Youngstown State University

College of Engineering & Technology

Civil & Environmental/Chemical Engineering Program

ENGR6924: Computer Based Tools for Engineers

This session deals with the evaluation improper integrals using the Residue theorem and complex path integrals.

Trigonometric Integrals via Contour Integrals

8.2  Trigonometric Integrals

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

    Suppose we wish to evaluate an integral of the form  

(8-3) ,    

where F(u,v) is a function of the two real variables u and v.  Consider the unit circle with parametrization  

,    for    ;   

which gives us the following symbolic differentials.

,    and   
(8-4)
.  

Combining      with   ,   we can obtain  

(8-5)     and    .  

Using the substitutions for  ,   , and    in Expression (8-3) transforms the definite integral into a contour integral

,  

where the new integrand is  .  

    Suppose that f(z) is analytic inside and on the unit circle , except at the points    that lie interior to  .  Then the residue theorem gives

(8-6) .  

The situation is illustrated in Figure 8.2.

Figure 8.2  The change of variables from a definite integral on to a contour integral around C.

Some of these problems can be solved using Mathematica's table of integrals.

We need to use the following substitution procedure.

Example 8.10.    Evaluate    by using complex analysis.  

Solution.  Using Substitutions (8-4) and (8-5), we transform the integral to  



where  .  The singularities of f(z) are poles located at the points where    or equivalently, where  .  Using the quadratic formula, we see that the singular points satisfy the relation    which implies that    or  .  Hence the only singularities that lie inside the unit circle are simple poles corresponding to the solutions of  ,  which are the two points    and  .  We use Theorem 8.2 and L'Hôpital's rule to get the residues at and :  

As    and  ,  the residues are given by  .  We now use Equation (8-6) to compute the value of the integral:

Maxima solution

/*example problem no. 1 */

I:'integrate((1/(1+3*(cos(t))^3)),t,0,2*%pi);

f1:1/(1+3*((cos(t))^2));

f2:ev(f1,cos(3*t)=(z^3+1/z^3)/2,cos(t)=(z+1/z)/2)/(%i*z);

f3:ratsimp(f2);

f4:factor(f3);

res1:residue(f4,z,%i/sqrt(3));

res2:residue(f4,z,-%i/sqrt(3));

answer1:2*%pi*%i*(res1+res2);

answer2:ev(I,integrate);

Example 8.11.  Evaluate    by using a computer algebra system.  

Example 8.12.    Evaluate  .  

Solution.  For values of z that lie on the unit circle , we have

   and   

We solve for    to obtain the substitutions  

    and    .  

Using the identity for along with Substitutions (8-4) and (8-5), we rewrite the integral as  




where  .  The singularities of lying inside are poles located at the points and .  We use Theorem 8.2 to get the residues:

Therefore we conclude that  

Maxima Solution

/* example problem no. 2 */

I1:'integrate(cos(2*t)/(5-4*cos(t)),t,0,2*%pi);

f1:cos(2*t)/(5-4*cos(t));

f2:ev(f1,cos(2*t)=(z^2+1/z^2)/2,cos(t)=(z+1/z)/2)/(%i*z);

f3:ratsimp(f2);

f4:factor(f3);

res1:residue(f4,z,0);

res2:residue(f4,z,0.5);

answer1:2*%pi*%i*(res1+res2);

answer2:ev(I1,integrate);

Improper Integrals of Rational Functions

8.3  Improper Integrals of Rational Functions

    An important application of the theory of residues is the evaluation of certain types of improper integrals.  Let f(x) be a continuous function of the real variable x on the interval  .  Recall from calculus that the improper integral of f(x) over  [)  is defined by

,  

provided that the limit exists.  If f(x)  is defined for all real x, then the integral of f(x) over is defined by

(8-7) ,  

provided both limits exist.  If the integral in Equation (8-7) exists, we can obtain its value by taking a single limit:

(8-8) .  

For some functions the limit on the right side of Equation (8-8) exists, but the limit on the right side of Equation (8-7) doesn't exist.

Example 8.13. ,  

but Equation (8-7) tells us that the improper integral of    over   doesn't exist, because the calculation  

    is undefined .  

Therefore we can use Equation (8-8) to extend the notion of the value of an improper integral, as Definition 8.2 indicates.  

Definition 8.2 (Cauchy Principal Value - P.V.).  Let f(x) be a continuous real valued function for all x.  The Cauchy principal value (P.V.) of the

integral    is defined by

,  

provided the limit exists

 Example 8.13 shows that    .  

Example 8.14.  Find the Cauchy principal value of    .

Solution.  

