1674 CHAPTER 53. RESIDUES

and it is this last equation which is established. Near α j,

f (z) = g j (z)+m j

∑r=1

b jr

(z−α j)r ≡ g j (z)+Q j (z) .

where g j is analytic at and near α j. Now define

G(z)≡ f (z)−N

∑j=1

Q j (z) .

It follows that G(z) has a removable singularity at each α j. Therefore, by Corollary51.7.20,

0 =m

∑k=1

∫γk

G(z)dz =m

∑k=1

∫γk

f (z)dz−N

∑j=1

m

∑k=1

∫γk

Q j (z)dz.

Now

m

∑k=1

∫γk

Q j (z)dz =m

∑k=1

∫γk

(b j

1(z−α j)

+m j

∑r=2

b jr

(z−α j)r

)dz

=m

∑k=1

∫γk

b j1

(z−α j)dz≡

m

∑k=1

n(γk,α j) res( f ,α j)(2πi) .

Therefore,

m

∑k=1

∫γk

f (z)dz =N

∑j=1

m

∑k=1

∫γk

Q j (z)dz

=N

∑j=1

m

∑k=1

n(γk,α j) res( f ,α j)(2πi)

= 2πiN

∑j=1

res( f ,α j)m

∑k=1

n(γk,α j)

= (2πi) ∑α∈A

res( f ,α)m

∑k=1

n(γk,α)

which proves the theorem.The following is an important example. This example can also be done by real variable

methods and there are some who think that real variable methods are always to be preferredto complex variable methods. However, I will use the above theorem to work this example.

Example 53.0.4 Find limR→∞

∫ R−R

sin(x)x dx

Things are easier if you write it as

limR→∞

1i

(∫ −R−1

−R

eix

xdx+

∫ R

R−1

eix

xdx

).