Solutions to Problem Sheet 2,

1. i) Draw the graphs for [x] and {x}.

ii) Show that for α ∈ R,∫ α+1

α

[t] dt = α and

∫ α+1

α

{t} dt = 1

2.

Hint Split these integrals at the integer which must lie in any intervalof length 1, such as [α, α + 1] .

iii) Prove that for x > 1∫ x

1

[t] dt =1

2[x] ([x]− 1) + {x} [x]

=1

2x (x− 1) +

1

2{x} (1− {x}) .

This is often written as

∫ x

1

[t] dt =1

2x (x− 1) +O (1) ,

though the error term is zero when x ∈ Z. The result will be used inthe course in the form

∫ x

1

[t] dt =1

2x2 +O (x) ,

(where the error is not now zero when x is an integer.)

Solution ii) Since the interval [α, α + 1] is of length 1 it must containan integer, at which point the integrand has a jump.

In fact this integer is N = [α + 1] then

∫ α+1

α

[t] dt =

∫ N

α

[t] dt+

∫ α+1

N

[t] dt

=

∫ N

α

(N − 1) dt+

∫ α+1

N

Ndt

= (N − 1) (N − α) + (α + 1−N)N

= α.

36

Using t = [t] + {t} then

∫ α+1

α

{t} dt =

∫ α+1

α

(t− [t]) dt =

∫ α+1

α

tdt−∫ α+1

α

[t] dt

=1

2

(

(α + 1)2 − α2)

− α,

by the last result on the last integral

=1

2.

iii) For x > 1 write N = [x] and split the interval [1, x] into a union ofintervals of length 1 with the residual [N, x], an interval of length < 1.Thus

∫ x

1

[t] dt =N−1∑

n=1

∫ n+1

n

[t] dt+

∫ x

N

[t] dt.

Applying the first result in Part ii gives

N−1∑

n=1

∫ n+1

n

[t] dt =N−1∑

n=1

n =(N − 1)N

2=

([x]− 1) [x]

2,

on recalling the formula for the sum of the first N − 1 integers. In theremaining interval [N, x] we have [t] = N and so

∫ x

N

[t] dt = N

∫ x

N

dt = N (x−N) = [x] (x− [x]) = [x] {x} .

Combining gives the required result.

2. How fast does the logarithm grow?.

i) Recall the fundamental idea from Chapter 1,

(

inft∈[y,x]

f (t)

)

(x− y) ≤∫ x

y

f (t) dt ≤(

supt∈[y,x]

f (t)

)

(x− y) . (25)

Use (25) to show that, for all real y > 1,

log y < y − 1 < y.

37

ii) By an appropriate choice of y in part i show that for all n ≥ 1 wehave

log x < nx1/n (26)

for x > 1.

iii) Deduce that for all ε > 0, we have log x ≪ε xε. This means that

the logarithm of x grows slower than any power of x.

Here log x ≪ε xε means there exists a constant C, though it doesdepend on ε, for which log x ≤ C (ε) xε for all x ≥ 1.

iv) In the notes we make use of both of

x1/3 log x < Cx1/2 and xδ < Cx

logℓ x,

for any constant C > 0, δ < 1 and ℓ ≥ 1. Prove these inequalities bothhold for sufficiently large x.

Solution i) For y > 1,

log y =

∫ y

1

dt

t<y − 1

1= y − 1 < y.

ii) Choose y = x1/n in log y < y to get log x1/n < x1/n which rearrangesto the required result.

iii) Let ε > 0 be given. Choose y = xε in log y < y to get log xε < xε

which rearranges to log x < (1/ε) xε, the required result..

38

iv)x1/3 log x < Cx1/2 iff log x < Cx1/6

By part ii, log x ≤ 12x1/12 for all x > 1. (Choose any n > 6, and Ichoose 12 only since 12 = 2× 6). Then demand that 12x1/12 < Cx1/6,i.e. x > (12/C)12, for then we deduce that log x ≤ 12x1/12 < Cx1/6 forall x sufficiently large.

Similarly,

xδ < Cx

logℓ xiff log x < C ′x(1−δ)/ℓ,

where C ′ = C1/ℓ, which is again true by part ii for all x sufficientlylarge.

3. A technical result used in more advanced results on the Prime NumberTheorem.

For a function whose growth lies between that of xδ for any δ < 1 andx/ logℓ x for any ℓ ≥ 1, consider

x exp (−C (log x)α)

with constants C > 0 and α.

