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VII. Simple Analytic Proof of the Prime Number Theorem
                        68
                           In fact we give two proofs. This first one is the shorter and
                        simpler of the two, but we pay a price in that we obtain one of
                        Landau’s equivalent forms of the theorem rather than the standard
                        form πàN) ∼ N/ log N. Our second proof is a more direct assault
                        on πàN) but is somewhat more intricate than the first. Here we find
                        some of Tchebychev’s elementary ideas very useful.
                           Basically our novelty consists in using a modified contour integral,

                                                        1     z
                                              fàz)N  z     +       dz,
                                                        z     R 2
                                            A

                                                             z −1
                        rather than the classical one,  fàz)N z dz. The method is rather
                                                     C
                        flexible, and we could use it to directly obtain πàN) by choosing
                        fàz)   log ζ(z). We prefer, however, to derive both proofs from the
                        following convergence theorem. Actually, this theorem dates back
                        to Ingham, but his proof is ´ a la Fourier analysis and is much more
                        complicated than the contour integral method we now give.


                        Theorem. Suppose |a n |≤ 1, and form the series      a n n −z  which
                        clearly converges to an analytic function Fàz) for  z> 1.If,in

                        fact, Fàz) is analytic throughout  z ≥ 1, then   a n n −z  converges
                        throughout  z ≥ 1.

                        Proof of the conveàgence theoremø             Fix a w in  w ≥ 1.
                        Thus Fàz + w) is analytic in  z ≥ 0. We choose an R ≥ 1 and
                                                       1
                        determine δ   δ(R) > 0, δ ≤      and an M   M(R) so that
                                                       2
                          Fàz + w) is analytic and bounded by M in − δ ≤ z, |z|≤ R.
                                                                                      (2)
                        Now form the counterclockwise contour A bounded by the arc |z|
                        R,  z> −δ, and the segment  z  −δ, |z|≤ R. Also denote by
                        A and B, respectively, the parts of A in the right and left half planes.
                           By the residue theorem,


                                                                 1     z

                                  2πiF(w)         Fàz + w)N   z     +   2   dz.       (3)
                                                A                z     R
                        Now on A, Fàz + w) is equal to its series, and we split this into
                        its partial sum S N (z + w) and remainder r N (z + w). Again by the
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