Complex varibles:Contour integration examples
1
Problem 1: sinx/x
Integration of sin x/x from −∞ to ∞ is an interesting problem
1.1
Method 1
In the first method let us consider
Z
∞
−∞
eiax
dx =
x
Z
∞
−∞
cos(ax)
dx + i
x
Z
∞
−∞
sin(ax)
dx
x
So when Z is real the integral is the imaginary part of the above integral. So if we select a contour
as shown below, then on the real axis, Z = x and if we can find the values of the other integrals we
can find the answer as the imaginary part of the real line integral. Thus, we replace the integral
with
Z −ǫ iaz
Z
Z ∞ iaz
Z
I
e
e
eiaz
eiaz
eiaz
dz =
dz +
dz +
dz +
dz
z
−∞ z
ǫ
Cǫ z
CR z
C z
As ǫ tends to zero we find that the first and the third part give us the real line integral
Z
∞
−∞
eiax
dx
x
As for CR integral, using Jordan’s lemma, it goes to zero for a > 0. With regard to Cǫ we have
for
Z
eiaz
dz
Cǫ z
z = 0 is a simple pole. Thus, one of the theorems says
Z
Cǫ
eiaz
dz = iφC−1 = −iπ
z
where C−1 is the residue at the pole. φ = −π because we go in the CW direction. Thus we have
I
C
eiaz
dz =
z
Z
∞
−∞
eiax
dx − iπ = 0
x
since no pole is enclosed. Thus,
Z
Matching real to real etc., we have
Z
∞
−∞
∞
−∞
eiax
dx = iπ
x
sinax
dx = π
x
1
CR
C
3
-R
R
z=0
Figure 1: Contour for problem 1.
1.2
Method 2
In the second method, we will include the pole within the contour. Thus,
I
C
eiaz
dz =
z
Or
I
C
Z
−ǫ
−∞
eiaz
dz =
z
Z
eiaz
dz +
z
∞
−∞
Z
CǫCC
eiaz
dz +
z
Or
Z
∞
−∞
Or
Z
Z
eiaz
dz +
z
Cǫ CC
Z
∞
ǫ
eiaz
dz +
z
Z
CR
eiaz
dz
z
eiaz
dz = 2πi(ResF (z = 0))
z
eiaz
dz + iπ = 2πi
z
∞
−∞
eiaz
dz = πi
z
And the result follows.
1.3
Method 3
Z
∞
−∞
sin(ax)
dx =
x
Z
∞
−∞
eiax − e−iax
dx →
2ix
Z
∞
−∞
eiaz − e−iaz
dz
2iz
The logic here is this: The first integral (what we want) is exactly equal to the second representation. We make it into a complex integral (with real limits) such that the complex integral will
equal the two prior integrals when z = x. Hence the complex integral (with real limits) must be a
branch of any contour we choose. Next, since the complex integral is a sum then it has to be split
into its components and then each integral has a singularity at z = 0. We must then calculate
2
the penalty for integrating through the singularity.
The first term is
Z ∞ iaz
e
dz
−∞ 2iz
If this is integrated over the same contour as in figure 2,
I
eiaz
dz =
2iz
C
Z
eiaz
dz +
2iz
∞
−∞
Z
Cǫ Cw
eiaz
dz +
2iz
Z
CR
The CR integral is zero by Jordan lemma. Or
eiaz
dz =
2iz
I
C
Or
I
C
Z
∞
−∞
eiaz
dz =
2iz
Or
Z
eiaz
dz +
2iz
Z
∞
−∞
∞
−∞
Z
Cǫ Cw
eiaz
dz = 0
2iz
eiaz
−iπ
dz +
=0
2iz
2i
π
eiaz
dz =
2iz
2
The second term is
−
Z
∞
−∞
e−iaz
dz
2iz
If this is integrated over the same contour
I
C
e−iaz
dz =
2iz
Z
e−iaz
dz +
2iz
∞
−∞
Z
Cǫ Cw
eiaz
dz +
2iz
Z
→∞
CR
The CR integral is infinity because for a > 0 the Jordan lemma holds for the upper half of the
complex plane. So we take the bottom contour and include the pole.
