Can anyone solve integral with variable limits

1. Apr 1, 2005

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.

2. Apr 1, 2005

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.

3. Apr 1, 2005

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?

4. Apr 1, 2005

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.

5. Apr 1, 2005

sid_galt

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?

6. Apr 1, 2005

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.

7. Apr 3, 2005

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?

8. Apr 4, 2005

dextercioby

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

Daniel.