An integral with residues

Question

I am trying to calculate the following integral using the residual method $$ \int_0^{2\pi}dx\frac{n \cos(x)(-a + 2r\cos(x))\sin(n(x+t))-(a-2)\cos(n(x+t))\sin(x)}{(a-2r\cos(x))^2} $$ where $n$ is a positive integer, $a$ is a positive real and $t \in(0,2\pi]$. The integral is done on the circle of radius 1.

Calling $z=e^{ix}$ I tried to calculate the pole contained in the circle $|z|=1$ and use the method of residuals. The poles are identified by the condition $$ (a-r(z^{-1}+z) )^2=0 $$ and that contained in the unit circle is $z_0 = \frac{a-\sqrt{a^2-4r^2}}{2r}$ with the condition $a > 2r$. How can I calculate the residue at this point?

Answer 1

One has to use the transformation formula for residues over the sum of fractions with simple roots. But the pole is order 2, so there is no residue

    Residue[expr,{x,ArcCos[a/(2r)]}]

But for numeric integers, with a bit of preparing sums of integrands of simple exponentials in indefinite integrals, there are results

  Itg[m_, r_, a_] := 
   Itg[m] = 
   FullSimplify[
   Subtract @@ 
   (((Integrate[#, x] &) /@ 
     ExpandAll@
      TrigToExp@(
        (1/(a -2 r Cos[x])^2  * 
            (-2 a r Cos[x]^2 Sin[m ( t + x)] + 
            2 Cos[m ( t + x)] Sin[x] - 
            a Cos[m ( t + x)] Sin[x]) )) // 
    Simplify) /. {{x -> 2 \[Pi]}, {x -> 0}})]

Here are some

   {{0,n==0},
   {-((I a^2 E^(-I t) \[Pi])/r^2),n==1},
   {(I a E^(-2 I t) \[Pi] (4-2 a-3 a^2+2 r^2))/(2 r^3),n==2},
   {-((I E^(-3 I t) \[Pi] (a^2 (-6+a (3+4 a))-2 (-2+a+a^2) r^2))/(2 r^4)),n==3}}

Answer 2

Here's an example computing integral 3 ways: (1): Numerically, (2): Using the built-in Residue command, and (3): Using the built-in Series command:

a = 3;
r = 1;
n = 3;
t = Pi/4; 
theInt = (
  n Cos[x] (-a + 2 r Cos[x]) Sin[n (x + t)] - (a - 2) Cos[
     n (x + t)] Sin[x])/(a - 2 r Cos[x])^2;

poles = x /. Solve[(a - 2 r Cos[x]) == 0, x]
poles // N

The poles are in this case $\pm\arccos(3/2)+2k\pi,\quad k\in\mathbb{Z}$ and $\pm \arccos(3/2)\approx \pm 0.962424i$ which are in the unit circle.

(1): First numerically integrate it over the unit circle via $x=e^{i\theta}$:

NIntegrate[(theInt  I Exp[I theta]) /. x -> Exp[I theta], {theta, 0, 2 Pi}]
(* 8.56637*10^-11 + 106.629 I *)

which gives approx $106.629i$ and a small residual real part.

(2): Use Residue to compute the residues at these poles and the Residue Theorem to compute the integral:

r1 = Residue[theInt, {x, -ArcCos[3/2]}] // N
r2 = Residue[theInt, {x, ArcCos[3/2]}] // N
2 Pi I (r1 + r2) // N
 (* 8.48528 - 8.53815 I
    8.48528 + 8.53815 I
    0. + 106.629 I *)

(3): Expand the integrand around each pole using Series and extract the $\displaystyle \frac{1}{x\pm \arccos(3/2)}$ terms as the residues:

series1 = Series[theInt, {x, ArcCos[3/2], 4}];
residue1 = Coefficient[series1, 1/(x - ArcCos[3/2])];
series2 = Series[theInt, {x, -ArcCos[3/2], 4}];
residue2 = Coefficient[series2, 1/(x + ArcCos[3/2])];
residue1 // N
residue2 // N

(*
Out[94]= 8.48528 + 8.53815 I
Out[95]= 8.48528 - 8.53815 I
(*)