I
C
e−iaz
dz =
2iz
Z
∞
−∞
e−iaz
dz +
2iz
Z
Cǫ Cw
eiaz
dz = −2πiRes(F (z = 0))
2iz
where the negative on the residue is because the direction of the contour is clockwise. Or
I
C
e−iaz
dz =
2iz
Z
e−iaz
−π
dz +
= (−1)2πiRes(F (z = 0))
2iz
2
∞
−∞
Or
Z
∞
−∞
Thus
−
So
Z
∞
−∞
Z
π
e−iaz
dz = + − π
2iz
2
∞
−∞
e−iaz
π
dz = +
2iz
2
eiaz − e−iaz
dz = π =
2iz
3
Z
∞
−∞
sin(ax)
dx
x
1.4
Method 4
At times, a point z seems like a singularity, but when the limits are computed, then it is not a
singularity. If an integral has to go through a singularity then there is no choice but to go through
it using an ǫ contour and see if this is a integrable singularity by taking ǫ to zero. There is no
avoiding it by deforming the contour. Avoiding the ’singularity’ by starting with an indented
contour can be done only when the seeming ‘singularity’ is not a singularity, i.e., the function is
analytic in the domain. With a singularity it is not. However, in this case, there is no singularity
in sin(ax)/x to begin with at x = 0. So when we break it into
Z
∞
−∞
sin(ax)
dx =
x
Z
∞
−∞
eiax − e−iax
dx →
2ix
Z
∞
−∞
eiaz − e−iaz
dz
2iz
the complex integral is continuous at z = 0 and analytic all the way along the real axis. First let
us observe that around the origin
lim
I
eiaz − e−iaz
dz,
2iz
lim
I
eiaǫ(cosθ+isinθ) − e−iaǫ(cosθ+isinθ) iθ
iǫ dθ = 0.
2iǫiθ
ǫ→0 ǫ
withz = ǫiθ ,
dz = iǫiθ dθ
we get
ǫ→0 ǫ
If f is continuous in a domain D and integrals round closed contours are zero, then the function
is analytic in D. Thus, the function is not singular and has no integral contributions from z = 0.
So nothing new can come integrating through the z = 0 point. Hence, we choose an indented
contour to begin with as shown in figure below (figure 2). The value of the integral must be the
same passing through the origin or on the indented contour.
C
3
R
−R
z=0
Figure 2: Contour for problem 3 method 4.
First Term:
Z
∞
−∞
eiaz
dz
2iz
We have two terms in the integral above. We take the first term. We now choose to deform the
already indented path. If we continuously deform the path now to +∞ this function is analytic
throughout and follows Jordan’s Lemma. Given a > 0 and 1/z goes to zero uniformly as z goes
to infinity. Thus,
Z ∞ iaz
e
dz = 0.
−∞ 2iz
4
Second Term:
Z
∞
−∞
e−iaz
dz
2iz
This term because of the −iaz = −ia(x + iy) = −iax + ay goes to infinity as y becomes large
positive in the upper half plane. This function is non-analytic at z = ∞ in the upper half. (If
the value of a function becomes infinity then it has a singularity at that point). So we deform it
the contour downward as shown in the figure. So for the second integral we have the downward
C
3
R
−R
z=0
CR
Figure 3: Contour for problem 3 method 4a.
contour
Z
∞
−∞
e−iaz
dz
2iz
If this is integrated over the new downward contour
I
C
e−iaz
dz =
2iz
Z
+
CR
Z
Cǫ Cw
e−iaz
dz +
2iz
Z
+
−∞ to 0
Z
0 to −∞
e−iaz
dz + +
2iz
Z
CR
We will hold onto the end points, i.e., +∞ and −∞, and continuously deform the contour downward, into CR at −∞. The value of this CR integral by Jordan’s Lemma is zero. The two
upward and downward integrals sum to zero, since the function is analytic and uniquely defined.
The integral limits are opposite. Thus, we are left with the integral round the singularity in the
clockwise direction. This gives a residue
Z
Cǫ Cw
Thus
Z
e−iaz
1
dz = −2πi = −π
2iz
2i
∞
−∞
eiaz − e−iaz
dz = π
2iz
as before.
