I had this integral pop up that I can't solve:

$\displaystyle \int_C \frac{x}{(x^2+z^2)^{3/2}}dz$

How do you write a base vector in Latex by the way?

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- Mar 27th 2010, 04:43 AMMondreusElectromagnetism: Integral
I had this integral pop up that I can't solve:

$\displaystyle \int_C \frac{x}{(x^2+z^2)^{3/2}}dz$

How do you write a base vector in Latex by the way? - Mar 27th 2010, 05:37 AMHallsofIvy
I don't know what you mean by a "base" vector but you can do vectors by \vec{v} : $\displaystyle \vec{v}$

That's a contour integral isn't it? I had started doing the integral with the substitution $\displaystyle z= x tan(\theta)$ before I recognized that! The integrand is analytic everywhere except at z= ix and z= -ix which are poles. If the contour does not inclose either of those, then the integral is 0. If it does inclose either or both, then you can look at its residues at those points. - Mar 27th 2010, 05:59 AMMondreus
Sorry I meant basis vector. It's a line integral where I'm supposed to calculate the electric field at $\displaystyle x\bold{e_x}$ due to a constant charge density $\displaystyle \rho_\ell$ along the line $\displaystyle C: -b \leq z \leq a$.

The original integral looks like this:

$\displaystyle \bold{E} = \int_C \frac{1}{4\pi \epsilon_0}\frac{x\bold{e_x}-z\bold{e_z}}{(x^2+z^2)^{3/2}}\rho_\ell dz$ - Mar 28th 2010, 01:51 AMMondreus
This is supposely a very common integral since it's in our formula collection, but I don't know how to solve it.

The answer is:

$\displaystyle \frac{z}{x^2(x^2+z^2)^{1/2}}$