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Topic 9 Notes
9 Definite integrals using the residue theorem
9.1 Introduction
In this topic we’ll use the residue theorem to compute some real definite integrals. ∫ (^) b
a
f (x) dx
The general approach is always the same
- Find a complex analytic function g(z) which either equals f on the real axis or which is closely connected to f , e.g. f (x) = cos(x), g(z) = eiz^.
- Pick a closed contour C that includes the part of the real axis in the integral.
- The contour will be made up of pieces.∫ It should be such that we can compute g(z) dz over each of the pieces except the part on the real axis.
- Use the residue theorem to compute
C
g(z) dz.
- Combine the previous steps to deduce the value of the integral we want.
9.2 Integrals of functions that decay
The theorems in this section will guide us in choosing the closed contour C described in the introduction.
The first theorem is for functions that decay faster than 1/z.
Theorem 9.1. (a) Suppose f (z) is defined in the upper half-plane. If there is an a > 1 and M > 0 such that
|f (z)| <
M
|z|a
for |z| large then
lim R→∞
CR
f (z) dz = 0,
where CR is the semicircle shown below on the left.
Re(z)
Im(z)
−R R
CR
Re(z)
Im(z) −R R
CR
Semicircles: left: Reiθ, 0 < θ < π right: Reiθ, π < θ < 2 π.
(b) If f (z) is defined in the lower half-plane and
|f (z)| <
M
|z|a^
where a > 1 then
lim R→∞
CR
f (z) dz = 0,
where CR is the semicircle shown above on the right.
Proof. We prove (a), (b) is essentially the same. We use the triangle inequality for integrals and the estimate given in the hypothesis. For R large ∣ ∣∣ ∣
CR
f (z) dz
CR
|f (z)| |dz| ≤
CR
M
|z|a^ |dz| =
∫ (^) π
0
M
Ra^ R dθ = M π Ra−^1
Since a > 1 this clearly goes to 0 as R → ∞. QED
The next theorem is for functions that decay like 1/z. It requires some more care to state and prove.
Theorem 9.2. (a) Suppose f (z) is defined in the upper half-plane. If there is an M > 0 such that |f (z)| <
M
|z|
for |z| large then for a > 0
x lim 1 →∞, x 2 →∞
C 1 +C 2 +C 3
f (z)eiaz^ dz = 0,
where C 1 + C 2 + C 3 is the rectangular path shown below on the left.
Re(z)
Im(z)
−x 2 x 1
i(x 1 + x 2 ) C 1
C 2
C 3
Re(z)
Im(z) −x 2 x 1
−i(x 1 + x 2 )
C 1
C 2
C 3
Rectangular paths of height and width x 1 + x 2.
(b) Similarly, if a < 0 then
x lim 1 →∞, x 2 →∞
C 1 +C 2 +C 3
f (z)eiaz^ dz = 0,
where C 1 + C 2 + C 3 is the rectangular path shown above on the right.
Note. In contrast to Theorem 9.1 this theorem needs to include the factor eiaz^.
Proof. (a) We start by parametrizing C 1 , C 2 , C 3.
C 1 : γ 1 (t) = x 1 + it, t from 0 to x 1 + x 2
It is clear that for z large f (z) ≈ 1 /z^4.
In particular, the hypothesis of Theorem 9.1 is satisfied. Using the contour shown below we have, by the residue theorem, ∫
C 1 +CR
f (z) dz = 2πi
residues of f inside the contour. (1)
Re(z)
Im(z)
−R R
CR
C 1
i
We examine each of the pieces in the above equation. ∫
CR
f (z) dz: By Theorem 9.1(a),
lim R→∞
CR
f (z) dz = 0.
∫
C 1
f (z) dz: Directly, we see that
lim R→∞
C 1
f (z) dz = lim R→∞
∫ R
−R
f (x) dx =
−∞
f (x) dx = I.
So letting R → ∞, Equation 1 becomes
I =
−∞
f (x) dx = 2πi
residues of f inside the contour.
Finally, we compute the needed residues: f (z) has poles of order 2 at ±i. Only z = i is inside the contour, so we compute the residue there. Let
g(z) = (z − i)^2 f (z) =
(z + i)^2
Then Res(f, i) = g′(i) = −
(2i)^3
4 i
So,
I = 2πi Res(f, i) =
π 2
Example 9.4. Compute
I =
−∞
x^4 + 1
dx.
