C08-3.mws

COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

Complimentary software to accompany the textbook:

COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc., 40 Tall Pine Drive, Sudbury, MA 01776
Tele. (800) 832-0034; FAX: (508) 443-8000, E-mail: mkt@jbpub.com, http://www.jbpub.com/


CHAPTER 8 RESIDUE THEORY

Section 8.3 Trigonometric Integrals

Amazingly, we can evaluate certain definite real integrals with the aid of the residue theorem. One way to do this by interpreting the definite integral as the parametric form of an integral of an analytic function along a simple closed contour.

The method in this section is used to evaluate integrals of the form Int(F(cos(theta),sin(theta)),theta = 0 .. 2*pi)
by using the contour C:
abs(z) = 1 , and the change of variable

cos(theta) = (z+1/z)/2 , sin(theta) = (z-1/z)/(2*i) , d*theta = d*z/(i*z) ,

and the complex function f(z) = 1/(i*z) F((z+1/z)/2,(z-1/z)/(2*i)) ,

and evaluating the resulting contour integral Int(f(z),z = C .. ` `) .

Load Maple's "residue" procedure.
Make sure this is done only ONCE during a Maple session.

> readlib(residue):

These problems were also solved using Maple's table of integrals.
The method of solution is to make a solution and obtain a contour integral.

We need to use the following substitution procedure.

> Zsub := proc(F1)
local f0;
f0 := F1;
f0 := subs({cos(t)=(z+1/z)/2,
sin(t)=-I*(z-1/z)/2},f0);
f0 := subs({cos(2*t)=(z^2+1/z^2)/2,
sin(2*t)=-I*(z^2-1/z^2)/2},f0);
f0 := subs({cos(3*t)=(z^3+1/z^3)/2,
sin(3*t)=-I*(z^3-1/z^3)/2},f0);
f0 := f0/(I*z);
f0 := normal(f0);
end:


Example 8.10, Page 318. Use substitution and an equivalent contour integral to evaluate:
int(1/(1+3*cos(theta)^2),theta = 0 .. 2*pi) .

> f:='f': F:='F': t:='t': z:='z':
F := 1/(1 + 3*cos(t)^2):
f1 := Zsub(F):
`Given F(t) ` = F;
`Use f(z) ` = f1;

`Given  F(t) ` = 1/(1+3*cos(t)^2)

`Use  f(z) ` = -4*I*z/(10*z^2+3*z^4+3)

Find the singularities of f(z) = -4*i*z/(3*z^4+10*z^2+3) .

> Zn := sort([solve(denom(f1)=0, z)]):
`For f(z) ` = f1;
`The singularities are:`;
z1 := subs(z=Zn[1],z): z[1] = z1;
z2 := subs(z=Zn[2],z): z[2] = z2;
z3 := subs(z=Zn[3],z): z[3] = z3;
z4 := subs(z=Zn[4],z): z[4] = z4;

`For  f(z) ` = -4*I*z/(10*z^2+3*z^4+3)

`The singularities are:`

z[1] = 1/3*I*sqrt(3)

z[2] = -1/3*I*sqrt(3)

z[3] = I*sqrt(3)

z[4] = -I*sqrt(3)

Find out which singularities lie within a circle abs(z) < 1 .

> print(abs(z[1]),`< 1 `, abs(z1)<1, evalb(evalf(abs(z1))<1));
print(abs(z[2]),`< 1 `, abs(z2)<1, evalb(evalf(abs(z2))<1));
print(abs(z[3]),`< 1 `, abs(z3)<1, evalb(evalf(abs(z3))<1));
print(abs(z[4]),`< 1 `, abs(z4)<1, evalb(evalf(abs(z4))<1));

abs(z[1]), `< 1   `, 1/3*sqrt(3) < 1, true

abs(z[2]), `< 1   `, 1/3*sqrt(3) < 1, true

abs(z[3]), `< 1   `, sqrt(3) < 1, false

abs(z[4]), `< 1   `, sqrt(3) < 1, false

Remark. Sometimes Maple will form the list of values in a different order.

It is always necessary to visually inspect the above results before proceeding.

Compute the residues at z[2] and z[3] .

> r2 := residue(f1, z=z2): `Res[f`,z2,`] ` = r2;
r3 := residue(f1, z=z3): `Res[f`,z3,`] ` = r3;

`Res[f`, -1/3*I*sqrt(3), `] ` = -1/4*I

`Res[f`, I*sqrt(3), `] ` = 1/4*I

The value of the integral is computed by the residue calculus:

> f:='f':
`f(t)` = F;
val := 2*Pi*I*(r3 + r2):
print(int(f(t),t=0..2*pi) = val);

`f(t)` = 1/(1+3*cos(t)^2)

int(f(t),t = 0 .. 2*pi) = 0

Computer algebra systems such as Maple are capable of finding some of these difficult integrals.
The functon
F(t) = 1/(1-3*cos(t)^2) can be integrated directly to obtain:

> `f(t) ` = F;
print(int(f1(t),t) = int(F,t));
print(int(f1(t),t=0..2*pi) = int(F,t=0..2*Pi));

`f(t) ` = 1/(1+3*cos(t)^2)

int(-4*I*z(t)/(10*z(t)^2+3*z(t)^4+3),t) = -1/2*arctan(-2*tan(1/2*t)+sqrt(3))+1/2*arctan(2*tan(1/2*t)+sqrt(3))

int(-4*I*z(t)/(10*z(t)^2+3*z(t)^4+3),t = 0 .. 2*pi) = Pi

End of Section 8.3.