Improper integral help: coloumb's law

[Edwardo_Elric]

Homework Statement

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}}}

Homework Equations

where
{\rho}_L = linear charge density
{\rho} = direction perpendicular to z axis in cylindrical coordinates

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
the answer which is
\vec{E}_{\rho} = \frac{\rho_L}{2\pi\epsilon_o\rho}

 

[HallsofIvy]

Homework Statement

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}}}

Homework Equations

where
{\rho}_L = linear charge density
{\rho} = direction perpendicular to z axis in cylindrical coordinates

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

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

no matter how i use lhopitals rule in this equation:

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.

\lim_{z \infty} \frac{z}{({\rho}^2 + z^2)^{\frac{1}{2}}
it keeps going back
\lim_{z \to \infty} \frac{z}{({\rho}^2 + z^2)^{\frac{1}{2}}

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

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

 

[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 a lot halls

 

[HallsofIvy]

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.