Prove that

a) If δ < 1 then for any C > 0

xδ ≤ x exp (−C (log x)α)

for all sufficiently large x as long as α < 1.

b) If ℓ ≥ 1 then for any C > 0

x exp (−C (log x)α) ≤ x

logℓ x

for all sufficiently large x as long as α > 0.

39

Solution

a)

xδ ≤ x exp (−C (log x)α) iff exp (C (log x)α) ≤ x1−δ

iff C (log x)α ≤ (1− δ) log x,

true for sufficiently large x as long as α < 1.

b)

x exp (−C (log x)α) ≤ x

logℓ xiff logℓ x ≤ exp (C (log x)α)

iff ℓ log log x ≤ C (log x)α .

Writing y = log x, this becomes ℓ log y ≤ Cyα, true for sufficiently largey, (and so x), by part iii.

4. i) Estimates of integrals found in error terms. Show that for α ≥ 0 andℓ ≥ 1 we have.

∫ x

2

tα

logℓ tdt≪ℓ,α

x1+α

logℓ x.

Hint Split the integral at√x and use the fact that log t is an increasing

function of t.

ii) Show that for α > 1 and ℓ ≥ 1 we have

∫ ∞

x

logℓ t

tαdt≪ℓ,α

logℓ x

xα−1.

Hint Split the integral at x2 and use (25). In the shorter interval againuse that log t is an increasing function while in the longer interval uselog t≪ tε with some appropriately chosen ε.

iii) A result used in more advanced results on the Prime Number The-orem. Show that with α > 0 and C > 0 we have

∫ x

2

exp (−C (log t)α) dt≪ x exp (−C ′ (log x)α) ,

for some C ′ > 0.

40

Solution i) For α ≥ 0 and ℓ ≥ 1 split the integral as suggested,

∫ x

2

tα

logℓ tdt =

∫

√x

2

tα

logℓ tdt +

∫ x

√x

tα

logℓ tdt.

In the first integral use logℓ t ≥ logℓ 2, in the second logℓ t ≥ logℓ√x =

2−ℓ logℓ x. Thus

∫ x

2

tα

logℓ tdt ≤ 1

logℓ 2

∫

√x

2

tαdt +2ℓ

logℓ x

∫ x

√x

tαdt

≪ ℓ,αx(1+α)/2 +

x1+α

logℓ x

≪ ℓ,αx1+α

logℓ x

where the last step can be justified by using part iv of Question 2.

ii) For α > 1 and ℓ ≥ 1 split the integral as suggested,

∫ ∞

x

logℓ t

tαdt =

∫ x2

x

logℓ t

tαdt+

∫ ∞

x2

logℓ t

tαdt.

In the first integral logℓ t ≤ logℓ (x2) = 2ℓ logℓ x, so

∫ x2

x

logℓ t

tαdt ≤ 2ℓ logℓ x

∫ x2

x

dt

tα≪ℓ,α (log x)ℓ

1

xα−1.

For the second integral use log t≪ t(α−1)/2, from Question 2 for then

∫ ∞

x2

logℓ t

tαdt ≪

∫ ∞

x2

dt

t(α+1)/2=

2

(α− 1)

1

(x2)(α−1)/2≪ 1

xα−1,

which is smaller than the bound on the first integral.

Aside. The integral is split at x2 since is the simplest power greaterthan the first. If the integral is split at xβ, β > 1, you need uselog t≪ tγ, γ = (β − 1) (α− 1) /β to deduce the same result.

41

iii) Split the integral at√x. Then since α > 0 and C > 0 the integrand

is a decreasing function,

∫ x

2

exp (−C (log t)α) dt ≤∫

√x

2

exp (−C (log t)α) dt+

+

∫ x

√x

exp (−C (log t)α) dt

≤ exp (−C (log 2)α)√x+

+x exp(

−C(

log√x)α)

≪√x+ x exp

(

−C2α

(log x)α)

.

The result follows by Question 2v.

�

42

Generalising Euler’s constant.

Recall from lectures that a result for comparing sums with integralsis that if f has a continuous derivative, is non-negative and monotonicthen

∑

1≤n≤xf (n) =

∫ x

1

f (t) dt+O (max (f (1) , f (x))) , (27)

for all real x ≥ 1.