Solution: Let f (z) = 1/(1 + z^4 ). We use the same contour as in the previous example
Re(z)
Im(z)
−R R
CR
C 1
ei^3 π/^4 eiπ/^4
As in the previous example,
lim R→∞
CR
f (z) dz = 0
and
lim R→∞
C 1
f (z) dz =
−∞
f (x) dx = I.
So, by the residue theorem
I = lim R→∞
C 1 +CR
f (z) dz = 2πi
residues of f inside the contour.
The poles of f are all simple and at
eiπ/^4 , ei^3 π/^4 , ei^5 π/^4 , ei^7 π/^4.
Only eiπ/^4 and ei^3 π/^4 are inside the contour. We compute their residues as limits using L’Hospital’s rule. For z 1 = eiπ/^4 :
Res(f, z 1 ) = lim z→z 1 (z − z 1 )f (z) = lim z→z 1
z − z 1 1 + z^4 = lim z→z 1
4 z^3
4ei^3 π/^4
e−i^3 π/^4 4
and for z 2 = ei^3 π/^4 :
Res(f, z 2 ) = lim z→z 2 (z − z 2 )f (z) = lim z→z 2
z − z 2 1 + z^4
= lim z→z 2
4 z^3
4ei^9 π/^4
e−iπ/^4 4
So,
I = 2πi(Res(f, z 1 ) + Res(f, z 2 )) = 2πi
− 1 − i 4
1 − i 4
= 2πi
2 i 4
= π
Example 9.5. Suppose b > 0. Show ∫ (^) ∞
0
cos(x) x^2 + b^2
dx = πe−b 2 b
Solution: The first thing to note is that the integrand is even, so
I =
−∞
cos(x) x^2 + b^2
9.4 Trigonometric integrals
The trick here is to put together some elementary properties of z = eiθ^ on the unit circle.
- e−iθ^ = 1/z.
- cos(θ) =
eiθ^ + e−iθ 2
z + 1/z 2
- sin(θ) = eiθ^ − e−iθ 2 i
z − 1 /z 2 i
We start with an example. After that we’ll state a more general theorem.
Example 9.6. Compute (^) ∫ 2 π
0
dθ 1 + a^2 − 2 a cos(θ)
Assume that |a| 6 = 1.
Solution: Notice that [0, 2 π] is the interval used to parametrize the unit circle as z = eiθ. We need to make two substitutions:
cos(θ) =
z + 1/z 2 dz = ieiθ^ dθ ⇔ dθ = dz iz
Making these substitutions we get
I =
∫ (^2) π
0
dθ 1 + a^2 − 2 a cos(θ)
=
|z|=
1 + a^2 − 2 a(z + 1/z)/ 2
dz iz
=
|z|=
i((1 + a^2 )z − a(z^2 + 1)) dz.
So, let
f (z) =
i((1 + a^2 )z − a(z^2 + 1))
The residue theorem implies
I = 2πi
residues of f inside the unit circle.
We can factor the denominator:
f (z) =
ia(z − a)(z − 1 /a)
The poles are at a, 1 /a. One is inside the unit circle and one is outside.
If |a| > 1 then 1/a is inside the unit circle and Res(f, 1 /a) =
i(a^2 − 1) If |a| < 1 then a is inside the unit circle and Res(f, a) =
i(1 − a^2 )
We have
I =
2 π a^2 − 1 if^ |a|^ >^1 2 π 1 −a^2 if^ |a|^ <^1
The example illustrates a general technique which we state now.
Theorem 9.7. Suppose R(x, y) is a rational function with no poles on the circle
x^2 + y^2 = 1
then for
f (z) =
iz
R
z + 1/z 2
z − 1 /z 2 i
we have (^) ∫ (^2) π
0
R(cos(θ), sin(θ)) dθ = 2πi
residues of f inside |z| = 1.
Proof. We make the same substitutions as in Example 9.6. So,
∫ (^2) π
0
R(cos(θ), sin(θ)) dθ =
|z|=
R
z + 1/z 2
z − 1 /z 2 i
dz iz
The assumption about poles means that f has no poles on the contour |z| = 1. The residue theorem now implies the theorem.
9.5 Integrands with branch cuts
Example 9.8. Compute
I =
0
x^1 /^3 1 + x^2 dx.
Solution: Let
f (x) =
x^1 /^3 1 + x^2
Since this is asymptotically comparable to x−^5 /^3 , the integral is absolutely convergent. As a complex function
f (z) = z^1 /^3 1 + z^2
needs a branch cut to be analytic (or even continuous), so we will need to take that into account with our choice of contour.
