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
variation.
Example 52.2.7The first example we consider is the following integral.
∫ ∞ 1
1+-x4dx
− ∞
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.
PICT
Let γ_{r}
(t)
= re^{it},t ∈
[0,π]
and let σ_{r}
(t)
= t : t ∈
[− r,r]
. Thus γ_{r} parameterizes the
top curve and σ_{r} parameterizes the straight line from −r to r along the x axis. Denoting
by Γ_{r} the closed curve traced out by these two, we see from simple estimates
that
and it is only the last two which are found in the inside of Γ_{r}. 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
evaluating
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
this.
Example 52.2.8The integral is
∫ π cosθ
0 2-+-cos-θdθ
This integrand is even and so it equals
∫
1 π --cosθ-dθ.
2 −π 2+ cosθ
For z on the unit circle, z = e^{iθ}, z =
1
z
and therefore, cosθ =
1
2
(z + 1)
z
. Thus dz = ie^{iθ}dθ
and so dθ =
dz
iz
. 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
is
∫ 1( 1) ∫ 2
1- --2-z1(+-z1)-dz = 1- ----z-+-1----dz
2i γ2 + 2 z + z z 2i γz (4z + z2 + 1)
where γ is the unit circle. Now the integrand has poles of order 1 at those points where
z
( )
4z + z2 + 1
= 0. These points are
√- √ -
0,− 2+ 3,− 2− 3.
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
also.
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 52.2.9The integral here is
∫
∞ -lnx--dx.
0 1 + x4
The same curve used in the integral involving
sinx
x
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 y axis and use Theorem 51.2.3 to define L
(z)
on this set. Thus L
(z)
= ln
|z|
+ iarg _{1}
(z)
where arg _{1}
(z)
will be the angle, θ, between
−
π
2
and
3π-
2
such that z =
|z|
e^{iθ}. Now the only singularities contained in this curve
are
1 √- 1√ - 1√- 1 √-
2 2 + 2i 2,− 2 2 + 2i 2
and the integrand, f has simple poles at these points. Thus using the same procedure as
in the other examples,
( 1√ - 1√ -)
Res f,- 2+ -i 2 =
2 2
√ - √ -
1- 2π− -1i 2π
32 32
and
( )
Res f, − 1√2-+ 1 i√2 =
2 2
3-√- 3- √-
32 2π + 32i 2π.
Consider the integral along the small semicircle of radius r. This reduces to
∫ 0
ln|r|+-it4-(rieit)dt
π 1+ (reit)
which clearly converges to zero as r → 0 because r lnr → 0. Therefore, taking the limit as
r → 0,
∫ ∫
-L(z)- −r ln-(−-t)+-iπ-
large semicircle1 +z4 dz + rli→m0+ − R 1 + t4 dt+
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
many examples.
Example 52.2.10Mellin transformationsare of the form
∫
∞ f (x)xαdx.
0 x
Sometimes it is possible to evaluate such a transform in terms of the constant,α.
Assume f is an analytic function except at isolated singularities, none of which are on
(0,∞ )
. Also assume that f has the growth conditions,
C
|f (z)| ≤-b,b > α
|z|
for all large
|z|
and assume that
-C′-
|f (z)| ≤ |z|b1,b1 < α
for all
|z|
sufficiently small. It turns out there exists an explicit formula for this Mellin
transformation under these conditions. Consider the following contour.
PICT
In this contour the small semicircle in the center has radius ε which will converge to 0.
Denote by γ_{R} the large circular path which starts at the upper edge of the slot and
continues to the lower edge. Denote by γ_{ε} the small semicircular contour and denote by
γ_{εR+} the straight part of the contour from 0 to R which provides the top edge of the slot.
Finally denote by γ_{εR−} 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
f
(z)
z^{α−1}. Let
zα−1 ≡ e(ln|z|+iarg(z))(α−1) = e(α−1)log(z)
where arg
(z)
is the angle of z in
(0,2π)
. Thus you use a branch of the logarithm which
is defined on ℂ ∖
(0,∞ )
. Then it is routine to verify from the assumed estimates
that
∫
Rlim→∞ f (z) zα− 1dz = 0
γR
and
∫
lim f (z)zα−1dz = 0.
ε→0+ γε
Also, it is routine to verify
∫ ∫ R
lim f (z)zα−1dz = f (x)xα−1dx
ε→0+ γεR+ 0
and
∫ ∫
α−1 i2π(α−1) R α−1
lε→im0+ γεR− f (z)z dz = − e 0 f (x)x dx.
Therefore, letting Σ_{R} denote the sum of the residues of f
(z)
z^{α−1} which are contained in
the disk of radius R except for the possible residue at 0,
( )∫ R
e(R)+ 1 − ei2π(α− 1) f (x)xα− 1dx = 2πiΣR
0
where e
(R)
→ 0 as R →∞. Now letting R →∞,
∫ R 2πi πe−πiα
lim f (x)xα−1dx = ----i2π(α−1)Σ = ------Σ
R→ ∞ 0 1− e sin(πα)