In fact we deduced (27) from the Euler Summation: Let f have acontinuous derivative x > 0. Then

∑

1≤n≤xf (n) =

∫ x

1

f (t) dt+ f (1)− {x} f (x) +∫ x

1

{t} f ′ (t) dt

for all real x ≥ 1.

As an application of (27) we showed that there exists a constant γ suchthat

∑

1≤n≤x

1

n= log x+ γ +O

(

1

x

)

, (28)

for x ≥ 1. The essential idea here is that we come across a convergentintegral

∫ x

1

{t}t2dt.

This could simply be bounded as O (1), but for a better result it iscompleted to infinity and the tail end, the integral from x to ∞, isbounded (often using the results of Question 4). This is a standardmethod and is used in the following question in which we generalise(28) to a sum of (log n)ℓ /n for any integer ℓ ≥ 1. The result of Question5 is used later when examining the Laurent Expansion of ζ (s) .

5. Prove that for all ℓ ≥ 1 there exists a constant Cℓ such that

∑

n≤x

logℓ n

n=

1

ℓ+ 1logℓ+1 x+ Cℓ +O

(

logℓ x

x

)

,

for all real x ≥ 1.

The result when ℓ = 0 has C0 = γ, Euler’s constant. Notice how wehave the best possible error term.

43

Solution Apply Euler Summation with f (n) = (log n)ℓ /n to get

∑

1≤n≤xf (n) =

∫ x

1

logℓ t

tdt− {x} log

ℓ x

x+

logℓ 1

1+

∫ x

1

{t} ddt

(

logℓ t

t

)

dt

=1

ℓ+ 1logℓ+1 x+O

(

logℓ x

x

)

+

∫ x

1

{t}(

ℓ logℓ−1 t− logℓ t

t2

)

dt.

(29)

The integral here converges; it is the difference of integrals of the form∫ x

1

{t} logm t

t2dt,

with m = ℓ or ℓ− 1. The fractional part {t} is bounded so the integralcan be compared with the larger

∫ x

1

logm t

t2dt,

which converges. Thus the integral in (29) can be completed to infinitygiving the constant

Cℓ =

∫ ∞

1

{t}(

ℓ logℓ−1 t− logℓ t

t2

)

dt.

The tail end can be estimated as∣

∣

∣

∣

∫ ∞

x

{t}(

ℓ logℓ−1 t− logℓ t

t2

)

dt

∣

∣

∣

∣

≤ ℓ

∫ ∞

x

logℓ−1 t

t2dt+

∫ ∞

x

logℓ t

t2dt

≪ logℓ−1 x

x+

logℓ x

x≪ logℓ x

x,

having used Question 4.

More Sums of logs.

6. Generalise a result in lectures written (in a weakened form) as

∑

1≤n≤xlog n = x log x+O (x)

44

i) With integer ℓ ≥ 1 justify

∑

1≤n≤xlogℓ n =

∫ x

1

logℓ tdt+O(

logℓ x)

(30)

for all real x ≥ 1.

ii) Prove that∫ x

1

logℓ tdt = x logℓ x+Oℓ

(

x logℓ−1 x)

.

Deduce∑

1≤n≤xlogℓ n = x logℓ x+Oℓ

(

x logℓ−1 x)

.

for all real x ≥ 1.

Note the best error term here would be O(

logℓ x)

, far smaller than theone here.

Solution i) Apply (27) .

ii) Integrate by parts so∫ x

1

logℓ tdt =[

t logℓ t]x

1− ℓ

∫ x

1

logℓ−1 tdt.

Bound the resulting integral simply by∫ x

1

logℓ−1 tdt ≤ logℓ−1 x

∫ x

1

dt < x logℓ−1 x.

7. Improve the result of Qu 6 to the best possible error term.

Change the variable of integration in (30) to u = log t so

∫ x

1

logℓ tdt =

∫ log x

0

euuℓdu

i) Prove by induction that

∫ y

0

euuddu = eyd

∑

r=0

(−1)r d!

(d− r)!yd−r − (−1)d d! (31)

45

for all d ≥ 0.

ii) Prove that for any integer ℓ ≥ 0 we have

∑

n≤xlogℓ n = xPℓ (log x) +O

(

logℓ x)

,

where

Pd (y) =d

∑

r=0

(−1)r d!

(d− r)!yd−r,

a polynomial of degree d.

