# Insane Integral

1. Mar 29, 2009

### roeb

1. The problem statement, all variables and given/known data
http://img4.imageshack.us/img4/4224/landau.png [Broken]
http://g.imageshack.us/img4/landau.png/1/ [Broken]

Note: do = 2*pi*sin(x) dx

Well, as you can see this is an extremely painful integral.

2. Relevant equations

3. The attempt at a solution

I have tried u = cos(1/2x) resulting in:
du = -sin(1/2 x) (1/2) dx
sin(x) = 2*sin(x/2)*cos(x/2)

(n^2 u - nu^2 - n + u)/(n^4 + 2n^2 - 4n^3 u + 4n^2 u^2 - 4nu) (by expanding both the top and bottom, but as you can see it's messy and useless, I'm also dropping the constant because I don't care about them right now).

I have no idea what kind of substitution to use for these beast... It evaluates to pi*a^2 with the integration limits 0 to X_max..

Anyone have any idea how to do this by hand? I am tempted to use Matlab, but I really am supposed to do it manually.

Last edited by a moderator: May 4, 2017
2. Mar 30, 2009

### Cyosis

The substitution you used is fruitful and if you don't work out the parentheses you should have gotten the following integral. (note that as x goes from 0 to xmax u goes from 1 to 1/n)

$$-2 a^2 n^2 \pi \int_1^{1/n} \frac{(n u-1)(n-u)}{(n^2+1-2 n u)^2}\,du$$

Now note that $$\frac{d}{du}(n u-1)(n-u) = n^2+1-2 n u$$ then use integration by parts. After some algebra you should get the value $$a^2 \pi$$.

Last edited: Mar 30, 2009