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Can anyone solve integral with variable limits

  1. Apr 1, 2005 #1
    Can anyone solve this?

    [tex]
    \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
    [/tex]

    where [tex]\sigma[/tex] and [tex]y[/tex] are constants.
     
  2. jcsd
  3. Apr 1, 2005 #2

    dextercioby

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    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.
     
  4. Apr 1, 2005 #3
    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 [tex]\sigma[/tex] and dimensions [tex]L_{\substack{1}}[/tex] (z-axis) and [tex]2 \times L_{\substack{4}}[/tex] (x and y axis) with the particle moving along the z axis and its position along z axis at any time given by [tex]L_{\substack{2}}[/tex].

    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?
     
  5. Apr 1, 2005 #4

    dextercioby

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

    Make the subsitution

    [tex] x=z\sinh u [/tex]

    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.
     
  6. Apr 1, 2005 #5
    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

    [tex]
    \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
    [/tex]

    Is this correct?
    Is there any way to integrate it further?
     
  7. Apr 1, 2005 #6

    dextercioby

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    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

    [tex]\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[/tex]

    Daniel.
     
  8. Apr 3, 2005 #7
    Thanks a lot for the help.

    Could you help me in this one?

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

    I tried the substitution [tex] x = \cosh u [/tex] which got me

    [tex] \displaystyle\frac{dx}{du} = \sinh u[/tex]
    [tex] \Rightarrow du = \displaystyle\frac{dx}{\sinh u}
    [/tex]

    Substituting

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

    Can this one be done?
     
  9. Apr 4, 2005 #8

    dextercioby

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

    Daniel.
     
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