COMPLEX ANALYSIS: Maple Worksheets, 2009
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

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CHAPTER 8 RESIDUE THEORY

Section 8.6 Indented Contour Integrals

If is continuous on the interval , then the improper integral of over is defined by

= , provided that the limit exists.

Similarly, if is continuous on the interval , then the improper integral of over is defined by

= , provided that the limit exists.

Theorem 8.5 Let where and are polynomials with real coefficients, of degree and , respectively, where . If has simple zeros at the points , , ..., on the axis, then

P. V. ,

, , ..., are the poles of that lie in the upper half plane.

Theorem 8.6 Let and be polynomials of degree and , respectively, where , and let have simple zeros at the points , , ..., on the - axis. If is a positive real number and if , then

,
and

,

, , ..., are the poles of that lie in the upper half plane.

Remark The formulas in these theorems give the Cauchy principal value of the integral, which pays special attention to the manner in which any limits are taken. They are similar to those in Sections 8.4 and 8.5, except here we add one-half of the value of each residue at the points on the -axis.

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

> readlib(residue):

Example 8.20, Page 334. Use the residue calculus to evaluate the integral:
.

> f:='f': F:='F': z:='z':
f := z -> z/(z^3 - 8):
`f(z) ` = f(z);

Find the singularities of .

> Zn := sort([solve(denom(f(z))=0, z)]):
`For f(z) ` = f(z);
`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;

Which poles lie on the -axis and which poles lie in the upper half plane ?

> print(`0 =`,Im(z[1]),` `,0=Im(z1), evalb(evalf(Im(z1))=0));
print(`0 =`,Im(z[2]),` `,0=Im(z2), evalb(evalf(Im(z2))=0));
print(`0 =`,Im(z[3]),` `,0=Im(z3), evalb(evalf(Im(z3))=0));
print(` `);
print(`0 <`,Im(z[1]),` `,Im(z1)>0, evalb(evalf(Im(z1))>0));
print(`0 <`,Im(z[2]),` `,Im(z2)>0, evalb(evalf(Im(z2))>0));
print(`0 <`,Im(z[3]),` `,Im(z3)>0, evalb(evalf(Im(z3))>0));

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 residue at on the real axis and in the upper half plane.

Remark. Maple does not seem to sor things the same each time it is run.

Be sure to use either or for this example, depending on the sort.

> r1 := residue(f(z), z=z1): `Res[f`,z1,`] ` = r1;
r2 := residue(f(z), z=z2): `Res[f`,z2,`] ` = r2;
r3 := residue(f(z), z=z3): `Res[f`,z3,`] ` = r3;

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

> `F(x)` = f(x);
val := Pi*I*r1 + 2*Pi*I*r3:
print(int(F(x),x=-infinity..infinity) = val);
val := evalc(Pi*I*r1 + 2*Pi*I*r3):
print(int(F(x),x=-infinity..infinity) = val);

Remark. Was the answer positive, if not, Maple did not sort things as expected.

Go back and use the other point if necessary.

Let's see if Maple can do it with anti-derivatives. It picks up an imaginary part in the anti-derivative !
It is necessary to take the real part to obtain the desired soultion !

> g := z -> subs(Z=z,int(f(Z),Z)):
`F(x) ` = f(x);
`G(x) = `, int(F(x),x) = g(x);
print(`Do the computation with limits!`);
L1 := limit(g(x),x=-infinity):
L2 := limit(g(x),x=infinity):
`L1 = limit G(x) as x->-oo ` = L2;
`L2 = limit G(x) as x->+oo ` = L2;
`G(-oo) ` = L1;
`G(oo) ` = L2;
`G(oo) - G(-oo) ` = L2 - L1;
`Re[G(oo) - G(-oo)]` = Re(L2 - L1);
print(int(F(x),x=-infinity..infinity) = Re(L2 - L1));

Although Maple can pull it out directly, taking the real part of a complex answer
leaves something to be desired. You are basically on your own if you trust the
computer to do this integral.

Example 8.22, Page 335. Use the residue calculus to evaluate the integral:
.

> f:='f': F:='F': g:='g': z:='z':
f := z -> exp(I*z)/((z-1)*(z^2 + 4)):
F0 := z -> sin(z)/((z-1)*(z^2 + 4)):
`F(z) ` = F0(z);
`f(z) ` = f(z);

Find the singularities of .

> Zn := sort([solve(denom(f(z))=0, z)]):
`For f(z) ` = f(z);
`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;

Which poles lie on the x-axis and which poles lie in the upper half plane ?

> print(`0 =`,Im(z[1]),` `,0=Im(z1), evalb(evalf(Im(z1))=0));
print(`0 =`,Im(z[2]),` `,0=Im(z2), evalb(evalf(Im(z2))=0));
print(`0 =`,Im(z[3]),` `,0=Im(z3), evalb(evalf(Im(z3))=0));
print(` `);
print(`0 <`,Im(z[1]),` `,Im(z1)>0, evalb(evalf(Im(z1))>0));
print(`0 <`,Im(z[2]),` `,Im(z2)>0, evalb(evalf(Im(z2))>0));
print(`0 <`,Im(z[3]),` `,Im(z3)>0, evalb(evalf(Im(z3))>0));

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 residue at on the real axis and in the upper half plane.

> r1 := residue(f(z), z=z1): `Res[f`,z1,`] ` = r1;
r2 := residue(f(z), z=z2): `Res[f`,z2,`] ` = r2;

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

> `F(z) = ` = F0(z);
val := 2*Pi*Re(r2) + Pi*Re(r1):
print(int(F(x),x=-infinity..infinity) = val);

Maple can find an anti-derivative, but unfortunately, the limiting values at infinity
do NOT produce the correct value.

g := z -> subs(Z=z,int(f(Z),Z)):
`F(x) ` = f0(x);
`g(x) = Integral of f(x) ` = g(x);
`G(oo) - G(-oo)` = g(infinity) - g(-infinity);
`G(oo) - G(-oo)` = simplify(g(infinity) - g(-infinity));
print(`Now do the above computation with limits!`);
L1 := limit(g(x),x=-infinity):
L2 := limit(g(x),x=infinity):
`L1 = limit G(x) as x->-oo ` = L1;
`L2 = limit G(x) as x->+oo ` = L2;
`G(-oo) ` = L1;
`G(oo) ` = L2;
`G(oo) - G(-oo) ` = L2 - L1;
`Re[G(oo) - G(-oo)]` = Re(L2 - L1);
print(int(F(x),x=-infinity..infinity) = Re(L2 - L1));
print(`Unfortunately this answer is wrong !`);

>

This does NOT agree with the computation obtained with the residue calculus.

> `F(z) = ` = F0(z);
val := 2*Pi*Re(r2) + Pi*Re(r1):
print(int(F(x),x=-infinity..infinity) = val);
print(`This answer obtained with residues is correct !`);

End of Section 8.6.