Integral for infinite resistor grid problem

Question

I'm studying the infinite grid resistor problem, and I've noticed that at some point Mathematica solution diverges with the one that's on the theorical study

$$\frac{1}{i\pi}\int_{0}^{\pi}\frac{1-e^{i \arccos(2-\cos(\alpha))}\cos(\alpha)}{\sin( \arccos(2-\cos(\alpha)))}d\alpha$$

The problem is that the site gives following answer:

$$\frac{2}{\pi}$$

While Mathematica gives this to me:

$$-\frac{2}{\pi}$$

The sign is the opposite. So I'd like to know what's I didn't understood.

This is the command that I've used for making the integral:

R11 = 1/(Pi I) Integrate[(1 - Exp[I (ArcCos[2 - Cos[a]])] Cos[a])/Sin[ArcCos[2 - Cos[a]]], {a, 0, Pi}]

I'd like to know what I'm doing wrong and how can I fix it.

Answer 1

The solution starting from the indefinite integral seems to match:

indef = 1/(I Pi) FullSimplify[
  Integrate[(1 - Exp[I (ArcCos[2 - Cos[a]])] Cos[a])/Sin[ArcCos[2 - Cos[a]]], a], 
     Assumptions -> {0 <= a <= Pi}];
Limit[indef, a -> Pi] - Limit[indef, a -> 0]
(* -(2/\[Pi]) *)