Solution i) Apply (27) .

ii) If d = 0 then∫ y

0

eudu = ey − 1

which is the right hand side of (31). Assume the result holds for d = k.Starting with an integration by parts

∫ y

0

euuk+1du =[

euuk+1]y

0− (k + 1)

∫ y

0

euukdu

= eyyk+1 − (k + 1)

(

eyk

∑

r=0

(−1)r k!

(k − r)!yk−r − (−1)k k!

)

by inductive hypothesis

= ey

(

yk+1 −k

∑

r=0

(−1)r (k + 1)!

(k − r)!yk−r

)

− (−1)k+1 (k + 1)!

Knowing we are going to change the variable of summation from r to

46

r + 1 write k − r as k + 1− (r + 1) so

∫ y

0

euuk+1du = ey

(

yk+1 +k

∑

r=0

(−1)r+1 (k + 1)!

(k + 1− (r + 1))!yk+1−(r+1)

)

− (−1)k+1 (k + 1)!

= ey

(

yk+1 +k+1∑

m=1

(−1)m (k + 1)!

(k + 1−m)!yk+1−m

)

+

− (−1)k+1 (k + 1)!

having changed the label of summation to m = r + 1. The free termyk+1 is the missing m = 0 term from the sum, so

∫ y

0

euuk+1du = ey

(

k+1∑

m=0

(−1)m (k + 1)!

(k + 1−m)!yk+1−m

)

−(−1)k+1 (k + 1)!

This shows that (31) holds for d = k + 1. Hence by induction it holdsfor all d ≥ 1.

iii) The result follows from (30). For the main term we use (31) , withthe (−1)ℓ ℓ! term being absorbed into the error.

47

Partial Summation

Partial Summation is often no more than an exercise in interchangingSums and Integrals.

8. Let x ≥ 1 be real, {an}n≥1 a sequence of complex numbers and A (x) =∑

1≤n≤x an.

a) Show that∑

1≤n≤xan (x− n) =

∫ x

1

A (t) dt.

b) Show that

∑

1≤n≤xan log

x

n=

∫ x

1

A (t)

tdt =

∫ log x

0

A (ey) dy.

c) Show that

∑

1≤n≤xan (e

x − en) =

∫ x

1

etA (t) dt =

∫ ex

e

A (log y) dy.

Hint Write x− n, log (x/n) and ex − en as integrals.

Solution a) Perhaps the easiest way to answer this question is to lookat the integral

∫ x

1

A (t) dt =

∫ x

1

(

∑

n≤tan

)

dt =∑

n≤xan

∫ x

n

dt,

interchanging the sum and integral, where the condition on n, n ≤ t inthe sum becomes the condition on t, t ≥ n in the integral. Continuing

∑

n≤xan

∫ x

n

dt =∑

n≤xan (x− n) .

b) The right hand side equality comes from the change of variabley = log t. For the other equality

∫ x

1

A (t)

tdt =

∫ x

1

(

∑

n≤tan

)

dt

t=

∑

n≤xan

∫ x

n

dt

t=

∑

n≤xan (log x− log n) ,

48

which gives the stated result.

c) The right hand side equality comes from the change of variable y =et. For the other equality

∑

1≤n≤xan (e

x − en) =∑

1≤n≤xan

∫ x

n

etdt =

∫ x

1

etA (t) dt.

9. The last question concerned sums over n ≤ x, this will be for sumsover n > x.

For f with a continuous derivative for x > 0 satisfying f (x) → 0 asx→∞ and

∫∞1|f ′ (t)| dt <∞, then

∑

n>x

anf(n) = −A (x) f(x)−∫ ∞

x

A (t) f ′(t) dt.

for sequences {an} for which the sum and integral converge.

Hint Write f (n) as an integral from n to ∞.

Solution Use

f(n) = −∫ ∞

n

f ′(t) dt,

which holds because f has a continuous derivative for x > 0 and satisfiesf (x)→ 0.

Next

∑

n>x

anf(n) = −∑

n>x

an

∫ ∞

n

f ′(t) dt = −∫ ∞

x

(

∑

x<n≤tan

)

f ′(t) dt,

where the interchange of summation and integral is justified because∫∞1|f ′ (t)| dt <∞. Continuing

∑

n>x

anf(n) = −∫ ∞

x

(A (t)− A (x)) f ′(t) dt

= A (x)

∫ ∞

x

f ′(t) dt−∫ ∞

x

A (t) f ′(t) dt

= −A (x) f(x)−∫ ∞

x

A (t) f ′(t) dt.

