{"id":1177,"date":"2018-11-26T10:40:02","date_gmt":"2018-11-26T10:40:02","guid":{"rendered":"http:\/\/msp01.epgpbooks.inflibnet.ac.in\/?post_type=chapter&p=1177"},"modified":"2019-04-30T12:09:26","modified_gmt":"2019-04-30T12:09:26","slug":"calculus-of-residues-ii","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/chapter\/calculus-of-residues-ii\/","title":{"rendered":"Calculus of residues-II"},"content":{"raw":"
\"epgp<\/a>\r\n<\/span><\/div>\r\n
\r\n\r\nTABLE OF CONTENTS<\/strong>\r\n\r\n\"\"\r\n\r\n \r\n\r\n \r\n\r\n<\/div>\r\n
\r\n\r\nLEARNING OBJECTIVES<\/strong>\r\n\r\n\"\"\r\n\r\n<\/div>\r\n \r\n
\r\n

C a l c u l u s\u00a0 o f\u00a0 r e s i d u e s\u00a0 -\u00a0 I I<\/strong><\/p>\r\n \r\n\r\n1. Evaluation of definite integrals<\/strong><\/span>\r\n

One of the first applications of the Cauchy\u2019s residue theorem was made in the evaluation of definite integrals. The integrals that can be evaluated by the method of residues can also be evaluated by other more traditional methods. However in many cases the method of residues turns out to be much simpler to apply.<\/p>\r\n

In the last unit we laid down the groundwork for the evaluation of definite integrals by the use of residue theorem.<\/p>\r\n

We now take up the actual task of evaluating such integrals.<\/p>\r\n

The integrals that can be evaluated by this method fall into four main categories:<\/p>\r\n\"\"\r\n

Sometimes integrals can be transformed by a complex change of variable to a form which can then be evaluated by these methods.<\/p>\r\n \r\n\r\n2. Integrals of the type\u00a0\u00a0\"\"<\/strong><\/span>\r\n

The method is useful when the function f<\/em> is a rational function of sin\u00a0 \u00a0and cos\u00a0 . To evaluate the integral we make the substitution<\/p>\r\n\"\"\r\n\r\nHence the integral becomes\r\n\r\n\"\"\r\n\r\nThe contour C<\/em> is the unit circle |z<\/em>| = 1 traversed in the positive (anticlockwise) direction.\r\n\r\n \r\n\r\nExample-1<\/span>\r\n\r\nLet us evaluate the integral\r\n\r\n\"\"\r\n\r\nHere a<\/em> is real and greater than 1.\r\n\r\nFirst of all we notice that the integral is invariant under \"\", so that\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\nExample-2<\/span><\/span>\r\n\r\nEvaluate the integral<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\n3. Integrals of the type\u00a0\"\"<\/span><\/strong><\/span>\r\n

We first consider integrals in which the function f(x)<\/em> is a real valued rational function in the real variable <\/span>x<\/em>. We assume that the degree of the denominator is at least two more than that of the numerator and <\/span>f<\/em> has no poles on the real axis, so that the integral exists.<\/span><\/p>\r\n

First of all we extend the definition of the function to the whole complex plane, so that we have function of a complex variable f(z)<\/em>. Now consider the integral over the closed contour <\/span>C<\/em>, which consists of the real axis from\u00a0<\/span>\u2013<\/span>R<\/em> to <\/span>R<\/em> and the semicircle <\/span>\u0393<\/em> in the upper half plane, as shown in the figure below:<\/span><\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\nExample<\/span><\/span>\r\n\r\n\"\"\u00a0\u00a0\u00a0\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n

We expect the integral over the semicircle to vanish in the limit of large <\/span>R<\/em>. Let us verify it rigorously. On the semicircle<\/span><\/p>\r\n\"\"\r\n

Hence the integral over the semicircle is indeed zero, so that<\/p>\r\n\"\"\r\n\r\n \r\n\r\n4. Integrals of the type\u00a0\"\"<\/span><\/strong>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\nExample-1<\/span><\/span>\r\n\r\nLet us evaluate the integral<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\nExample-2<\/span>\r\n

By use of Jordan\u2019s lemma, we can evaluate integrals in which the degree of the denominator is only one more that of the numerator. As an example, now evaluate the integral<\/p>\r\n\"\"\r\n\r\nConsider instead the integral<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\n4.1 An alternative method- the rectangular contour<\/span><\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nsay. Then<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n

Thus we see that <\/span>I<\/em> tends to zero as <\/span>R<\/em> tends to infinity and the result is proved.<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n4.2 Indenting of contour<\/span>\r\n

