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

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COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
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CHAPTER 8 RESIDUE THEORY

Section 8.5 Improper Integrals Involving Trigonometric Functions

Let and be polynomials of degree and , respectively, where . We can show (but omit the proof) that if for all real , then

and

are convergent improper integrals. Integrals of this type are sometimes encountered in the study of Fourier transforms and Fourier integrals. We now show how to evaluate them.

Particularly important is our use of the identities

and , where is a positive real number.

The crucial step in the proof of our next theorem will not hold if and are used instead of , as you will see when we get to Jordan's Lemma.

Theorem 8.4 Let and be polynomials, of degree and , where and for all real .

If and , then
P. V. , and
P. V. ,
where , , ..., are the poles of that lie in the upper half plane.

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

> readlib(residue):

Example 8.17, Page 329. Use the residue calculus to evaluate the integral:
.

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

Which poles lie in the upper half plane?

> 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));

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 .

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

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

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

Remark. The integral for this problem is g(x).
It involves a complex component which does NOT seem to cancel out.
Taking the real part gives the correct real answer.

> g := z -> subs(Z=z,int(Z*sin(Z)/(Z^2 + 4),Z)):
`F(x) ` = F0(x);
`G(x) = `, int(F(x),x) = g(x);
`G(oo) - G(-oo)` = g(infinity) - g(-infinity);
`G(oo) - G(-oo)` = simplify(g(infinity) - g(-infinity));
v := int(x*sin(x)/(x^2 + 4),x=-infinity..infinity):
print(int(F(x),x=-infinity..infinity) = v);
print(Re(int(F(x),x=-infinity..infinity)) = v);
print(Re(int(F(x),x=-infinity..infinity)) = evalf(v));
print(`Unfortunately this answer is wrong !`);

The latter method does NOT agree with the answer obtained via residues.

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

Example 8.18, Page 329. Use the residue calculus to evaluate the integral:
.

> f:='f': F:='F': z:='z':
F0 := z -> cos(z)/(z^4 + 4):
f := z -> exp(I*z)/(z^4 + 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;
z4 := subs(z=Zn[4],z): z[4] = z4;

Which poles lie in the upper half plane?

> 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));
print(`0 <`,Im(z[4]),` `, Im(z4)>0, evalb(evalf(Im(z4))>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 residues at and .

> 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)` = F0(x);
val := -2*Pi*(Im(r2)+Im(r3)):
print(int(F(x),x=-infinity..infinity) = val);
val := factor(-2*Pi*(Im(r2)+Im(r3))):
print(int(F(x),x=-infinity..infinity) = val);

Remark. The integral for this problem is g(x), but it CANNOT be used in any meaningful way
to solve the problem at hand.

> g := z -> subs(Z=z,int(cos(Z)/(Z^4 + 4),Z)):
`G(x) ` = cos(x)/(x^4 + 4);
`G(x) = `, int(F(x),x) = g(x);
`G(oo) - G(-oo)` = g(infinity) - g(-infinity);
`G(oo) - G(-oo)` = simplify(g(infinity) - g(-infinity));
v := int(cos(x)/(x^4 + 4),x=-infinity..infinity):
print(int(F(x),x=-infinity..infinity) = v);
print(Re(int(F(x),x=-infinity..infinity)) = evalf(evalc(Re(v))));
print(`Unfortunately this answer is wrong !`);

Which does NOT agree with the answer obtained via residues.

> `F(x)` = F0(x);
val := -2*Pi*(Im(r2)+Im(r3)):
print(int(F(x),x=-infinity..infinity) = val);
val := factor(-2*Pi*(Im(r2)+Im(r3))):
print(int(F(x),x=-infinity..infinity) = val);
val := evalf(-2*Pi*(Im(r2)+Im(r3))):
print(int(F(x),x=-infinity..infinity) = val);
print(`This answer obtained with residues is correct !`);

Lemma 8.1 (Jordan's lemma) Suppose and are polynomials of degreee and , respectively, where .

If is the upper semicircle , for , then

.

End of Section 8.5.