35.3. CONTOUR INTEGRALS 703

Lemma 35.3.6 Let f : [a,b]→ R be continuous. Then there exists c ∈ (a,b) such that

f (c)(b−a) =∫ b

af (x)dx

Proof: Let F (x)≡∫ x

a f (t)dt. Then by the mean value theorem,

F (b)−F (a) = F ′ (c)(b−a)

for some c ∈ (a,b). But F ′ (x) = f (x) and so this proves the lemma. ■

Lemma 35.3.7 Let γ,η be parametrizations for two smooth curves, γ ([a,b]) and η ([c,d])and let f : γ∗×η∗→ R be continuous. Then∫

η

∫γ

f (z,w)dzdw =∫

γ

∫η

f (z,w)dwdz

In other words, you can switch the contour integrals.

Proof: That on the left is by definition,∫η

∫γ

f (z,w)dzdw =∫ d

c

∫ b

af (γ (t) ,η (s))γ

′ (t)η′ (s)dtds

Let P be a partition for [a,b] and Q a partition for [c,d] . Then the above is

n

∑i=1

m

∑j=1

∫ si

si−1

∫ t j

t j−1

f (γ (t) ,η (s))γ′ (t)η

′ (s)dtds

=n

∑i=1

m

∑j=1

∫ si

si−1

f (γ (t̂ j) ,η (s))γ′ (t̂ j)

(t j− t j−1

)η′ (s)ds

=n

∑i=1

m

∑j=1

f (γ (t̂ j) ,η (ŝi))γ′ (t̂ j)η

′ (ŝi)(t j− t j−1

)(si− si−1)

by an application of the mean value theorem for integrals from calculus. Here (t̂ j, ŝi) ∈(t j−1, t j

)× (si−1,si). Similarly,∫

γ

∫η

f (z,w)dwdz =m

∑j=1

n

∑i=1

f (γ (t̃ j) ,η (s̃i))γ′ (t̃ j)η

′ (s̃i)(t j− t j−1

)(si− si−1)

=n

∑i=1

m

∑j=1

f (γ (t̃ j) ,η (s̃i))γ′ (t̃ j)η

′ (s̃i)(t j− t j−1

)(si− si−1)

where (t̃ j, s̃i) ∈(t j−1, t j

)× (si−1,si). By uniform continuity, if ∥P∥ ,∥Q∥ are small enough,

then ∣∣ f (γ (t̃ j) ,η (s̃i))γ′ (t̃ j)η

′ (s̃i)− f (γ (t̂ j) ,η (ŝi))γ′ (t̂ j)η

′ (ŝi)∣∣< ε

and so ∣∣∣∣∫η

∫γ

f (z,w)dzdw−∫

γ

∫η

f (z,w)dwdz∣∣∣∣< ε (b−a)(d− c) .

Since ε is arbitrary, the two contour integrals must be equal. ■