\"\"<\/p>\r\n \r\n\r\nExample<\/span>\r\n\r\nLet us evaluate the integral\r\n\r\n\"\"\r\n\r\nConsider the function\r\n\r\n\"\"\r\n

This function has a pole at the origin which lies on the path of integration. One way out is to consider instead the function<\/p>\r\n\"\"\r\n

which has a removable singularity at the origin.\u00a0 However we wish to study the method of indenting the contour. We use the contour shown above where the point c<\/em> is the origin. The integral now consists of four parts:<\/p>\r\n\"\"\r\n

Since there are no singularities inside the contour C<\/em>, by Cauchy\u2019s residue theorem<\/p>\r\n\"\"\r\n

As we have already seen, on use of Jordan\u2019s lemma, the integral on the part \u0393<\/em> tends to zero as R<\/em> tends to infinity. Combining all this together, we obtain<\/p>\r\n\"\"\r\n\r\nOn comparing real and imaginary parts we obtain<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\n5. Integrals of the type\u00a0\"\"<\/span><\/strong><\/span>\r\n

Let the function\u00a0f(x)<\/em> be a rational function which has either no poles or simple poles on the positive real axis. If further<\/span><\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\nThere are three main methods to evaluate such integrals by the method of residues.<\/span>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n5.1 Method-1<\/span>\r\n\r\n\"\"\r\n

Here the contour C<\/em> consists of the real axis from \u2013R<\/em> to R<\/em> and the usual semicircle in the upper half plane. If there are poles on the real axis, they are indented in the manner described above.<\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n5.2 Method-2<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nThe value of the intended integral is obtained by equating imaginary parts.\r\n\r\n \r\n\r\n5.3 Method-3<\/span>\r\n\r\n\"\"\r\n\r\nExample<\/span>\r\n

We illustrate all the three methods by evaluating the following integral<\/p>\r\n\"\"\r\n\r\nMethod-1<\/span>\r\n\r\nWrite the integral as\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nConsider therefore the integral<\/span>\r\n\r\n\"\"\r\n

where <\/span>C<\/em> is the rectangular contour of Figure-4. The integrand has only one pole inside the contour, a simple pole at the point<\/span><\/p>\r\n\"\"\r\n\r\nThe integral over the four parts of the contour is<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\nMethod-2<\/span><\/span>\r\n\r\n\"\"\r\n\r\nThe contour consists of six parts; hence<\/span>\r\n\r\n\"\"\r\n\r\nNow<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n

The principle value is with respect to the singularity at <\/span>z<\/em> = 1. On equating the imaginary part we obtain the desired integral<\/span><\/p>\r\n\"\"\r\n\r\nIn addition, on comparing the real parts we get<\/span>\r\n\r\n\"\"\r\n\r\nThe second method has provided the value of two definite integrals in one go.<\/span>\r\n\r\n \r\n\r\nMethod-3<\/span>\r\n

In this case the contour is as shown alongside.\u00a0Inside the contour there is only one singularity,\u00a0a simple pole at z<\/em> = 1. The residue at the pole is<\/p>\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\n6. Integrals of the type\u00a0\"\"<\/span><\/strong><\/span>\r\n

Here log refers to the principal value of the logarithm. The function <\/span>f<\/em> is a rational function of the real variable <\/span>x<\/em>. We assume that the degree of the denominator of <\/span>f<\/em> is at least two more than that of the numerator. We further assume that <\/span>f<\/em> has no poles on the nonnegative real axis, so that the integrals are convergent. We will also assume that <\/span>f<\/em> is an even function of <\/span>x<\/em>.<\/span><\/p>\r\nLet us consider the function<\/span>\r\n\r\n\"\"\r\n\r\nNow<\/span>\r\n\r\n\"\"\r\n\r\nFollowing the usual routine, we can easily show that<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\nExample<\/span><\/span>\r\n\r\nLet us find the integrals<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nOn substituting in the general result obtained above, we have<\/span>\r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\nOn substituting this value above and comparing the real and imaginary parts, we obtain<\/span>\r\n\r\n\"\"\r\n\r\n \r\n\r\n \r\n\r\nSUMMARY<\/strong>\r\n\r\n\"\"\r\n\r\n\r\n<\/div>\r\n\r\n\r\n\r\n
you can view video on Calculus of residues-II<\/strong><\/td>\r\n\"\"<\/a><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>","rendered":"
\"epgp<\/a>
\n<\/span><\/div>\n
\n

