# Improper integral help: coloumb's law

1. Jul 5, 2008

### Edwardo_Elric

1. The problem statement, all variables and given/known data
i was deriving an infinite line of charge formula by coloumb's law:
so i got stuck with this integral (since it is in the maths forum)
$$\vec{E}_{\rho} = \int_{-\infty}^{\infty} \frac{\rho_L \rho dz}{4\pi\epsilon_o ({\rho}^2 + z^2)^{\frac{3}{2}}}$$

2. Relevant equations
where
$${\rho}_L =$$ linear charge density
$${\rho} =$$ direction perpendicular to z axis in cylindrical coordinates

3. The attempt at a solution
so when integrating (using trigo substitution):
$$\vec{E}_{\rho} = \frac{\rho_L\rho}{4\pi\epsilon_o} (\frac{1}{{\rho}^2} \frac{z}{p^2 + z^2})_{-\infty}^{\infty}$$
this is where i got stuck
no matter how i use lhopitals rule in this equation:
\lim_{z \infty} \frac{z}{({\rho}^2 + z^2)^{\frac{1}{2}}
it keeps going back

because the infinite is supposed to add to equate to 2
$$\vec{E}_{\rho} = \frac{\rho_L}{2\pi\epsilon_o\rho}$$

Last edited: Jul 5, 2008
2. Jul 5, 2008

### HallsofIvy

Staff Emeritus
I think you need to show your work here. This is not at all what I got.

You don't need to use L'Hopital's rule- just divide numberator and denominator by z2. Of course, then the limits are both 0 which is not what you want.

I would prefer to change the limits of integration with each substitution.

3. Jul 5, 2008

### Edwardo_Elric

yeah typo:
$$(\frac{z}{\sqrt{p^2 + z^2}})_{-\infty}^{\infty}$$

the problem is differentiating $$\sqrt{p^2 + z^2}$$ when i keep using LHR it
keeps on going back

edit:
wait did you mean this?
$$\frac{\frac{z}{z}}{\frac{\frac{\sqrt{p^2+z^2}}{z^2}}}}$$
i tihink i get it now

just divide both sides by z
the denominator you divide by sqrt(z^2)

thx alot halls

Last edited: Jul 5, 2008
4. Jul 5, 2008

### HallsofIvy

Staff Emeritus
You say you used a trig substitution so why worry about the integral as a function of z at all? Change the limits of integration as you change the variable.

When I did it, my first substitution was $\rho tan(\theta)= z$. z will go to $-\infty$ and $\infty$ as $\theta$ goes from $-\pi/2$ to $\pi/2$. I then integrated the resulting trig function by letting $u= sin(\theta)$. While theta goes from $-\pi/2$ to $\pi/2$, u goes from -1 to 1 so I only had to evaluate the "u" formula between -1 and 1 without having to go back to the original variable.