Number Theory: An Introduction to Mathematics

2 Chebyshev’s Functions 369

by Lemma 3, it follows that

π(x)=ψ(x)/logx+O(x/log^2 x).

Suppose next thatπ(x)=O(x/logx).Foranyx>2wehave

θ(x)=

∫x+

2 −

logtdπ(t)

=π(x)logx−

∫x

2

π(t)/tdt=O(x),

and hence alsoψ(x)=O(x), by Lemma 3.

It follows at once from Lemma 4 thatthe prime number theorem,π(x)∼x/logx,
is equivalent toψ(x)∼x.
The method of argument used in Lemma 4 can be carried further. Put

θ(x)=x+R(x), π(x)=

∫x

2

dt/logt+Q(x).

Subtracting

∫x

2

dt/logt=x/logx− 2 /log 2+

∫x

2

dt/log^2 t

from

π(x)=θ(x)/logx+

∫x

2

θ(t)/tlog^2 tdt,

we obtain

Q(x)=R(x)/logx+

∫x

2

R(t)/tlog^2 tdt+ 2 /log 2. (3) 1

Also, adding

∫x

2

(∫t

2

du/logu

)

dt/t=

∫x

2

(∫x

u

dt/t

)

du/logu

=

∫x

2

(logx−logu)du/logu

=logx

∫x

2

dt/logt−x+ 2

to

θ(x)=π(x)logx−

∫x

2

π(t)/tdt

we obtain

R(x)=Q(x)logx−

∫x

2

Q(t)/tdt− 2. (3) 2