Processing math: 100%
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

7.3: Getting Closer to the Proof of the Prime Number Theorem

( \newcommand{\kernel}{\mathrm{null}\,}\)

We know prove a theorem that is related to the defined functions above. Keep in mind that the prime number theorem is given as follows: limxπ(x)logxx=1. We now state equivalent forms of the prime number theorem.

The following relations are equivalent limxπ(x)logxx=1 limxθ(x)x=1 limxψ(x)x=1.

We have proved in Theorem 86 that (???) and (???) are equivalent, so if we show that (???) and (???) are equivalent, the proof will follow. Notice that using the integral representations of the functions in Theorem 85, we obtain θ(x)x=π(x)logxx1xx2π(t)tdt and π(x)logxx=θ(x)x+logxxx2θ(t)tlog2tdt. Now to prove that ([3]) implies (???), we need to prove that limx1xx2π(t)tdt=0. Notice also that (???) implies that π(t)t=O(1logt) for t2 and thus we have 1xx2π(t)tdt=O(1xx2dtlogt) Now once you show that (Exercise 1) x2dtlogtxlog2+xxlogx, then (???) implies (???) will follow. We still need to show that (???) implies (???) and thus we have to show that limxlogxxx2θ(t)dttlog2t=0. Notice that θ(x)=O(x) and hence logxxx2θ(t)dttlog2t=O(logxxx2dtlog2t). Now once again we show that (Exercise 2) x2dtlog2txlog22+xxlog2x then (???) implies (???) will follow.

Define l1=lim infxπ(x)x/logx,     L1=lim supxπ(x)x/logx, l2=lim infxθ(x)x,     L2=lim supxθ(x)x, and l3=lim infxψ(x)x,     L3=lim supxψ(x)x, then l1=l2=l3 and L1=L2=L3.

Notice that ψ(x)=θ(x)+θ(x1/2)+θ(x1/3)+...θ(x1/m)θ(x) where mlog2x

Also, ψ(x)=px[logxlogp]logppxlogxlogplogp=logxπ(x). Thus we have θ(x)ψ(x)π(x)logx As a result, we have θ(x)xψ(x)xπ(x)x/logx and we get that L2L3L1. We still need to prove that L1L2.

Let α be a real number where 0<α<1, we have θ(x)=pxlogpxαpxlogp>xαpxαlogx   (logp>αlogx)=αlogx{π(x)π(xα)} However, π(xα)xα. Hence θ(x)>αlogx{π(x)xα} As a result, θ(x)x>απ(x)x/logxαxα1logx Since limxαlogx/x1α=0, then L2αlim supxπ(x)x/logx As a result, we get that L2αL1 As α1, we get L2L1.

Proving that l1=l2=l3 is left as an exercise.

We now present an inequality due to Chebyshev about π(x).

There exist constants a<A such that axlogx<π(x)<Axlogx for sufficiently large x.

Put l=lim infxπ(x)x/logx,     L=lim supxπ(x)x/logx, It will be sufficient to prove that L4log2 and llog2. Thus by Theorem 2, we have to prove that lim supxθ(x)x4log2 and lim infxψ(x)xlog2 To prove (???), notice that N=C(2n,n)=(n+1)(n+2)...(n+n)n!<22n<(2n+1)N Suppose now that p is a prime such that n<p<2n and hence pN. As a result, we have Nn<p<2np. We get Nθ(2n)θ(n). Since N<22n, we get that θ(2n)θ(n)<2nlog2. Put n=1,2,22,...,2m1 where m is a positive integer. We get that θ(2m)<2m1log2. Let x>1 and choose m such that 2m1x2m, we get that θ(x)θ(2m)2m+1log24xlog2 and we get (???) for all x.

We now prove (???). Notice that by Lemma 9, we have that the highest power of a prime p dividing N=(2n)!(n!)2 is given by sp=μpi=11{[2npi]2[npi]}. where μp=[log2nlogp]. Thus we have N=p2npsp. If x is a positive integer then [2x]2[x]<2, It means that [2x]2[x] is 0 or 1. Thus spμp and we get Npe2npμp. Notice as well that ψ(2n)=p2n[log2nlogp]logp=p2nμplogp. Hence we get logNψ(2n). Using the fact that 22n<(2n+1)N, we can see that ψ(2n)>2nlog2log(2n+1). Let x>2 and put n=[x2]1. Thus x21<n<x2 and we get 2nx. So we get ψ(x)ψ(2n)>2nlog2log(2n+1)>(x2)log2log(x+1). As a result, we get lim infxψ(x)xlog2.

Exercises

  1. Show that l1=l2=l3 in Theorem 88.
  2. Show that x2dtlogtxlog2+xxlogx,
  3. Show that x2dtlog2txlog22+xxlog2x
  4. Show that N=C(2n,n)=(n+1)(n+2)...(n+n)n!<22n<(2n+1)N
  5. Show that 22n2n<N=C(2n,n)<22n2n.
    Hint: For one side of the inequality, write N2n=(2n)!22n(n!)2=1.3.5....(2n1)2.4.6....(2n).2.4.6.....(2n)2.4.6...(2n), then show that 1>(2n+1).N224n>2n.N224n. The other side of the inequality will follow with similar arithmetic techniques as the first inequality.

Contributors and Attributions

  • Dr. Wissam Raji, Ph.D., of the American University in Beirut. His work was selected by the Saylor Foundation’s Open Textbook Challenge for public release under a Creative Commons Attribution (CC BY) license.


This page titled 7.3: Getting Closer to the Proof of the Prime Number Theorem is shared under a CC BY license and was authored, remixed, and/or curated by Wissam Raji.

  • Was this article helpful?

Support Center

How can we help?