Electric field due to infinite line charge: Bounds of integration? $$\vec{E} = \frac{\rho _{o}}{4\pi\epsilon}\int_{-\infty}^{+\infty}\frac{dz'(\rho\vec{a_{\rho}} + (z-z')\vec{a_{z}}}{(\sqrt{\rho^{2} + (z-z')^{2}})^{3}}$$
I am confused about what the bounds of integration in calculating the electric field of an infinite line charge would be. When I try integrating it I cannot come up with $\vec{E}=\frac{\rho_{o}}{2\pi\epsilon\rho}\vec{a_{\rho}}$
Edit:
By using an indefinite integral I got the answer:
$$\vec{E} = \frac{\rho _{o}}{4\pi\epsilon}(\frac{-(z-z')}{\rho\sqrt{\rho^{2} + (z-z')^{2}}} \vec{a_{\rho}} + \frac{1}{\sqrt{\rho^{2} + (z-z')^{2}}}\vec{a_{z}})$$
When I sub in $-\infty$ and $+\infty$ the $\vec{a_{z}}$ is equal to $0$. 
The $\vec{a_{\rho}}$ component is 
$$\vec{E} = \frac{\rho _{o}}{4\pi\epsilon\rho}(\frac{-(z-(\infty))}{\sqrt{\rho^{2} + (z-(\infty))^{2}}} \vec{a_{\rho} - \frac{-(z-(\infty))}{\sqrt{\rho^{2} + (z-(\infty))^{2}}} \vec{a_{\rho}})}$$
I don't know how to go about simplifying the expression above.
 A: When you ask these questions, you should define the symbols you use. That will make it clear for people reading it. Took me a few moments to realise you're using $\rho$ for radius away from the line. 

Regardless, you cannot take infinite limits by simply substituting $\infty$ into your expression. That's not how limits work. In this case, you want to write $$\frac{-(z-z')}{\rho\sqrt{\rho +(z-z')^2}} = \frac{1-\left(\frac{z}{z'}\right)}{\rho\sqrt{\frac{\rho^2}{z'^2} + \left(\frac{z}{z'}-1\right)^2}}$$ which you get by dividing top and bottom by $z'$. Now you can think about what happens as $z'$ gets large and positive. The expression will tend to $\frac{1}{\rho}$. 
When we are looking at what happens as $z'$ tends to negative infinity, we have to be more careful. Since $z'$ is negative, we can't just put it into our square root as we did in the expression above. Instead, we put $-z'$ into the square root and keep a negative out front, i.e. we work with 
$$\frac{-(z-z')}{\rho\sqrt{\rho +(z-z')^2}} = - \frac{1-\left(\frac{z}{z'}\right)}{\rho\sqrt{\frac{\rho^2}{z'^2} + \left(\frac{z}{z'}-1\right)^2}}$$ and this tends to $-\frac{1}{\rho}$ as $z' \rightarrow -\infty$. 

Now, if we feed this into the integral, we have $$\vec{E} = \frac{\rho_0}{4\pi\varepsilon_0} \left(\frac{1}{\rho} - \left(-\frac{1}{\rho}\right)\right) = \frac{\rho_0}{2\pi\varepsilon_0\rho}$$ in the direction $\vec{a_p}$. 
Moral: deal with limits properly!
