Double integral with variable limits and denominator with nested roots

[sid_galt]

Can anyone solve this?

$$\int_{-L{\substack{2}}}^{L{\substack{1}}-L{\substack{2}}} \left[ \int_{-L{\substack{2}}+R{\substack{x}}}^{L{\substack{4}}+R{\substack{x}}} \frac{{\sigma}xz}{4\pi\varepsilon{\substack{0}}(x^2+y^2+z^2){\sqrt{(x^2+z^2)(y^2+z^2)}}} dx \right] dz$$

where $$\sigma$$ and $$y$$ are constants.

 

[dextercioby]

Is that integral with variable limits ??If it weren't,you'd be able to integrate wrt "x" quite easily.Though the integral wrt "z" is quite difficult.

Daniel.

 

[sid_galt]

The limits with z are variable.
Actually I was trying to analyze the motion of a charged particle in a charged hollow cuboid with charge density $$\sigma$$ and dimensions $$L_{\substack{1}}$$ (z-axis) and $$2 \times L_{\substack{4}}$$ (x and y axis) with the particle moving along the z axis and its position along z axis at any time given by $$L_{\substack{2}}$$.

This equation I made up for the electric field the charged particle would experience along the x-axis at any point in the hollow cuboid.

Is there any way to solve the integral?

 

[dextercioby]

Shouldn't the integral be a constant...?I mean,should it depend on "x"...?

Make the subsitution

$$x=z\sinh u$$

and solve the integral wrt "x" and then write the one wrt "z" (here),so we can see whether it's doable or not.

Daniel.

 

[sid_galt]

Shouldn't the integral be a constant...?I mean,should it depend on "x"...?

Sorry, my bad. It should be constant.


I tried doing the integration (performed part of integration at site integrals.wolfram.com as it had got very tough). Here's what I got

$$\int_{-L_2}^{L_1-L_2} \displaystyle\frac{\sigma}{4\pi\varepsilon_0} \times \displaystyle\frac{z}{\sqrt{y^2+z^2}} \times \left[\displaystyle\frac{\ln(x^2+z^2)}{2y^2}-\displaystyle\frac{\ln(x^2+y^2+z^2)}{2y^2}\right]dz$$

Is this correct?
Is there any way to integrate it further?

 

[dextercioby]

I don't know whether it is correct or not,i didn't do the calculations...I think your integral can be done.

Use this

$$\int \frac{x\ln \left( 1+x^2\right) }{\sqrt{1+x^2}}dx=\allowbreak \sqrt{\left( 1+x^2\right) }\ln \left( 1+x^2\right) -2\sqrt{\left( 1+x^2\right) } +C$$

Daniel.

 

[sid_galt]

Thanks a lot for the help.

Could you help me in this one?

$$\int \sinh u \arctan \left (\sqrt{\displaystyle\frac{a}{b} + \sinh^2 u}\right) du$$
$$\Rightarrow \int \sinh u \arctan \left (\sqrt{\displaystyle\frac{a}{b} - 1 + \cosh^2 u}\right) du$$

I tried the substitution $$x = \cosh u$$ which got me

$$\displaystyle\frac{dx}{du} = \sinh u$$
$$\Rightarrow du = \displaystyle\frac{dx}{\sinh u}$$

Substituting

$$\int \arctan (\sqrt{c + x^2}) dx \quad where \quad c = \displaystyle\frac {a}{b} - 1$$

Can this one be done?

 

[dextercioby]

I don't think it's expressible as a combo of familiar special functions,else wolfram's site would have done it...

Daniel.