TABLE OF CONTENTS<\/strong><\/p>\n

\"\"<\/p>\n

 <\/p>\n

 <\/p>\n<\/div>\n

\n

LEARNING OBJECTIVES<\/strong><\/p>\n

\"\"<\/p>\n<\/div>\n

 <\/p>\n

\n

C a l c u l u s\u00a0 o f\u00a0 r e s i d u e s\u00a0 –\u00a0 I I<\/strong><\/p>\n

 <\/p>\n

1. Evaluation of definite integrals<\/strong><\/span><\/p>\n

One of the first applications of the Cauchy\u2019s residue theorem was made in the evaluation of definite integrals. The integrals that can be evaluated by the method of residues can also be evaluated by other more traditional methods. However in many cases the method of residues turns out to be much simpler to apply.<\/p>\n

In the last unit we laid down the groundwork for the evaluation of definite integrals by the use of residue theorem.<\/p>\n

We now take up the actual task of evaluating such integrals.<\/p>\n

The integrals that can be evaluated by this method fall into four main categories:<\/p>\n

\"\"<\/p>\n

Sometimes integrals can be transformed by a complex change of variable to a form which can then be evaluated by these methods.<\/p>\n

 <\/p>\n

2. Integrals of the type\u00a0\u00a0\"\"<\/strong><\/span><\/p>\n

The method is useful when the function f<\/em> is a rational function of sin\u00a0 \u00a0and cos\u00a0 . To evaluate the integral we make the substitution<\/p>\n

\"\"<\/p>\n

Hence the integral becomes<\/p>\n

\"\"<\/p>\n

The contour C<\/em> is the unit circle |z<\/em>| = 1 traversed in the positive (anticlockwise) direction.<\/p>\n

 <\/p>\n

Example-1<\/span><\/p>\n

Let us evaluate the integral<\/p>\n

\"\"<\/p>\n

Here a<\/em> is real and greater than 1.<\/p>\n

First of all we notice that the integral is invariant under \"\", so that<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Example-2<\/span><\/span><\/p>\n

Evaluate the integral<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

3. Integrals of the type\u00a0\"\"<\/span><\/strong><\/span><\/p>\n

We first consider integrals in which the function f(x)<\/em> is a real valued rational function in the real variable <\/span>x<\/em>. We assume that the degree of the denominator is at least two more than that of the numerator and <\/span>f<\/em> has no poles on the real axis, so that the integral exists.<\/span><\/p>\n

First of all we extend the definition of the function to the whole complex plane, so that we have function of a complex variable f(z)<\/em>. Now consider the integral over the closed contour <\/span>C<\/em>, which consists of the real axis from\u00a0<\/span>\u2013<\/span>R<\/em> to <\/span>R<\/em> and the semicircle <\/span>\u0393<\/em> in the upper half plane, as shown in the figure below:<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Example<\/span><\/span><\/p>\n

\"\"\u00a0\u00a0\u00a0\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

We expect the integral over the semicircle to vanish in the limit of large <\/span>R<\/em>. Let us verify it rigorously. On the semicircle<\/span><\/p>\n

\"\"<\/p>\n

Hence the integral over the semicircle is indeed zero, so that<\/p>\n

\"\"<\/p>\n

 <\/p>\n

4. Integrals of the type\u00a0\"\"<\/span><\/strong><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Example-1<\/span><\/span><\/p>\n

Let us evaluate the integral<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n<\/div>\n

\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Example-2<\/span><\/p>\n

By use of Jordan\u2019s lemma, we can evaluate integrals in which the degree of the denominator is only one more that of the numerator. As an example, now evaluate the integral<\/p>\n

\"\"<\/p>\n

Consider instead the integral<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

4.1 An alternative method- the rectangular contour<\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

say. Then<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Thus we see that <\/span>I<\/em> tends to zero as <\/span>R<\/em> tends to infinity and the result is proved.<\/span><\/p>\n<\/div>\n

\n

 <\/p>\n

4.2 Indenting of contour<\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

Example<\/span><\/p>\n

Let us evaluate the integral<\/p>\n

\"\"<\/p>\n

Consider the function<\/p>\n

\"\"<\/p>\n

This function has a pole at the origin which lies on the path of integration. One way out is to consider instead the function<\/p>\n

\"\"<\/p>\n

which has a removable singularity at the origin.\u00a0 However we wish to study the method of indenting the contour. We use the contour shown above where the point c<\/em> is the origin. The integral now consists of four parts:<\/p>\n

\"\"<\/p>\n

Since there are no singularities inside the contour C<\/em>, by Cauchy\u2019s residue theorem<\/p>\n

\"\"<\/p>\n

As we have already seen, on use of Jordan\u2019s lemma, the integral on the part \u0393<\/em> tends to zero as R<\/em> tends to infinity. Combining all this together, we obtain<\/p>\n

\"\"<\/p>\n

On comparing real and imaginary parts we obtain<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

5. Integrals of the type\u00a0\"\"<\/span><\/strong><\/span><\/p>\n

Let the function\u00a0f(x)<\/em> be a rational function which has either no poles or simple poles on the positive real axis. If further<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

There are three main methods to evaluate such integrals by the method of residues.<\/span><\/p>\n<\/div>\n

\n

 <\/p>\n

5.1 Method-1<\/span><\/p>\n

\"\"<\/p>\n

Here the contour C<\/em> consists of the real axis from \u2013R<\/em> to R<\/em> and the usual semicircle in the upper half plane. If there are poles on the real axis, they are indented in the manner described above.<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n<\/div>\n

\n

 <\/p>\n

5.2 Method-2<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

The value of the intended integral is obtained by equating imaginary parts.<\/p>\n

 <\/p>\n

5.3 Method-3<\/span><\/p>\n

\"\"<\/p>\n

Example<\/span><\/p>\n

We illustrate all the three methods by evaluating the following integral<\/p>\n

\"\"<\/p>\n

Method-1<\/span><\/p>\n

Write the integral as<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Consider therefore the integral<\/span><\/p>\n

\"\"<\/p>\n

where <\/span>C<\/em> is the rectangular contour of Figure-4. The integrand has only one pole inside the contour, a simple pole at the point<\/span><\/p>\n

\"\"<\/p>\n

The integral over the four parts of the contour is<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Method-2<\/span><\/span><\/p>\n

\"\"<\/p>\n

The contour consists of six parts; hence<\/span><\/p>\n

\"\"<\/p>\n

Now<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

The principle value is with respect to the singularity at <\/span>z<\/em> = 1. On equating the imaginary part we obtain the desired integral<\/span><\/p>\n

\"\"<\/p>\n

In addition, on comparing the real parts we get<\/span><\/p>\n

\"\"<\/p>\n

The second method has provided the value of two definite integrals in one go.<\/span><\/p>\n

 <\/p>\n

Method-3<\/span><\/p>\n

In this case the contour is as shown alongside.\u00a0Inside the contour there is only one singularity,\u00a0a simple pole at z<\/em> = 1. The residue at the pole is<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

6. Integrals of the type\u00a0\"\"<\/span><\/strong><\/span><\/p>\n

Here log refers to the principal value of the logarithm. The function <\/span>f<\/em> is a rational function of the real variable <\/span>x<\/em>. We assume that the degree of the denominator of <\/span>f<\/em> is at least two more than that of the numerator. We further assume that <\/span>f<\/em> has no poles on the nonnegative real axis, so that the integrals are convergent. We will also assume that <\/span>f<\/em> is an even function of <\/span>x<\/em>.<\/span><\/p>\n

Let us consider the function<\/span><\/p>\n

\"\"<\/p>\n

Now<\/span><\/p>\n

\"\"<\/p>\n

Following the usual routine, we can easily show that<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

 <\/p>\n

Example<\/span><\/span><\/p>\n

Let us find the integrals<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

On substituting in the general result obtained above, we have<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

On substituting this value above and comparing the real and imaginary parts, we obtain<\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

 <\/p>\n

SUMMARY<\/strong><\/p>\n

\"\"<\/p>\n<\/div>\n\n\n\n
you can view video on Calculus of residues-II<\/strong><\/td>\n\"\"<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"author":3,"menu_order":18,"template":"","meta":{"_acf_changed":false,"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1177","chapter","type-chapter","status-publish","hentry"],"part":3,"_links":{"self":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/pressbooks\/v2\/chapters\/1177","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/wp\/v2\/users\/3"}],"version-history":[{"count":8,"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/pressbooks\/v2\/chapters\/1177\/revisions"}],"predecessor-version":[{"id":1416,"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/pressbooks\/v2\/chapters\/1177\/revisions\/1416"}],"part":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/pressbooks\/v2\/chapters\/1177\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/wp\/v2\/media?parent=1177"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/pressbooks\/v2\/chapter-type?post=1177"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/wp\/v2\/contributor?post=1177"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/msp01\/wp-json\/wp\/v2\/license?post=1177"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}