Oct 14 '12 at 2:28

The perhaps unexpectedly simple reason why you can't evaluate this integral is that it diverges.

You're trying to calculate the potential generated by an infinite charged cylinder from first principles, but this diverges because it's basically the integral over $1/z$. You can get the field using Gauß' law and then integrate it to find a potential, but that potential differs by an infinite additive constant from the integral over the potential of the charge distribution. The situation is similar to that of an infinite charged sheet.

If you want to evaluate integrals instead of using Gauß' law, you can evaluate the integral for the field using the Weierstraß substitution:

\begin{align} f'(r) &=\int_{-\infty}^{\infty}\mathrm dz\int_0^{2\pi}\mathrm d\phi\frac{R\cos\phi-r}{\sqrt{z^2+r^2+R^2-2rR\cos\phi}^3} \\ &= 2\int_0^{2\pi}\mathrm d\phi\frac{R\cos\phi-r}{r^2+R^2-2rR\cos\phi} \\ &= 4\int_0^\pi\mathrm d\phi\frac{R\cos\phi-r}{r^2+R^2-2rR\cos\phi} \\ &= 4\int_0^{\infty}\frac{R(1-t^2)/(1+t^2)-r}{r^2+R^2-2rR(1-t^2)/(1+t^2)}\frac{2\mathrm dt}{1+t^2} \\ &= 4\int_{-\infty}^{\infty}\frac{R(1-t^2)-r(1+t^2)}{(r^2+R^2)(1+t^2)-2rR(1-t^2)}\frac{\mathrm dt}{1+t^2} \\ &= -4\int_{-\infty}^{\infty}\frac{r-R+(r+R)t^2}{(r-R)^2+(r+R)^2t^2}\frac{\mathrm dt}{1+t^2} \\ &= -\frac4{r+R} \int_{-\infty}^{\infty}\frac{\lambda+t^2}{\lambda^2+t^2}\frac{\mathrm dt}{1+t^2} \end{align}

with $\lambda=(r-R)/(r+R)$. We can close the contour with a semicircle at infinity since the integrand decays as $t^{-2}$. The poles are at $\pm\mathrm i$ and $\pm\lambda\mathrm i$. The residue at $\mathrm i$ is

$$\frac{\lambda-1}{\lambda^2-1}\frac1{2\mathrm i}=\frac1{\lambda+1}\frac1{2\mathrm i}=\frac{r+R}{4r\mathrm i}\;,$$

and the one at $\lambda\mathrm i$ is also

$$\frac{\lambda-\lambda^2}{2\lambda\mathrm i}\frac1{1-\lambda^2}=\frac1{2\mathrm i}\frac1{1+\lambda}=\frac{r+R}{4r\mathrm i}\;.$$

For $R\gt r$, we have $\lambda\lt0$, so we need the residue at $-\lambda\mathrm i$, so the contributions cancel and the integral vanishes; whereas for $R\lt r$ we have $\lambda\gt0$, so the contributions add up and the integral is

$$2\pi\mathrm i\frac{-4}{r+R}2\frac{r+R}{4r\mathrm i}=-\frac{4\pi}r\;.$$

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You are right in every detail. Thanks. – Kostya Oct 16 '12 at 17:54