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:


While Mathematica gives this to me:


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.

  • $\begingroup$ The article you sited is wrong, because the integrand is strictly negative. By the way, NIntegrate gives the same answer as Integrate to several significant figures. $\endgroup$ – bbgodfrey Sep 16 '19 at 18:55

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]) *)
  • $\begingroup$ It works also without the limit, (indef /. a -> Pi) - (indef /. a -> 0). So I'm doing the definite integral in the wrong way? $\endgroup$ – Jepessen Sep 16 '19 at 9:05

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