# Integration of arctan

1. Jul 6, 2011

### AllanGH

Hi!

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

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

my problem (so far) is for 2 dimensions:

approximately the problem reduces to

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

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

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

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

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

or

$\int\frac{\text{d}x \sqrt{x}}{1+x^2}$

will be grateful for some help! :)

Al

Last edited: Jul 6, 2011
2. Jul 7, 2011