1.1. The method of Chebyshev 5
The inequality
ψ(x)−2ψ(x1/2)
ϑ(x) ψ(x) follows since ϑ is an increasing
and nonnegative function, and then
log(2) lim inf
x→+∞
ϑ(x)
x
lim sup
x→+∞
ϑ(x)
x
2 log(2)
by our estimates on ψ(x).
Clearly ϑ(x) π(x) log(x), and so π(x)/(x/ log(x)) ϑ(x)/x. Finding
an upper bound for π(x) is only slightly more challenging. The inequality
ϑ(x)
yp≤x
log(p) (π(x) π(y)) log(y)
yields
π(x)
x/ log(y)

ϑ(x)
x
+
π(y)
x/ log(y)

ϑ(x)
x
+
y
x/ log(y)
.
To obtain the desired upper bound, we must let y increase fast enough with
x so that the left-hand side of the inequality is close to π(x)/(x/ log(x))
for large x, while the second term on the right-hand side should become
negligible in comparison. The choice y = x/
log2(x)
works well, giving
π(x)
x/(log(x) 2 log log(x))

ϑ(x)
x
+
x/
log2(x)
x/(log(x) 2 log log(x))
.
The second term on the right-hand side tends to zero, and
π(x)
x/(log(x)−2 log log(x))
π(x)
x/ log(x)
=
log(x) 2 log log(x)
log(x)
1
as x +∞. Thus
lim inf
x→+∞
ϑ(x)
x
lim inf
x→+∞
π(x)
x/ log(x)
lim sup
x→+∞
π(x)
x/ log(x)
lim sup
x→+∞
ϑ(x)
x
and so
log(2) lim inf
x→+∞
π(x)
x/ log(x)
lim sup
x→+∞
π(x)
x/ log(x)
2 log(2).
Since limx→+∞ li(x)/(x log(x)) = 1, the last inequality shows that π(x) and
li(x) have the same order of growth.
Choosing y =
xa
with 0 a 1 and a otherwise, arbitrarily yields
ϑ(x)
x

π(x)
x/ log(x)

1
a
ϑ(x)
x
+
xa
x/ log(x)
;
thus we see that π(x) x/ log(x), and π(x) li(x), and ϑ(x) x, and
ψ(x) x, and a = A = 1, are equivalent formulations of the Prime Number
Theorem.
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