    If  ,  where P(x) and Q(x) are polynomials, then f(x) is called a rational function.  Techniques in calculus were developed to integrate rational functions.  We now show how the Residue Theorem can be used to obtain the Cauchy principal value of the integral of f(x) over  .  

Theorem 8.3 (Contour Integration for Rational Functions).  Let    where P(x) and Q(x) are polynomials, of degree m and n , respectively.  If for all real x and  ,  then the Cauchy Principal Value (P.V.) of the integral is

.  

where    are the poles of    that lie in the upper half plane.   The situation is illustrated in Figure 8.4.  

Figure 8.4  The poles    of    that lie in the upper half-plane.

Example 8.15.  Evaluate  .  

Solution.  We write the integrand as  .  We see that f(z) has simple poles at the points and ,  and that the points    and  ,  are the only singularities of f(z) in the upper half-plane.  Computing the residues, we obtain  

Using Theorem 8.3, we conclude that  

Maxima Solution

/* improper integral rational functions

example no. 1 */

I:'integrate(1/((x^2+1)*(x^2+4)),x,-inf,inf);

f1:1/((z^2+1)*(z^2+4));

res1:residue(f1,z,%i);

res2:residue(f1,z,2*%i);

answer1:2*%pi*%i*(res1+res2);

answer2:ev(I,integrate);

Example 8.16.  Evaluate  .  

Solution.  The integrand    has a poles of order 3 at the points , and    is the only singularity of f(z) in the upper half-plane.  Computing the residue at   , we get  

Therefore

.  

Maxima Solution

/* improper integral rational functions

example no. 2 */

I:'integrate(1/(x^2+4)^3,x,-inf,inf);

f1:1/((z^2+4)^3);

res1:residue(f1,z,2*%i);

answer1:2*%pi*%i*(res1);

answer2:ev(I,integrate);

Maxima Example where limits of integration are from 0 to Inf

Example

Maxima Solution

/* improper integral rational functions

example no. 3 */

I:'integrate((2*x^2-1)/(x^4+5*x^2+4),x,0,inf);

f1:(2*z^2-1)/((z^4+5*z^2+4));

f2:factor(f1);

res1:residue(f2,z,%i);

res2:residue(f2,z,2*%i);

answer1:%pi*%i*(res1+res2);

answer2:ev(I,integrate);

Improper Integrals Involving Trigonometric Functions

8.4  Improper Integrals Involving Trigonometric Functions

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

    and    

are convergent improper integrals.  You may encounter integrals of this type 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 Theorem 8.4 wouldn't hold if we were to use    and    are used instead of  , as you will see when you get to Lemma 8.1.

Theorem 8.4 (Contour Integration for Improper Trig. Integrals).  Let P(x) and Q(x) be polynomials with real coefficients, of degree m and n, respectively, where    and    for all real x .  If    and  

(8-12) ,    then

(8-13) ,    and  

(8-14) ,

where    are the poles of f(z) that lie in the upper half plane and    and    are the real and imaginary parts of  ,  respectively.  

Figure 8.4  The poles    of     or     that lie in the upper half-plane.

Example 8.17.  Evaluate    .  

Solution.  The function f(z) in Equation (8-12) is  ,  which has a simple pole at the point    in the upper half-plane.  Calculating the residue yields

Using Equation (8-14) gives

Maxima Solution

I:'integrate(x*sin(x)/(x^2+4),x,-inf,inf);

f1:z*exp(%i*z)/(z^2+4);

allroots(denom(f1));

res1:realpart(residue(z*exp(%i*z)/(z^2+4),z,2*%i));

I1:2*%pi*(res1);

answer1:float(I1);

answer2:float(ev(I,integrate));

Example 8.18.  Evaluate      .  

Solution.  The function f(z) in Equation (8-12) is  ,  which has simple poles at the points    and    in the upper half-plane.  We get the residues with the aid of L'Hôpital's rule:  

Similarly,

Using Equation (8-13), we get  

Maxima Solution

I:'integrate(cos(x)/(x^4+4),x,-inf,inf);

f1:exp(%i*z)/(z^4+4);

allroots(denom(f1));

res1:imagpart(residue(exp(%i*z)/(z^4+4),z,1+%i));

res2:imagpart(residue(exp(%i*z)/(z^4+4),z,-1+%i));

I1:-2*%pi*(res1+res2);

ratsimp(I1);

answer1:float(I1);

answer2:float(ev(I,integrate));

References

http://math.fullerton.edu/mathews/c2003/IntegralsTrigMod.html

http://math.fullerton.edu/mathews/c2003/IntegralsRationalMod.html

http://math.fullerton.edu/mathews/c2003/IntegralsTrigImproperMod.html