First, choose the following branch cut along the positive real axis. That is, for z = reiθ^ not on the axis, we have 0 < θ < 2 π.
Next, we use the contour C 1 + CR − C 2 − Cr shown below.
Letting r → 0 and R → ∞ the analysis above shows
(1 − ei^2 π/^3 )I = 2πi(Res(f, i) + Res(f, −i))
All that’s left is to compute the residues using the chosen branch of z^1 /^3
Res(f, −i) =
(−i)^1 /^3 − 2 i
(ei^3 π/^2 )^1 /^3 2ei^3 π/^2
e−iπ 2
Res(f, i) =
i^1 /^3 2 i
eiπ/^6 2eiπ/^2
e−iπ/^3 2
A little more algebra gives
(1 − ei^2 π/^3 )I = 2πi · −1 + e−iπ/^3 2
= πi(−1 + 1/ 2 − i
3 /2) = −πieiπ/^3.
Continuing
I =
−πieiπ/^3 1 − ei^2 π/^3
πi eiπ/^3 − e−πi/^3
π/ 2 (eiπ/^3 − e−iπ/^3 )/ 2 i
π/ 2 sin(π/3)
π √ 3
Whew! (Note: a sanity check is that the result is real, which it had to be.)
Example 9.9. Compute
I =
1
dx x
x^2 − 1
Solution: Let f (z) =
z
z^2 − 1
The first thing we’ll show is that the integral ∫ (^) ∞
1
f (x) dx
is absolutely convergent. To do this we split it into two integrals ∫ (^) ∞
1
dx x
x^2 − 1
1
dx x
x^2 − 1
2
dx x
x^2 − 1
The first integral on the right can be rewritten as
∫ (^2)
1
x
x + 1
x − 1
dx ≤
1
x − 1
dx =
x − 1
2
1
This shows the first integral is absolutely convergent.
The function f (x) is asymptotically comparable to 1/x^2 , so the integral from 2 to ∞ is also absolutely convergent.
We can conclude that the original integral is absolutely convergent.
Next, we use the following contour. Here we assume the big circles have radius R and the small ones have radius r.
Re(z)
Im(z)
R
r r
C 1
C 2 −C 3
−C 4
C 5
−C 6
−C 7
− 1 1
C 8
We use the branch cut for square root that removes the positive real axis. In this branch
0 < arg(z) < 2 π and 0 < arg(
w) < π.
For f (z), this necessitates the branch cut that removes the rays [1, ∞) and (−∞, −1] from the complex plane.
The pole at z = 0 is the only singularity of f (z) inside the contour. It is easy to compute that
Res(f, 0) =
i = −i.
So, the residue theorem gives us ∫
C 1 +C 2 −C 3 −C 4 +C 5 −C 6 −C 7 +C 8
f (z) dz = 2πi Res(f, 0) = 2π. (2)
In a moment we will show the following limits
lim R→∞
C 1
f (z) dz = lim R→∞
C 5
f (z) dz = 0
lim r→ 0
C 3
f (z) dz = lim r→ 0
C 7
f (z) dz = 0.
We will also show
lim R→∞, r→ 0
C 2
f (z) dz = lim R→∞, r→ 0
−C 4
f (z) dz
= lim R→∞, r→ 0
−C 6
f (z) dz = lim R→∞, r→ 0
C 8
f (z) dz = I.
Using these limits, Equation 2 implies 4I = 2π, i.e.
I = π/ 2.
All that’s left is to prove the limits asserted above.
Thus,
lim R→∞, r→ 0
−C 6
f (z) dz =
∞
−f (x) dx =
1
f (x) dx = I.
We can parameterize C 2 by z = −x + i where x goes from ∞ to 1. Thus, on C 2 , we have
arg(z^2 − 1) ≈ 2 π,
so (^) √ z^2 − 1 ≈ −
x^2 − 1.
This implies
f (z) ≈
(−x)(−
x^2 − 1)
= f (x).
Thus,
lim R→∞, r→ 0
C 2
f (z) dz =
∞
f (x) (−dx) =
1
f (x) dx = I.
The last curve −C 4 is handled similarly.
9.6 Cauchy principal value
First an example to motivate defining the principal value of an integral. We’ll actually compute the integral in the next section.
Example 9.10. Let
I =
0
sin(x) x dx.
This integral is not absolutely convergent, but it is conditionally convergent. Formally, of course, we mean
I = lim R→∞
∫ R
0
sin(x) x dx.
We can proceed as in Example 9.5. First note that sin(x)/x is even, so
I =
−∞
sin(x) x
dx.
