I am interested in the elastic theory of lipids and using geometric methods to model them so I've been reading Geometric Methods in Elastic Theory of Membranes in Liquid Crystal Phases. Within the first few pages, they talk about calculating the orientation order in liquid crystal states by using the order parameter S:
$\ S = 2\pi\int_0^\pi P_2(\cos\theta)f(\theta)\sin\theta d\theta$
where $P_2(\cos\theta) = (3/2)\cos^2\theta - 1/2$. Now they said that when $ f(\theta) = 1/(4\pi)$, S = 0. That's fine and I was able to get that using u-v substitution. Then they specified that when $ f(\theta) = \delta(\cos\theta-1)/(2\pi)$ with the delta being the Dirac function, S = 1. I've been having trouble replicating this result. I read up on delta functions and tried a couple scenarios. The obvious method to me was to pick out the single point where $\delta(x) = 0$, i.e. when $\theta = 0$. This gave me S = 0, so incorrect. Then I tried using change of variables where $ x = \cos\theta - 1$, giving me $dx = -\sin\theta d\theta $, and leaving me with
$ S = \int_0^\pi (\frac{1}2-\frac{3}2(x+1)^2)\delta(x)dx $
This led me to the answer of S = -1, based on my rudimentary understanding that I am pulling out the singular $1/2-3/2(x+1)^2 $ from the integral, setting $ x =0 $ when I do so. Either way, I got frustrated and used Wolfram Alpha which told me the answer was $0$. What is the right way to go about this problem physically and mathematically?
When you adapt your bounds of integration, as @amsmath suggested, you’ll get: $$- \int_{0}^{-2} \left( \frac{3}{2}(x+1)^2 -\frac{1}{2} \right) \delta(x) dx =$$
$$\int_{-2}^{0} \left( \frac{3}{2}(x+1)^2 -\frac{1}{2} \right) \delta(x) dx = 1$$
I hesitate to put “$= 1$”, since $x=0$ on the boundary of integration, instead of the interior, and I’m not sure the convention for that.