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Integration of arctan

  1. Jul 6, 2011 #1
    Hi!

    I am trying to calculate undulations for a smectic A liquid crystal for 1,..,4 dimensions. The general equation for [itex]d[/itex] dimensions are

    [itex]\langle u^2(\mathbf{x})\rangle = \frac{k_BT}{(2\pi)^dB} \int_\frac{1}{L}^{q_c}\frac{\text{d}q_\parallel \text{d}^dq_\perp}{q_\parallel+\lambda^2q_\perp^4}
    [/itex]

    my problem (so far) is for 2 dimensions:

    approximately the problem reduces to

    [itex]
    \langle u^2(\mathbf{x})\rangle = \frac{k_BT}{(2\pi)^dB} \int_\frac{1}{L}^{q_c}\text{d}q_\perp\, \arctan\left(\frac{q_c}{\lambda q_\perp^2}\right)\frac{1}{\lambda q_\perp^2}
    [/itex]

    or, eventually, is there a more clever way to use the first equation? I have tried to use subs with the arctan function and partial integration.

    With help from Abramowitz' Handbook of Mathical Function the second equation can be written as

    [itex]
    \langle u^2(\mathbf{x})\rangle =-\frac{\lambda}{q_c}\left(\frac{q_c}{\lambda}\right)^{3/2} \frac{k_BT}{(2\pi)^dB} \left[2\sqrt{q_\perp} \arctan{q_c} - 2\int\frac{\sqrt{q_\perp}\text{d}q_\perp}{1+ q_\perp^2}\right]
    [/itex]

    but I think I am moving in circles now.. so my problem is basically to calculate either

    [itex]
    \int\text{d}x\arctan(a/x^2)\frac{1}{x^2}
    [/itex]

    or

    [itex]
    \int\frac{\text{d}x \sqrt{x}}{1+x^2}
    [/itex]

    will be grateful for some help! :)

    Al
     
    Last edited: Jul 6, 2011
  2. jcsd
  3. Jul 7, 2011 #2
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