# An integral with residues

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?

• Please post your code in Mathematica syntax ?. Nobody feels like retyping your code. May 21 at 10:51
• may be the Residue command could be of use here. May 21 at 10:55
• The integral written is not around the unit circle. So either what was written is incorrect, or the statement that this is a contour integral is incorrect. May 21 at 14:16
• Problem is confussing (1): If integration is around circle then $x=re^{it}$, (2): The statement $t\in (0,2\pi)]$ conflicts with (1) above or else $t$ in integrand is a different $t$, (3): Integrand has no poles if $a>2r$ so integral would be zero.
– josh
May 21 at 14:46
• People here generally like users to post code as copyable Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find the meta Q&A, How to copy code from Mathematica so it looks good on this site, helpful May 22 at 0:45

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}}


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
(*)