49

10. a) Use Partial Summation to prove that if f has a continuous derivativeon [1, x] then for a sum over primes we have

∑

p≤xf(p) = −

∫ x

2

π (t) f ′(t) dt+ π (x) f(x) . (32)

b) Recalling that θ (x) =∑

p≤x log p, deduce that

θ (x) = π (x) log x+O

(

x

log x

)

.

Compare this with Theorem 2.20.

Hint for b. You may have to use Chebyshev’s bound π (x) = O (x/ log x)in the integral that arises along with Question 4

Proof a)∑

p≤xf(p) =

∑

p≤x

(

f(x)− (f(x)− f(p)))

= f(x)∑

p≤x1−

∑

p≤x

(∫ x

p

f ′(t) dt

)

= f(x) π (x)−∫ x

2

(

∑

p≤t1

)

f ′(t) dt,

the required result. Note that the integral when interchanged with thesum runs from 2 since 2 is the smallest value the prime variable p cantake.

b) Take f (x) = log x, for then

θ (x) =∑

p≤xlog p = π (x) log x−

∫ x

2

π (t)

tdt

by (32). To estimate this integral use Chebyshev’s bound in the formπ (t)≪ t/ log t, so

∫ x

2

π (t)

tdt ≪

∫ x

2

t

log t

dt

t=

∫ x

2

dt

log t≪ x

log x

by Question 4.

50

Deductions from Merten’s Theorem.

11. Prove that∑

2≤n≤x

1

n log n= log log x+O (1) .

This last result could be compared with Merten’s result

∑

p≤x

1

p= log log x+O (1) .

This may lead you to think that the n-th prime is “of size” n log n.

Solution The summand is decreasing so we can apply (27) to get

∑

2≤n≤x

1

n log n=

∫ x

2

dt

t log t+O (1) = log log x+O (1) .

12. On the previous Problem Sheet you were asked to show that

∞∑

n=2

1

n (log n)1+α

converges for all α > 0.

After Question 11 it might be thought that the n-th prime pn is “ap-proximately” of size n log n, which we might write as pn ≈ n log n. Inwhich case log pn may be thought of as of size

log pn ≈ log(n log n) = log n+ log log n ≈ log n.

Thus pn(log pn)α would be “approximately” of the size

pn(log pn)α ≈ (n log n)(log n)α = n(log n)1+α.

Hence part i might then suggest that the sum over primes

∑

p

1

p (log p)α.

51

converges for all α > 0. Prove that this is so.

Hint Use Partial Summation to remove the 1/ (log p)α so you canapply Merten’s Theorem, Theorem 2.22, and in particular (16) .

Solution Remove the 1/ (log p)α by partial summation,

∑

p≤x

1

p (log p)α=

∑

p≤x

1

p

(

1

(log x)α−

(

1

(log x)α− 1

(log p)α

))

=1

(log x)α∑

p≤x

1

p+ α

∑

p≤x

1

p

∫ x

p

dt

t (log t)1+α

=1

(log x)α∑

p≤x

1

p+ α

∫ x

2

(

∑

p≤t

1

p

)

dt

t (log t)1+α .

Now insert Merten’s result on the sum of the inverses of primes,

∑

p≤x

1

p (log p)α=

1

(log x)α(log log x+O (1))

+α

∫ x

2

(log log t+O (1))dt

t (log t)1+α by (16)

= α

∫ x

2

log log t

t (log t)1+αdt +log log x

(log x)α

+O

(

1

(log x)α

)

+O

(∫ x

2

dt

t (log t)1+α

)

.

All the non-integral terms→ 0 while the integrals converge as x→∞.For the integrals this can most easily seen by changing variables y =log t when they become

∫ x

2

log log t

t (log t)1+αdt =

∫ log x

log 2

log y

y1+αdy and

∫ x

2

dt

t (log t)1+α =

∫ log x

log 2

dy

y1+α.

Hence the given sum over primes converges.

52

13. Prove that for a fixed c > 1,

∑

x<n≤cx

Λ (n)

n≪ 1,

i.e. this sum is bounded for all x > 1.

(What is more difficult, and thus of more interest, is to show that thissum is bounded below and thus non-zero. For this would then showthe existence of n ∈ [x, cx] for which Λ (n) 6= 0. This n would be apower of a prime, and since powers greater or equal to 2 are rare, thiswould lead to the existence of a prime in [x, cx] for any c > 1.)

