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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:

I found this example in a textbook dated 1937 which I use as supplementary material for M828. It gave me some hard time but finally sorted out many fine tricks of contour integration. Some inspiration was provided by the discussion of the Pochhammer's extension of the Eulerian integral of the first kind by Whittaker and Watson.

Evaluate the following integral:

Consider the integral in the complex plane:

where contour C is constructed as follows. Make a cut along the segment of the real axis. Let on the upper edge of the cut. Integrate along the upper edge of the cut in the positive direction. Follow around the point along a small semi-circle in the clockwise direction. The argument of the second factor in the numerator will decrease by . Then proceed along the real axis till and further round the circle in the counter-clockwise direction.

This circle will enclose both branch point of the integrand and , however since the exponents add up to unity the function will return to its initial value:

This is the reason why we only need to make the cut along the segment and not along the entire real positive semi-axis. Then integrate from in the negative direction, then along the second small semi-circle around the point where the argument of the second factor will again decrease by . Finally integrate along the lower edge of the cut and along a small circle around the origin where the argument of the first factor will decrease by .

As the result of this construction the integral is split into the following parts:

Two integrals around the small circles add up to an integral over a circle:

Similarly, the integral around a small circle around the origin vanishes:

Integrals along the segment cancel out:

The integral over the large circle also tends to 0 as R increases. This can be shown using the Jordan's lemma, or by direct calculation:

Finally the only two terms that survive allow us to express the contour integral in terms of the integral along the segment of the real axis:

The contour encloses the only singularity of the integrand which is the pole of the third order at . Hence, by the residue theorem:

The residue can be calculated using the standard formula:

Calculation of the derivative can be facilitated by taking logarithm first: