Edited by Valentin Fadeev, Sunday, 18 Sep 2011, 23:23
Now I got really fascinated with this topic. The most exciting part of evaluating integrals using residues is constructing the right contour. There are general guidelines for certain types of integrals, but in most cases the contour has to be tailored for a particular problem. Here is another example. Evaluate the following integral:
Where are real, are roots of the integrand.
The approach used in the previous example does not work as the integral along a large circle centered at the origin does not tend to 0. This, in fact, is a clue to the solution, as it prompts to have a look what indeed is happening to the function for large . But first consider the integrand in the neighborhood of the origin (simple pole):
Now the integrand is regular for large hence it can be expanded in the Laurent series convergent for where is large:
The residue at infinity is defined to be the coefficient at with the opposite sign:
Now we construct contour C by cutting the real axis along the segment and integrating along the upper edge of the cut in the negative direction, then along a small circle and along the lower edge of the cut. As the result of that one of the factors of the integrand increases its argument by . Finally integrate along a small circle
Integrating along this contour amounts to integrating in the positive direction along a very large circle centered at infinity. Hence the outside of C is the inside of this large circle which therefore includes both the residue at the origin and at infinity. Therefore, by the residue theorem:
Around the infinity
Now I got really fascinated with this topic. The most exciting part of evaluating integrals using residues is constructing the right contour. There are general guidelines for certain types of integrals, but in most cases the contour has to be tailored for a particular problem. Here is another example. Evaluate the following integral:
Where are real,
are roots of the integrand.
The approach used in the previous example does not work as the integral along a large circle centered at the origin does not tend to 0. This, in fact, is a clue to the solution, as it prompts to have a look what indeed is happening to the function for large . But first consider the integrand in the neighborhood of the origin (simple pole):
Now the integrand is regular for large hence it can be expanded in the Laurent series convergent for where is large:
The residue at infinity is defined to be the coefficient at with the opposite sign:
Now we construct contour C by cutting the real axis along the segment and integrating along the upper edge of the cut in the negative direction, then along a small circle and along the lower edge of the cut. As the result of that one of the factors of the integrand increases its argument by
. Finally integrate along a small circle
Integrating along this contour amounts to integrating in the positive direction along a very large circle centered at infinity. Hence the outside of C is the inside of this large circle which therefore includes both the residue at the origin and at infinity. Therefore, by the residue theorem: