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Electric field of "half" an infinite charged sheet

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1. The problem statement, all variables and given/known data
A charged sheet with charge density ##\sigma## is described by ##-\infty<x<0,-\infty<y<\infty, z = 0##. Find the electric field at ##(0,0,z)##.

2. Relevant equations
Electric field of continuous density charged body from the Coulomb law:
$$E = \frac{1}{4\pi \epsilon_0}\int_V\frac{\vec{r} − \vec{r_0}}{|\vec{r} − \vec{r_0}|^3}\rho(\vec{r}')dV'$$

3. The attempt at a solution
First I tried to do a double integral but handling symmetry of x y at the same time seems very messy. So I did integral on y dimension first. It is infinite line of charge with density ##\sigma dx##, and the answer is $$\frac{\sigma dx}{2\pi \epsilon_0\sqrt{x^2+z^2}}$$
Then integrate the answer from ##x=-\infty## to ##x=0##.
What I did is integrate z component and x component separately, and do vector addition and take mod in the very end.
So I ended up with 2 integrals:
x component$$\frac{\sigma}{2\pi \epsilon_0}\int_{-\infty}^{0}\frac{x}{x^2+z^2}dx$$
and z component$$\frac{\sigma}{2\pi \epsilon_0}\int_{-\infty}^{0}\frac{z}{x^2+z^2}dx$$
These integral look reasonably simple, but the result I get is strange, so I guess I am wrong at some point. The first one is ##\frac{ln(x^2+z^2)}{2}##, and the definite integral gives intinity. The second one is ##arctan(\frac{x}{z})## gives ##\frac{\pi}{2}##.
I think this kind of question should have a finite answer and the answer given is $$\frac{\sigma}{4\epsilon_0}$$ which is finite. What went wrong?
 
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Actually never mind, the question asked for z component only, I can't believe I overlooked it and spent 2 hours on this.
 

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