Hint Use Merten’s result (14) (twice).

Solution From (14) we have

∑

x<n≤cx

Λ (n)

n=

∑

n≤cx

Λ (n)

n−

∑

n≤x

Λ (n)

n

= (log cx+O (1))− (log x+O (1))

= log c+O (1) ,

equivalent to the stated result.

14. Prove that∫ x

1

ψ (u)

u2du = log x+O (1) .

Hint Interchange the integral and the summation within the definitionof ψ, use Merten’s Theorem and Chebyshev’s bound ψ (x)≪ x.

53

Solution∫ x

1

ψ (u)

u2du =

∫ x

1

∑

n≤uΛ (n)

du

u2=

∑

n≤xΛ (n)

∫ x

n

du

u2

=∑

n≤xΛ (n)

(

1

n− 1

x

)

=∑

n≤x

Λ (n)

n− 1

x

∑

n≤xΛ (n)

= log x+O (1) +O

(

1

xx

)

by (14) on the first term and Chebyshev’s inequality for ψ (x) on thesecond.

15. Prove that for a fixed constant c > 1,∫ x

1

ψ (cu)− ψ (u)

u2du = (c− 1) log x+Oc (1) .

Note, Since for sufficiently large x the right hand side is greater than0 we must have that the integrand is non-zero, thus ψ (cu)−ψ (u) > 0,and in particular, there is a prime in [u, cu], for some values of u.Unfortunately this does not tell us for which u these intervals containa prime (it is, in fact, for all u sufficiently large) and how many primesare in these intervals.

Solution Change variables w = cu,∫ x

1

ψ (cu)

u2du = c

∫ cx

c

ψ (w)

w2dw = c

∫ cx

1

ψ (w)

w2dw +Oc (1) ,

since the integral from 1 to c is simply a constant (depending on c).Then, using Question 14,∫ x

1

ψ (cu)− ψ (u)

u2du = c

∫ cx

1

ψ (w)

w2dw −

∫ x

1

ψ (u)

u2du+Oc (1)

= (c− 1)

∫ x

1

ψ (u)

u2du+ c

∫ cx

x

ψ (w)

w2dw +Oc (1) .

54

The first term here

(c− 1)

∫ x

1

ψ (u)

u2du = (c− 1) (log x+O (1)) = (c− 1) log x+Oc (1) .

For the second term,

c

∫ cx

x

ψ (w)

w2dw = c

(∫ cx

1

ψ (w)

w2dw −

∫ x

1

ψ (w)

w2dw

)

= c (log (cx)− log x+Oc (1))

= c log c+Oc (1)

= Oc (1)

Combine to get required result.

55

Prime Number Theorem

16. (Tricky) Assume the Prime Number Theorem in the form π (x) ∼x/ log x as x→∞. Prove that the n-th prime pn satisfies

pn ∼ n log n

as n→∞, i.e.

limn→∞

pnn log n

= 1.

Hint Note that π (pn) = n, apply the Prime Number Theorem in theform given in the question and take logarithms.

Solution Let x = pn in π (x) ∼ x/ log x as x→∞ to get n ∼ pn/ log pnwhich is equivalently

n log pnpn

→ 1 (33)

as n→∞. Take logarithms to get

log n+ log log pn − log pn → 0

as n→∞. Divide by log pn to get

log n

log pn+

log log pnlog pn

− 1→ 0

as n → ∞. For the second term use the idea that as y → ∞ then(log y) /y → 0. This leaves

log n

log pn− 1→ 0 (34)

as n→∞. Combine (33) and (34) ,

n log n

pn=n log pnpn

× log n

log pn→ 1× 1 = 1,

i.e.pn ∼ n log n

as n→∞, as required.

56

Stirling’s Theorem

Stirling’s Theorem relates to any asymptotic result for N ! as N →∞.In Problem Sheet 1 you were asked to show

e

(

N

e

)N

≤ N ! ≤ eN

(

N

e

)N

, . (35)

by examining

logN ! =∑

1≤n≤Nlog n.

In Chapter 2 we show that

∑

1≤n≤xlog x = x log x− x+O (log x) , (36)

for any real x > 1. Choosing x = N , an integer, and then exponenti-ating gives

N ! = e−NNNeO(logN).