2
The motivation problem
The following integral needs to be worked out
−iωF
v(x) =
2πB
Z
∞
−∞
eikx
dk.
k 4 − kp4
(1)
The evaluation of the above improper real integral will be done using complex analysis. We
now extend the function v(k) over the complex k plane. The motivation in doing so being that
5
in the complex plane using Cauchy integral theorem and Jordan’s lemma, infinite integrals of the
form in equation (1), can be easily computed. This solution methodology is more elegant and
powerful.
Cauchy integral formula states that for a function f (z) analytic except at a finite number of
points (z1 , z2 , . . . zN ) lying inside a closed contour C0 , the following relation holds
I
f (z)dz = i2π
C0
N
X
n=1
Residue [f (z)] z=zn
.
(2)
Jordan’s lemma states that if in the above contour integral, f (z) = g(z)eiaz with a > 0 and
C0 consists of a contour from −R to R on the real line followed by a semicircle centered at the
origin and lying in the upperhalf plane (lower half plane if a < 0), then under certain conditions
on g(z), the contribution from the semi-circular part to the contour integral approaces zero as
R → ∞. Thus, in such case we have
Z
∞
−∞
=
I
= i2π
C
X
n∈I
Residue [f (z)] z=zn
,
where I is an index set such that zn , n ∈ I is a pole of f (z) lying strictly over the real line. We
shall use Cauchy’s integral formula and Jordan’s lemma extensively.
We return to our problem of inverting v(k) for x > 0 through equation (1). The symmetry of
the problem suggests identical results for x < 0. The form of the integral in equation (1) satisfies
the conditions required by Jordan’s lemma, with g(k) having singularities at ±kp & ±ikp . The
pole ikp lies inside the semi-circle prescribed in Jordan’s lemma and hence gives a contribution
to the integral. This is not so with the pole −ikp as it lies outside the contour. The poles ±kp
lie on the contour and we need to deal with these separately. The contour and the pole locations
are indicated in figure 4 for both x > 0 and x < 0.
iK p
-K p
iK p
Kp
-K p
-iKp
Kp
-iKp
x<0
X>0
Figure 4: Schematic showing the contours along with the pole positions for the inversion integral
in equation 1.
Let us now displace the poles located on the real axis (±kp ) by a small distance ǫ > 0 along
the imaginary axis. In so doing, the Cauchy integral formula will be applicable as we will be able
to ascertain whether or not the poles lie inside the contour. This small perturbation of the poles
from the real axis may be carried in many ways as described below. We then find the limit as
6
ǫ → 0 in each case and judge the physical relevance of the solution obtained. Figure 5 depicts
pictorially all the possible pole shifts. The poles can be shifted in the following ways:
1. Both poles ±kp are shifted upward. In that case, for x > 0 three poles lie inside the contour
namely ikp , kp+iǫ, −kp+iǫ. This is depicted in case 1 of figure 5. Thus, by Cauchy integral
formula we expect v(x) for x > 0 to be of the form Ae−kp x + Beikp x−ǫx + Ce−ikp x−ǫx . In
the limit ǫ → 0, the third term in the above represents a negative travelling wave in the
region x > 0. Recall, the forcing is at x = 0. Thus, the incoming wave solution violates the
causality condition and hence is physically unacceptable.
2. Both poles ±kp are shifted downward. In that case, for x < 0 three poles lie inside the
contour namely −ikp , kp−iǫ, −kp−iǫ. This is depicted in case 2 of figure 5. Thus, by Cauchy
integral formula we expect v(x) for x < 0 to be of the form Aekp x + Beikp x+ǫx + Ce−ikp x+ǫx .
In the limit ǫ → 0, the second term in the above represents a positive travelling wave in the
region x < 0. As the force is at x = 0, this solution violates the causality condition and
hence is physically unacceptable.