Next, to avoid the problem that sin(z) goes to infinity in both the upper and lower half- planes we replace the integrand by e ix x. We’ve changed the problem to computing
I˜ =
−∞
eix x
dx.
The problems with this integral are caused by the pole at 0. The biggest problem is that the integral doesn’t converge! The other problem is that when we try to use our usual strategy of choosing a closed contour we can’t use one that includes z = 0 on the real axis. This is our motivation for defining principal value. We will come back to this example below.
Definition. Suppose we have a function f (x) that is continuous on the real line except at the point x 1 , then we define the Cauchy principal value as
p.v.
−∞
f (x) dx = lim R→∞, r 1 → 0
∫ (^) x 1 −r 1
−R
f (x) dx +
∫ R
x 1 +r 1
f (x) dx.
Provided the limit converges. You should notice that the intervals around x 1 and around ∞ are symmetric. Of course, if the integral ∫ (^) ∞
−∞
f (x) dx
converges, then so does the principal value and they give the same value. We can make the definition more flexible by including the following cases.
- If f (x) is continuous on the entire real line then we define the principal value as
p.v.
−∞
f (x) dx = lim R→∞
∫ R
−R
f (x) dx
- If we have multiple points of discontinuity, x 1 < x 2 < x 3 <... < xn, then
p.v.
−∞
f (x) dx = lim
∫ (^) x 1 −r 1
−R
f (x) dx +
∫ (^) x 2 −r 2
x 1 +r 1
∫ (^) x 3 −r 3
x 2 +r 2
∫ R
xn+rn
f (x) dx.
Here the limit is taken as R → ∞ and each of the rk → 0.
x x 1 x 2 [ ] [ ] [ ] −R x 1 −^ r 1 x 1 −^ r 1 x 2 −^ r 2 x 2 −^ r 2 R
Intervals of integration for principal value are symmetric around xk and ∞
The next example shows that sometimes the principal value converges when the integral itself does not. The opposite is never true. That is, we have the following theorem.
Theorem 9.11. If f (x) has discontinuities at x 1 < x 2 <... < xn and
−∞
f (x) dx
converges then so does p.v.
−∞
f (x) dx.
Proof. The proof amounts to understanding the definition of convergence of integrals as limits. The integral converges means that each of the limits
lim R 1 →∞, a 1 → 0
∫ (^) x 1 −a 1
−R 1
f (x) dx
lim b 1 → 0 , a 2 → 0
∫ (^) x 2 −a 2
x 1 +b 1
f (x) dx
... (3)
lim R 2 →∞, bn→ 0
∫ R 2
xn+bn
f (x) dx.
Proof. Since we take the limit as r goes to 0, we can assume r is small enough that f (z) has a Laurent expansion of the punctured disk of radius r centered at z 0. That is, since the pole is simple,
f (z) = b 1 z − z 0
- a 0 + a 1 (z − z 0 ) +... for 0 < |z − z 0 | ≤ r.
Thus, ∫
Cr
f (z) dz =
∫ (^) π
0
f (z 0 + reiθ) rieiθ^ dθ =
∫ (^) π
0
b 1 i + a 0 ireiθ^ + a 1 ir^2 ei^2 θ^ +...
dθ
The b 1 term gives πib 1. Clearly all the other terms go to 0 as r → 0. QED.
If the pole is not simple the theorem doesn’t hold and, in fact, the limit does not exist.
The same proof gives a slightly more general theorem.
Theorem 9.14. Suppose f (z) has a simple pole at z 0. Let Cr be the circular arc γ(θ) = z 0 + reiθ, with θ 0 ≤ θ ≤ θ 0 + α. Then
lim r→ 0
Cr
f (z) dz = αi Res(f, z 0 )
Re(z)
Im(z)
z 0
r
Cr
α
Small circular arc of radius r around z 0
Example 9.15. (Return to Example 9.10.) A long time ago we left off Example 9.10 to define principal value. Let’s now use the principal value to compute
I˜ = p.v.
−∞
eix x dx.
Solution: We use the indented contour shown below. The indentation is the little semicircle the goes around z = 0. There are no poles inside the contour so the residue theorem implies ∫
C 1 −Cr +C 2 +CR
eiz z
dz = 0.
Re(z)
Im(z)
0
C 1 C 2
CR
−Cr
−R −r^ r^ R
2 Ri
Next we break the contour into pieces.
lim R→∞, r→ 0
C 1 +C 2
eiz z
dz = I.˜
Theorem 9.2(a) implies
lim R→∞
CR
eiz z dz = 0.
