Anomalous value of electric field due to a uniformly charged disc at a point on its central axis Upon trying to find the electric field using integration (as done below) we get the following result :

According to me there are two problems with this result : 


*

*This function is not of odd parity. 
It should be of odd parity because the disc divides the space into two symmetric parts and therefore the value of electric field in the two parts must be of equal magnitudes and opposite sign (because of opposite direction).

*The value of electric field at center of disc is coming out to be finite according to the result which I feel intuitively must be zero.


So my question is : 
Are these arguments right? If yes, where was the mistake done while deriving the result. If no, why?
Thank you very much :)
Image source : hyperphysics
 A: The result is correct only for $z> 0$. If you follow the calculation carefully, you will see that it appears $\sqrt{{{z}^{2}}}=\left| z \right|$. If $z<0$, $\sqrt{{{z}^{2}}}=-z$ or $\frac{\sqrt{{{z}^{2}}}}{z}=\pm 1$
To be more precise : you have to make a change of variable :
$z\left( \int\limits_{0}^{R}{\frac{rdr}{{{\left( {{r}^{2}}+{{z}^{2}} \right)}^{3/2}}}} \right)=z\left( \int\limits_{{{z}^{2}}}^{{{R}^{2}}+{{z}^{2}}}{\frac{1}{2}\frac{du}{{{\left( u \right)}^{3/2}}}} \right)=-\frac{z}{\sqrt{{{R}^{2}}+{{z}^{2}}}}+\frac{z}{\sqrt{{{z}^{2}}}}=\pm 1-\frac{z}{\sqrt{{{R}^{2}}+{{z}^{2}}}}$
It is well known that the electric field is discontinuous at the crossing of a charged sheet. Your symetry argument (value 0) would be valuable for a volumic distribution of charges. For a surfacic distribution, the electric field is not defined on the sheet. You should abandon the surfacic model to pass to a volumic model and the field would be zero in 0 if the symmetry is preserved.
A: You are wrong in the intergation result. Instead of 1 it should be $\text{sign}(z)$. The electric field projection $E_z$ is proportional to an odd function of $z$, so it is odd too which is visible from your first integral. It is also intuitively comprehensible. And it is zero when $z=0$. As often happens in Physics, we use some approximations, in particular, in your case the notion of the surface charge is governed by the inequality $r\gg a$, where $a$ is the plate thickness.
