16.8 Evaluation of Improper Integrals
You can use the above method of residues to evaluate obnoxious integrals of the form
provided the degree of p
is two less than the degree of
. This can be done by using the contour
which goes from
along the real line and then on the semicircle of radius
Letting CR be the circular part of this contour, for large R,
which converges to 0 as R →∞. Therefore, it is only a matter of taking large enough R to enclose all the
roots of q
which are in the upper half plane, finding the residues at these points and then computing the
contour integral. Then you would let
and the part of the contour on the semicircle will disappear
leaving the Cauchy principal value integral which is desired. There are other situations which will work just
as well. You simply need to have the case where the integral over the curved part of the contour converges
to 0 as R →∞
Here is an easy example.
Example 16.8.1 Find ∫
You know from calculus that the answer is π. Lets use the method of residues to find this. The function
has poles at
. We don’t need to consider −i.
It seems clear that the pole at i
is of order 1 and
so all we have to do is take
Then the integral equals 2πi
That one is easy. Now here is a genuinely obnoxious integral.
Example 16.8.2 Find ∫
It will have poles at the roots of 1 + x4. These are
Using the above contour, we only need consider
Since they are all distinct, the poles at these two will be of order 1. To find the residues at these points, you
would need to have
factoring 1 + x4 and computing the limit, you could get the answer. Then applying L’Hospital’s rule to
identify the limit you know is there,
Similarly, the residue at
Then the contour integral is
You might observe that this is a lot easier than doing the usual partial fractions and trig substitutions etc.
Now here is another tedious example.
Example 16.8.3 Find ∫
The poles of interest are located at i,2i. The pole at 2i is of order 2 and the one at i is of order 1. In
this case, the partial fractions expansion is
The pole at i would be
Now consider the pole at 2i by consideration of the next two terms in the partial fractions expansion. You
must multiply it by
take the derivative and then take a limit as x →
. Multiplying and taking
the derivative yields
Then you have to take a limit as x → 2i which is
Finally, consider the last term which has a pole of order 1.
Then adding in the minus sign, we have the following for the integral.
Sometimes you 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 this.
Example 16.8.4 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 that this is done in order to get a complex integral which reduces to
the one of interest. It follows that a complex integral which reduces to the integral of interest
where γ is the unit circle oriented counter clockwise. Now the integrand has poles of order 1 at those points
= 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.
Other rational functions of the trig functions will work out by this method also.
Sometimes we have to be clever about which version of an analytic function that reduces to a real
function we should use. The following is such an example.
Example 16.8.5 The integral here is
It is natural to try and use the contour in the following picture in which the small circle has radius r
and the large one has radius R.
However, this will create problems with the log since the usual version of the log is not defined on the
negative real axis. This difficulty may be eliminated by simply using another branch of the logarithm as in
Example 16.1.3. Leave out the ray from 0 along the negative y axis and use this example to define L
this set. Thus
will be the angle
Then the function used is f
. Now the only singularities contained in this contour
and the integrand f has simple poles at these points. Thus res
Of course it is necessary to consider the integral along the small semicircle of radius r. This reduces
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.
Example 16.8.6 The Fresnel integrals are
To evaluate these integrals we will consider f
on the curve which goes from the origin to the
on the x
axis and from this point to the point r
along a circle of radius
and from there
back to the origin as illustrated in the following picture.
Thus the curve is shaped like a slice of pie. The angle is 45∘. Denote by γr the curved part. Since f is
0 as r →∞.
This used ∫
Now examine the first of these integrals.
which converges to zero as r →∞. Therefore, taking the limit as r →∞,
and so the Fresnel integrals are given by
The following example is one of the most interesting. By an auspicious choice of the contour it is
possible to obtain a very interesting formula for cotπz known as the Mittag Leffler expansion of
Example 16.8.7 Let γN be the contour which goes from −N −
−Ni horizontally to N
and from there, vertically to N
Ni and then horizontally to −N−
Ni and finally vertically
to −N −
− Ni. Thus the contour is a large rectangle and the direction of integration is in the
counter clockwise direction.