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Integration
Adrian Down
September 22, 2005
1 Review: Riemann integration of real-valued
functions
Assume we have a bounded function
f : [a, b] → R
for which we want to define the integral, ∫ (^) b
a
f (x)dx
Definition. A partition P of [a, b] is a set of points,
P = {t 0 , t 1 ,... , tN }
where
a = t 0 < t 1 < t 2 <... < tN = b
Definition. The mesh of a partition is the maximum length of a subinterval of that partition
|P | = max(tj − tj− 1 ), 1 ≤ j ≤ N
Definition. The Riemann sum of a given function f on a partition P is
S(P, f ) =
∑^ N
j=
f (sj )(tj − tj− 1 )
where sj ∈ [tj− 1 , tj ], ∀j
Note. • The points sj must be specified for a particular Riemann sum. We could include this dependence by writing
S(P, f, s 1 , s 2 ,... , sN )
but this would be tedious.
- Each one of the terms in the Riemann sum is the area of the rectan- gle with base tj − tj− 1 and height f (s(j)). Adding these rectangles approximates the area under the curve.
Definition. The integral of a function f on an interval [a, b] is
∫ (^) b
a
f (x)dx = lim |P |→ 0
S(P, f )
Note. This is a new kind of limit. It implies the usual − δ statement.
∀ > 0 , ∃δ > 0 st ∀P with |P | < δ, |S(P, f ) − L| <
Theorem. If f is continuous, then
∫ (^) b
a
f (x)dx
exists.
Note. It can be proven that the Darboux integral exists if and only if the Riemann integral exists, and that wherever these two integrals exist for a given function, they are equivalent.
2 Paths
Definition. A path is a continuous function
γ[a, b] :→ C
Definition. A smooth path is a path γ : [a, b] → C such that γ is continu- ously differentiable on [a, b].
3.2 Contour integration
Definition. Let g : [a, b] → C be bounded. Write
g = u + ıv
Then the contour integral of g along [a, b] is
∫ (^) b
a
g =
∫ (^) b
a
g(x)dx =
∫ (^) b
a
u + ı
∫ (^) b
a
v
provided that both integrals exist.
To compute contour integrals, we need the following theorem.
Theorem. Let γ : [a, b] → S be a smooth path, S open ⊂ C, and f : S → C continuous, then
∫
γ
f =
∫ (^) b
a
f ◦ γ(t) · γ′(t)dt
The proof of this theorem
Note. • This theorem is a practical tool for computing integrals.
- To compute these integrals, you can use all the tools of basic calculus.
- We use this theorem, but we will not prove it, nor will it appear on exams. The proof is very similar to a proof done in real analysis.
3.3 Path length
To define the length, we need the following definitions.
Definition. A path γ is polygonal if there is a partition P of [a, b] such that γ each line segment between successive points is parametrized by an affine function t 7 → ct + d.
Note. Geometrically, the path looks like a collection of straight lines between consecutive points. Each interval of the path is traversed at constant speed.
Definition. The length of a polygonal path γ : [a, b] → C is
L(γ) = |γ(t 1 ) − γ(t 0 )| + |γ(t 2 ) − γ(t 1 ) +...
∑^ N
j=
|γ(tj ) − γ(tj− 1 )|
Definition. The length of a general path γ is
L(γ) = sup L(γp)
where P = {t 0 < t 1 <.. .}, γp(tj ) = γ(tj ), ∀j, and γp affine on [tj− 1 , tj ], ∀t
Note. • Geometrically, we are making an approximation with a number of straight lines, the length of each one is approaching 0.
- Beware, ∃ continuous paths such that L(γ) = ∞. This can occur when a path is not smooth.
Theorem. If γ : [a, b] is a smooth path, then
L(γ) =
∫ (^) b
a
|γ′(t)|dt
Note. If we write γ = g + ıh, then
|γ′(t)| =
g′(t)^2 + h′(t)^2
Proof. Consider any partition P of [a, b]. We will use the mean value theorem from real analysis, so we write everything in terms of real and imaginary parts. We write
γ(t) = u(t) + ıv(t)
where u, v are real-valued and continuously differentiable functions.
where R is a remainder. From real analysis,
lim |P |→ 0
∑^ N
j=
(tj − tj− 1 )|γ′(sj )| =
∫ (^) b
a
|γ′(t)|dt
Thus we only have to prove
lim |P |→ 0
R = 0
From the triangle inequality,
|R| ≤
∑^ N
j=
(tj − tj− 1 )|v′(rj ) − v′(sj )|
From real analysis, v′^ : [a, b] → R continuous on a closed and bounded interval ⇒ v′^ is uniformly continuous. Therefor if |P | < δ, then |v′(sj ) − v′(rj )| < , ∀j. If |P | < δ, then
|R| ≤
∑^ N
j=
(tj − tj− 1 ) = (b − a)
Since this holds ∀ > 0
lim |P |→ 0
R = 0
We will address the problem of lim vs. lim sup next time.
3.4 Estimation Lemma
Lemma. Let γ be a smoothed path in S, S open ⊂ C, f : S → C continuous. Then ∣ ∣ ∣ ∣
γ
f
∣ ≤^ L(γ)^ ·^ max^ |f^ (γ(t))|, t^ ∈^ [a, b]
Note. • This lemma gives us an upper bound for the path integral for a given function.
- From the last theorem, L(γ) is finite. Since f and γ are both continu- ous, from real analysis, we know that their composition is continuous and thus finite. Hence both factors in the integrand are finite.
Proof. To prove the lemma, it is sufficient that ∀ partitions P of [a, b],
|S(P, γ, f )| ≤ L(γ) · max [a,b]
|f ◦ γ|
since if we have the above, then ∣ ∣ ∣ ∣ (^) |Plim |→ 0 S(P, γ, f^ )
∣ ≤^ L(γ)^ ·^ max [a,b]^ |f^ ◦^ γ|
By definition,
S(P, γ, f ) =
∑^ N
j=
f (γ(sj )) · (γ(tj ) − γ(tj− 1 )))
for some sj ∈ [tj− 1 , tj ]. Our basic tool for setting upper bounds is the triangle inequality. We apply it here to the summation, taking the absolute value of each term,
|S(P, γ, f )| ≤
∑^ N
j=
|f (γ(sj )) · (γ(tj ) − γ(tj− 1 )))|
≤ max [a,b]
|f ◦ γ| ·
∑^ N
j=
|γ(tj ) − γ(tj− 1 )|
= max [a,b]
|f ◦ γ| · L(γp)
≤ max [a,b]
|f ◦ γ|L(γ)