Equation 5 in Theorem 9.13 tells us that
lim r→ 0
Cr
eiz z dz = πi Res
eiz z
= πi
Combining all this together we have
lim R→∞, r→ 0
C 1 −Cr +C 2 +CR
eiz z dz = I˜ − πi = 0,
so I˜ = πi. Thus, looking back at Example 5, where I =
0
sin(x) x
dx, we have
I =
Im( I˜) =
π 2
There is a subtlety about convergence we alluded to above. That is, I is a genuine (con- ditionally) convergent integral, but I˜ only exists as a principal value. However since I is a convergent integral we know that computing the principle value as we just did is sufficient to give the value of the convergent integral.
9.8 Fourier transform
Definition. The Fourier transform of a function f (x) is defined by
f̂ (ω) =
−∞
f (x)e−ixω^ dx
This is often read as ‘f -hat’.
Theorem. (Fourier inversion formula.) We can recover the original function f (x) with the Fourier inversion formula
f (x) =
2 π
−∞
f (ω)eixω^ dω.
So, the Fourier transform converts a function of x to a function of ω and the Fourier inversion converts it back. Of course, everything above is dependent on the convergence of the various integrals.
Proof. We will not give the proof here. (We may get to it later in the course.)
Example 9.16. Let
f (t) =
e−at^ for t > 0 0 for t < 0
Theorem 9.2(b) implies that
x lim 1 →∞, x 2 →∞
C 1 +C 2 +C 3
g(z)eizt^ dz = 0
Clearly
lim x 1 →∞, x 2 →∞
2 π
C 4
g(z)eizt^ dz =
2 π
−∞
f (ω) dω
Since, there are no poles of g(z)eizt^ in the lower half-plane, applying the residue theorem to the entire closed contour, we get for large x 1 , x 2 : ∫
C 1 +C 2 +C 3 +C 4
g(z)eizt^ dz = − 2 πi Res
eizt a + iz
, ia
Thus, 1 2 π
−∞
f (ω) dω = 0 for t < 0
This shows the inversion formula holds for t < 0.
Finally, we give the promised argument that the inversion integral converges. By definition ∫ (^) ∞
−∞
f (ω)eiωt^ dω =
−∞
eiωt a + iω
dω
−∞
a cos(ωt) + ω sin(ωt) − iω cos(ωt) + ia sin(ωt) a^2 + ω^2
dω
The terms without a factor of ω in the numerator converge absolutely because of the ω^2 in the denominator. The terms with a factor of ω in the numerator do not converge absolutely. For example, since ω sin(ωt) a^2 + ω^2 decays like 1/ω, its integral is not absolutely convergent. However, we claim that the integral does converge conditionally. That is, both limits
lim R 2 →∞
∫ R 2
0
ω sin(ωt) a^2 + ω^2
dω and lim R 1 →∞
−R 1
ω sin(ωt) a^2 + ω^2
dω
exist and are finite. The key is that, as sin(ωt) alternates between positive and negative
arches, the function
ω a^2 + ω^2 is decaying monotonically. So, in the integral, the area under
each arch adds or subtracts less than the arch before. This means that as R 1 (or R 2 ) grows the total area under the curve oscillates with a decaying amplitude around some limiting value.
ω
Total area oscillates with a decaying amplitude.
9.9 Solving DEs using the Fourier transform
Let
D = d dt
Our goal is to see how to use the Fourier transform to solve differential equations like
P (D)y = f (t).
Here P (D) is a polynomial operator, e.g.
D^2 + 8D + 7I.
We first note the following formula:
̂ Df (ω) = iω f .̂ (9)
Proof. This is just integration by parts:
Df̂ (ω) =
−∞
f ′(t)e−iωt^ dt
= f (t)e−iωt
−∞
f (t)(−iωe−iωt^ dt
= iω
−∞
f (t)e−iωt^ dt
= iω f̂ (ω) QED
In the third line we assumed that f decays so that f (∞) = f (−∞) = 0.
It is a simple extension of Equation 9 to see
(P̂ (D)f )(ω) = P (iω)f .̂
We can now use this to solve some differential equations.
Example 9.17. Solve the equation
y′′(t) + 8y′(t) + 7y(t) = f (t) =
e−at^ if t > 0 0 if t < 0
Solution: In this case, we have
P (D) = D^2 + 8D + 7I,
so P (s) = s^2 + 8s + 7 = (s + 7)(s + 1).