But what does eO(logN) means? It represents ef(N) for some functionf (N) satisfying

f (N) = O (logN) , i.e. − C logN < f (N) < C logN,

for some C. Then

N−C = e−C logN < ef(N) < eC logN = NC ,

which we express as eO(logN) = NO(1). Thus we have a very weak form,N ! = e−NNN+O(1), of Stirling’s Theorem. To improve this we need toimprove (36), but only for integer x.

17. The function {t} is periodic, period 1. The result of Question 1

∫ α+1

α

{t} dt = 1

2,

can be interpreted as saying that {t} has average value 1/2.

Define

P (x) =

∫ x

1

(

{t} − 1

2

)

dt.

57

Prove that P (x) is periodic, period 1 and P (n) = 0 for all n ∈ Z.

Solution It suffices to prove that P (x) = P (x− 1) for all x ∈ R. Todo this we consider

P (x)− P (x− 1) =

∫ x

x−1

(

{t} − 1

2

)

dt

=

∫ x

x−1{t} dt− 1

2

=1

2− 1

2= 0.

Then for any n ∈ Z we have P (n) = P (1) = 0.

18. Euler Summation, Proposition 2.8, is for sums over n ≤ x with x real.Improved results can be given when x is an integer.

i) Prove∑

1≤n≤Nlog n = N logN −N + 1 +

∫ N

1

{t}tdt, (37)

for integers N ≥ 1.

ii) Prove, by integrating by parts,

∑

1≤n≤Nlog n = N logN −N +

1

2logN + 1 +

∫ N

1

P (t)

t2dt, (38)

for integers N ≥ 1, where P (t) is the function seen in Question 17.

iii) Prove that there exists a constant C such that

∑

1≤n≤Nlog n = N logN −N +

1

2logN + C +O

(

1

N

)

, (39)

for integers N ≥ 1.

58

Hint Use an idea seen in the proof of Theorem 2.9, replacing anyintegral over [1, x] of an integrable function by an integral over [1,∞)and then estimating the tail end integral over (x,∞).

This could be compared with Example 2.11,∑

1≤n≤xlog n = x log x− x+O (log x) ,

for real x ≥ 1.

iv) Deduce Stirling’s formula in the form

N ! = A

(

N

e

)N √N

(

1 +O

(

1

N

))

,

for some constant A.

It can be shown that A =√2π.

This shows that the true result for N ! lies ‘midway’ between the boundsin (35).

Solution i) Euler Summation states that if f is differentiable then

∑

1≤n≤xf(n) =

∫ x

1

f(t) dt+ f(1)− {x} f(x) +∫ x

1

{t} f ′(t) dt.

So if x = N an integer, then {N} f(N) = 0 while, with f(n) = log n,the other terms give

∑

1≤n≤Nlog n =

∫ N

1

log tdt+ log 1 +

∫ N

1

{t}tdt

= [t log t− t]N1 +

∫ N

1

{t}tdt,

leading to the stated result (37).

ii) Continuing∫ N

1

{t}tdt =

∫ N

1

{t} − 1/2

tdt+

1

2

∫ N

1

dt

t

=

[

P (t)

t

]N

1

+

∫ N

1

P (t)

t2dt+

1

2logN,

59

having integrated the first integral by parts. The [...]N1 term vanishessince P (n) = 0 at all integers, by Question 17.

iii) We can complete the integral to ∞ because |P (t)| is bounded by aconstant (by virtue of the fact that it is a continuous periodic function)and so the integrand is≪ 1/t2, a function integrable over [1,∞). Thuswe can write

∫ N

1

P (t)

t2dt =

∫ ∞

1

P (t)

t2dt−

∫ ∞

N

P (t)

t2dt.

The first integral on the right is a constant, while the second can bebounded as

∫ ∞

N

P (t)

t2dt = O

(∫ ∞

N

1

t2dt

)

= O

(

1

N

)

.

Then

C = 1 +

∫ ∞

1

P (t)

t2dt

in (39) .

iv) Exponentiating this last result gives

N ! = A

(

N

e

)N √NeE(N). (40)

where E (N) is an error function satisfying E (N) = O (1/N). Use theidea that |ex − 1| < 2|x| for |x| < 1/2, (a result seen in MATH20101)so

∣

∣eE(N) − 1∣

∣ < 2 |E (N)| = O

(

1

N

)

, i.e. eE(N) = 1 +O

(

1

N

)

.

Hence we get the stated result.

60