3. The pole kp is shifted to kp − iǫ and the pole −kp is shifted to −kp + iǫ. For x > 0, the
interior poles are ikp & −kp + iǫ. This is depicted in case 3 of figure 5. Thus, by Cauchy
integral formula we expect v(x) for x > 0 to be of the form Ae−kp x + Be−ikp x−ǫx . In the
limit ǫ → 0, the third term in the above represents negative travelling wave in the region
x > 0. As the force is at x = 0, this solution violates the causalty condition and hence is
physically unacceptable.
4. The pole kp is shifted to kp + iǫ and the pole −kp is shifted to −kp − iǫ. For x > 0, the
interior poles are ikp & kp + iǫ. This is depicted in case 4 of figure 5. Thus, by Cauchy
integral formula we expect v(x) for x > 0 to be of the form Ae−kp x + Beikp x−ǫx . The first
term represents an evanescent wave, the second term in the limit ǫ → 0 represent a positive
traveling wave. This solution is physically acceptable. Similarly for x < 0, the interior poles
are −ikp & −kp − iǫ. Thus, by Cauchy integral formula we expect v(x) for x < 0 to be of
the form Aekp x + Be−ikp x+ǫx . The first term represents an evanescent wave, the second term
with ǫ → 0 represent a negative travelling wave. In the limit ǫ → 0, for all x, the solution is
decomposed into a travelling and an evanescent wave contribution. Hence, out of the four
ways of perturbing the poles, this is the only one which gives physically meaningful results.
In dealing with integrals where the poles lie on the contour in complex analysis, the notion of
principal value of the integral is commonly used. The principal value interpretation of the integral
too fails to give a physically meaningful results in this case.
Generalizing this idea, we observe that for x > 0 we need to account for the pole on the
positive real axis and ignore the ones on the negative real axis. Similarly, for x < 0 we need to
account for the pole on the negative real axis and ignore the ones on the positive real axis. The
equivalent approach would be, instead of taking the integral from −∞ to ∞ on the real axis, we
deform the contour slightly upward in the left half plane and slightly downward in the right half
plane (the principle of deformation of contours aloows this). Thus, for x > 0 when we close the
contour in the upper half, the pole on the positive real axis will contribute to the integral and
that on the negative real axis would be left out. Similarly, for x < 0 when we close the contour
in the lower half, the pole on the negative real axis will contribute to the integral and that on the
positive real axis would be left out. This scheme is depicted in figure 6.
7
Case 1
Both poles displaced upwards
Case 3
Right Pole Shifted Downwards
Left Pole Shifted Upwards
Case 2
Both Poles Displaced downwards
Case 4
Right Pole Shifted upwards
Left Pole Shifted downwards
Figure 5: Different Strategies to shift poles by a small amount
For x>0
C
For x<0
Figure 6: Redefining the contour integral path C for computing the inverse Fourier transform.
For applying the Cauchy residue theorem with x > 0 (x < 0), the semicircular contour in the
upper (lower) half plane is chosen.
8
2.1
Structural Response
We return to the evaluation of equation (1). In light of the preceeding discussion, for x > 0 the
residue contributions corresponding to the poles at kp & ikp need to be considered. The residues
−kp x
ikp x
corresponding to kp and ikp are e4k3 and ie4k3 , respectively. Thus, for x > 0 we have
p
p
#
"
i
−iωF
ωF h ikp x
eikp x ie−kp x
−kp x
v(x) =
.
+
=
e
+
ie
2πi
2πB
4kp3
4kp3
4kp3 B
(3)
For x < 0 by taking the residue contribution from poles at −kp & −ikp we arrive at a similar
result,
"
#
i
e−ikp x iekp x
ωF h −ikp x
−iωF
2πi
+
= 3
e
+ iekp x .
v(x) =
3
3
2πB
4kp
4kp
4kp B
(4)
The above relations indicate that the velocity response of the line-driven plate can be decomposed in terms of free wavenumbers (from equation ??). The line excitation has no spatial extent
and hence enforces only the frequency. And the free wavenumbers at this frequency are generated
in the plate.
Jordan’s Lemma was easily invoked to determine the structural response. The theorem required the function under consideration to be single-valued and analytic except at the poles. For
the present case of uncoupled analysis the function eikx /(k 4 − kp4 ) satisfied these conditions.
9