12.3. INTEGRATING AND DIFFERENTIATING FOURIER SERIES 283

where Ak =1

2π

∫π

−πG(y)eikydy. Now from 12.10 and the definition of the Fourier coeffi-

cients for f ,

G(π) = F (π)−a02π = 0 = A0 + limn→∞

n

∑k=−n,k ̸=0

Ak (−1)k (12.11)

and so

A0 =− limn→∞

n

∑k=−n,k ̸=0

Ak (−1)k ≡−∞

∑k=−∞,k ̸=0

Ak (−1)k (12.12)

Next consider Ak for k ̸= 0.

Ak ≡ 12π

∫π

−π

G(x)e−ikxdx≡ 12π

∫π

−π

∫ x

−π

( f (t)−a0)dte−ikxdx

=1

2π

∫π

−π

e−ikx∫ x

−π

( f (t)−a0)dtdx

Now let ψ (x) ≡∫ x−π

( f (t)−a0)dt and φ k (x) =e−ikx

−ik . Then the above integral is of theform 1

2π

∫π

−πψ (x)dφ k (x) . Since ψ (π)φ k (π) = 0 = ψ (−π)φ k (−π) , integration by parts,

Theorem 9.4.1, says that the above equals

− 12π

∫π

−π

φ k (x)dψ (x) =1

2π

∫π

−π

e−ikx

ikdψ (x)

Use Corollary 9.3.18 to write as a sum of finitely many integrals

n

∑k=1

∫ xk

xk−1

e−ikx

ikdψ (x) .

It follows from Proposition 9.4.3 that this is

Ak =1

2π

∫π

−π

e−ikx

ik( f (x)−a0)dx =

1ik

12π

∫π

−π

f (x)e−ikxdx≡ ak

ik

Thus this shows from 12.12, that for all x,

G(x) =∞

∑k=−∞,k ̸=0

ak

ikeikx +A0, A0 =−

∞

∑k=−∞,k ̸=0

Ak (−1)k

and so, ∫ x

−π

f (t)dt−∫ x

−π

a0 =∞

∑k=−∞,k ̸=0

ak

ikeikx− ak

ikeik(−π)

which shows that ∫ x

−π

f (t)dt =∫ x

−π

a0 +∞

∑k=−∞,k ̸=0

ak

∫ x

−π

eiktdt

This proves the following theorem.

Theorem 12.3.2 Let f be 2π periodic and piecewise continuous. Then∫ x

−π

f (t)dt =∫ x

−π

a0dt + limn→∞

n

∑k=−n,k ̸=0

ak

∫ x

−π

eiktdt

where ak are the Fourier coefficients of f . This holds for all x ∈ R.