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.