Analytic Number Theory
Solutions
Solution to Problem 1. We prove the formula for Li(x) by the induction on k. By the
definition,
Li(x) =
Z
x
2
1
dt.
log t
R
R 0
We use the integration by parts f g 0 = f g
gf with f (t) = 1/ log t and g(t) = t to obtain
Z x
Z x
1
t x
1 1
x
2
dt =
+
· t dt =
+ I1 (x).
2
log
t
log
t
t
(log
t)
log
x
log
2
2
2
2
This implies the formula for k = 1. Now assume that the formula holds for k
1 and we
wish to prove it for k + 1. Again using the integration by parts with f (t) = 1/(log t)k+1 and
g(t) = t we have
Ik (x) =
t
(log t)k+1
x
2
Z
x
2
(k + 1)
x
· t dt =
t(log t)k+2
(log x)k+1
2
+ (k + 1)Ik+1 (x).
(log 2)k+1
Therefore,
Li(x) = Ck +
= Ck +
k 1
X
n=0
k 1
X
n=0
⇣
= Ck
k!
n!
x
+ k!Ik (x)
(log x)n+1
n!
⇣
x
x
+
k!
(log x)n+1
(log x)k+1
⌘
2
+
(k
+
1)I
(x)
k+1
(log 2)k+1
⌘ X
2
x
+
n!
+ (k + 1)!Ik+1 (x).
k+1
(log x)n+1
(log 2)
k
n=0
This finishes the proof of the formula by introducing the appropriate constant Ck+1 .
Since the function 1/(log t)2 is decreasing as t increases, we have
Z px
Z x
1
1
I1 (x) =
dt
+
dt
p (log t)2
2
(log
t)
2
x
p
p
1
1
 ( x 2)
+ (x
x)
2
(log 2)
(log x1/2 )2
p
x
4x

+
.
(log 2)2 (log x)2
Using the formula for Li(x) with k = 1 we have
Li(x) = C1 +
x
+ I1 (x).
log x
3
Then,
lim
x!1
Li(x)
C1 log x
I1 (x)
I1 (x)
= lim
+ lim 1 + lim
= 1 + lim
x!1
x!1 x/ log x
x!1 x/ log x
x/ log x x!1
x
However, by the above inequality
I1 (x)
 lim
x!1 x/ log x
x!1
lim
p
x
(log 2)2
x
log x
4x
(log x)2
x
x!1
log x
+ lim
= lim
x!1
log x
4
+ lim
=0
x(log 2)2 x!1 log x
Solution to Problem 2. We have
f1 (n)/f3 (n)
f1 (n)f4 (n)
f1 (n)
f4 (n)
= lim
= lim
· lim
= 1 · 1 = 1.
n!1 f2 (n)/f4 (n)
n!1 f2 (n)f3 (n)
n!1 f2 (n) n!1 f3 (n)
lim
and
lim
n!1
f1 (n) + f3 (n)
f1 (n) + f3 (n)
f4 (n)
f1 (n)/f3 (n) + 1
1+1
= lim
· lim
= lim
=
= 1.
f2 (n) + f4 (n) n!1 f2 (n) + f4 (n) n!1 f3 (n) n!1 f2 (n)/f4 (n) + 1
1+1
In the special case given, we have f1 (n)
lim
n!1
f3 (n) = n and f2 (n)
f1 (n)
f2 (n)
f4 (n) = 2n, thus,
f3 (n)
1
= 6= 1.
f4 (n)
2
Solution to Problem 3. Let us introduce the notation
G1 (N, M ) =
M
X
f1 (n),
G2 (N, M ) =
n=N
M
X
By the hypothesis, for every ✏ > 0 there is N0 such that for all n
f2 (n)(1
f2 (n).
n=N
N0 we have
✏)  f1 (n)  (1 + ✏)f2 (n).
This implies that F1 (N ) ! 1 if and only if F2 (N ) ! 1, when N ! 1. Also, summing up
these inequalities we conclude that for all M
(1
N0 we have
✏)G2 (N0 , M )  G1 (N0 , M )  (1 + ✏)G2 (N0 , M ).
4
Hence,
lim sup
M !1
F1 (M )
G1 (1, N0
= lim sup
F2 (M )
M !1 G2 (1, N0
G1 (1, N0
= lim sup
M !1
= lim sup
M !1
1) + G1 (N0 , M )
1) + G2 (N0 , M )
1)/G2 (N0 , M ) + G1 (N0 , M )/G2 (N0 , M )
G2 (1, N0 1)/G2 (N0 , M ) + 1
G1 (N0 , M )
G2 (N0 , M )
 1 + ✏.
Similarly, lim inf M !1
F1 (M )
F2 (M )
1
✏. As ✏ > 0 was arbitrary, we conclude that the limit exists
and is equal to +1.
Solution to Problem 4. a) the answer is no, for instance one can define
f (x) = g(x) + sin(eg(x) ) + sin(ex ).
