52.2.3 Contour Integrals And Evaluation Of Integrals
Here are some examples of hard integrals which can be evaluated by using residues.
This will be done by integrating over various closed curves having bounded
Example 52.2.7 The first example we consider is the following integral.
One could imagine evaluating this integral by the method of partial fractions and it
should work out by that method. However, we will consider the evaluation of this integral
by the method of residues instead. To do so, consider the following picture.
: t ∈
top curve and σr
parameterizes the straight line from −r
along the x
the closed curve traced out by these two, we see from simple estimates
This follows from the following estimate.
We compute ∫
using the method of residues.
The only residues
of the integrand are located at points, z
where 1 + z4
These points are
and it is only the last two which are found in the inside of Γr
. Therefore, we need to
calculate the residues at these points. Clearly this function has a pole of order one at
each of these points and so we may calculate the residue at α
in this list by
Similarly we may find the other residue in the same way
Thus, taking the limit we obtain
Obviously many different variations of this are possible. The main idea being that the
integral over the semicircle converges to zero as r →∞.
Sometimes we don’t blow up the curves and take limits. Sometimes the problem of
interest reduces directly to a complex integral over a closed curve. Here is an example of
Example 52.2.8 The integral is
This integrand is even and so it equals
For z on the unit circle, z = eiθ, z =
and therefore, cos
and so dθ
Note this is proceeding formally to get a complex integral which reduces
to the one of interest. It follows that a complex integral which reduces to the one desired
where γ is the unit circle. Now the integrand has poles of order 1 at those points where
= 0. These points are
Only the first two are inside the unit circle. It is also clear the function has simple poles
at these points. Therefore,
Other rational functions of the trig functions will work out by this method
Sometimes you have to be clever about which version of an analytic function that
reduces to a real function you should use. The following is such an example.
Example 52.2.9 The integral here is
The same curve used in the integral involving
earlier will create problems with
the log since the usual version of the log is not defined on the negative real axis. This
does not need to be of concern however. Simply use another branch of the logarithm.
Leave out the ray from 0 along the negative
axis and use Theorem 51.2.3
to define L
on this set. Thus
will be the angle,
Now the only singularities contained in this curve
and the integrand, f has simple poles at these points. Thus using the same procedure as
in the other examples,
Consider the integral along the small semicircle of radius r. This reduces to
which clearly converges to zero as r → 0 because r lnr → 0. Therefore, taking the limit as
r → 0,
Observing that ∫
0 as R →∞,
0 as R →∞.
From an earlier example this becomes
Now letting r → 0+ and R →∞,
which is probably not the first thing you would thing of. You might try to imagine how
this could be obtained using elementary techniques.
The next example illustrates the use of what is referred to as a branch cut. It includes
Example 52.2.10 Mellin transformations are of the form
Sometimes it is possible to evaluate such a transform in terms of the constant,
Assume f is an analytic function except at isolated singularities, none of which are on
Also assume that f
has the growth conditions,
for all large
and assume that
sufficiently small. It turns out there exists an explicit formula for this Mellin
transformation under these conditions. Consider the following contour.
In this contour the small semicircle in the center has radius ε which will converge to 0.
Denote by γR the large circular path which starts at the upper edge of the slot and
continues to the lower edge. Denote by γε the small semicircular contour and denote by
γεR+ the straight part of the contour from 0 to R which provides the top edge of the slot.
Finally denote by γεR− the straight part of the contour from R to 0 which provides the
bottom edge of the slot. The interesting aspect of this problem is the definition of
is the angle of
Thus you use a branch of the logarithm which
is defined on ℂ ∖
Then it is routine to verify from the assumed estimates
Also, it is routine to verify
Therefore, letting ΣR denote the sum of the residues of f
which are contained in
the disk of radius R
except for the possible residue at 0,
0 as R →∞.
Now letting R →∞,