Then, since g(x) is bounded from below fro x
1, then f (x) = O(g(x)) + O(1) + O(1) =
O(g(x)). On the other hand,
f 0 (x) = g 0 (x) + g 0 (x)eg(x) cos(eg(x) ) + ex cos(ex ).
Depending on whether g(x) = O(1) (and hence g 0 (x) ! 0 as x ! 1) or not, one can see that
f 0 (x) 6= O(g 0 (x)).
b) The answer is yes, since f (x)  Cg(x) implies that
Z X
Z X
Z
f (t) dt 
Cg(t) dt  C
2
2
X
g(t) dt.
2
Solution to Problem 5. By contradiction, assume that there is n0 such that for n
the di↵erences pn+1
n > n0 , pn+1
pn+1
pn
pn
1 + (pn
n0
pn are strictly increasing. Since the di↵erences are integers, then for
1 + (pn
pn
1)
pn
1 ).
2 + (pn
In particular, for large values of n > n0 we have
1
pn
2)
···
(n n0 ) + (pn0 +1
pn 0 )
(n n0 ) + 1.
Putting these together, we obtain
pn
pn 0 =
n
X1
i=n0
(pi+1
pi )
n
X1
(i
n0 + 1) =
i=n0
nX
n0
i=1
5
i=
(n
n0 ) 2 n n0
+
2
2
(n
n0 ) 2
.
2
By PNT we have ⇡(x) ⇠
x
log x .
For instance, for large values of x we must have ⇡(x)
pn
2 log pn .
Since, for large enough n, pn is large enough, we must have n = ⇡(pn )
On the other hand, for large enough x, we know that
must have
pn
log pn
2n
⇣ (n
p3/4
n
which is a contradiction.
x
log x
n0 )2 ⌘3/4
2
(n
x
2 log x .
x3/4 . Then, for large n we
n0 )3/2
23/4
,
Solution to Problem 6. By PNT
lim
x!1
⇡(x)
= 1.
x/ log x
In particular, we look at the limit along the sequence pn as n ! 1. By the definition
⇡(pn ) = n. Then,
lim
n!1
In other words,
pn
log pn
we have
n
= 1.
pn / log pn
⇠ n. Since log is a continuous function over the positive real numbers,
lim (log n
n!1
log pn + log(log pn )) = lim log
n!1
This implies that
⇣ log n
log n
= lim
n!1
n!1 log pn
lim
= 0 + 1 + 0 = 1.
⇣
⌘
n
= 0.
pn / log pn
log pn + log(log pn ) log pn log(log pn ) ⌘
+
+
log pn
log pn
log pn
To find the limit of the last ration in the above equation we may for instance use log(x)  x1/2 ,
that is, log(log pn )  (log pn )1/2 .
Finally,
⇣ n log n log p ⌘
n log n
n
= lim
⇥
n!1
n!1
pn
pn
log pn
⇣ n log p
log n ⌘
n
= lim
⇥
n!1
pn
log pn
n
log n
= lim
⇥ lim
n!1 pn / log pn
n!1 log pn
lim
=1⇥1=1
6
Solution to Problem 7. Let us define the sets
An = {d  n; d|n}, Bn = {d  2n
We have d(n) = #An and d(2n
1; d|2n
1) = #Bn . We claim that
1}.
(d) = 2d
1 maps An into Bn .
To see this, let n = dk, for some d 2 An . We have
2n
which implies 2d
1 = (2d )k
1|2n
1 = (2d
1) (2d )k
1
+ (2d )k
2
+ ··· + 1
: An ! Bn is one -to-one, we must have #An  #Bn .
1. Since,
Solution to Problem 8. Let N = [X].
Z X
S(t)F 0 (t) dt
1
=
=
=
=
=
=
=
=
=
N
X1 Z i+1
i=1
N
X1
i=1
N
X1
i=1
N
X1
i=1
N
X1
i=1
N
X2
S(t)F 0 (t) dt +
i
S(i)
Z
i+1
Z
X
S(t)F 0 (t) dt
N
0
F (t) dt + S(N )
i
S(i) F (i + 1)
S(i)F (i + 1)
S(i)F (i + 1)
Z
X
F 0 (t) dt
N
F (i) + S(N ) F (X)
N
X1
i=1
N
X2
F (N )
S(i)F (i) + S(N ) F (X)
F (N )
S(i + 1)F (i + 1) + S(N ) F (X)
F (N )
i=0
(S(i)
i=1
N
X2
i=1
N
X
i=1
N
X
S(i + 1))F (i + 1) + S(N
f (i + 1)F (i + 1) + S(N
f (i)F (i) + S(N
1)F (N )
1)F (N )
S(1)F (1) + S(N ) F (X)
S(1)F (1) + S(N
1)F (X) + f (N )F (X)
f (i)F (i) + S(X)F (X).
i=1
7
F (N )
1) + f (N ) F (X)
F (N )