# A comment about the integral representation of the Riemann xi-function

*1926*

#### Translator’s note

*This page is a translation into English of the following:*

Pólya, G. “Bemerkung Über die Integraldarstellung der Riemannschen *Acta Math.* **48** (1926), 305–317. DOI: 10.1007/BF02565336.

*The translator (Tim Hosgood) takes full responsibility for any errors introduced, and claims no rights to any of the mathematical content herein.*

Version: `94f6dce`

*[Translator.] The numbering of the footnotes in the original has not been replicated in this translation, since this would have resulted in multiple footnotes with the same number on one page. We also thank Juan Arias de Reyna for their Spanish translation of this paper, which helped greatly.*

The Riemann ^{1}

With regards to the Riemann hypothesis, one could ask the following question^{2}: does the function given by replacing

The answer is no (see §4): the resulting function has infinitely many imaginary zeros.
If, however, the right-hand side of Equation (5) is used, instead of the right-hand side of Equation (4), then we obtain the function
*has only real zeros*.
Incidentally,

In what follows, I provide a proof of the fact that all the zeros of *has only real zeros*.
From this, the same property then easily follows for

# 1

The most important property of the entire function

The proof of Equation (11) goes as follows: write

Now,

I will also state here, without proof, the two following representations, which will not be used in what follows:

# 2

The asymptotic representation and the estimates of

Different areas of the

## I

*For |y|\geqslant 1 and a\leqslant A,*

*where the absolute value of the function*\chi(z)=\chi(z;a) is bounded above by a value depending only on A .

We know, by Equation (11), that

## II

*Let \varepsilon and A be fixed positive numbers.*

*Set*

*Then, in the half-plane*x\geqslant\varepsilon ,

*and, in fact, if*a\leqslant A , then this convergence is uniform in z and a .

It follows, from Stirling’s formula, that there exists some constant ^{3}, it must tend uniformly to zero in the entire half-strip as

Of course, instead of using general theorems, Equation (18) could also be used on suitable curves intersecting the half-strip, e.g. to estimate on the straight lines

## III

*If a\leqslant\frac14, then there are no zeros of {\mathfrak{G}}(z;a) outside of the strip -\frac12<x<\frac12.*

Since

For what follows, it is useful to have in mind a certain partition of the plane (cf. the above figure): the strip

# 3

We will now study consequences of the difference equation in (9), and link them to those that we have just obtained from the representation in Equation (11).

## IV

There are two things that we have to consider:
firstly, the difference equation and the symmetry of

Firstly:
the function

Secondly:
it is not possible for

For real

## V

*There are no zeros of {\mathfrak{G}}(z) outside of the strip {\mathscr{S}}.*

This has already been proven for

## VI

*All the zeros of {\mathfrak{G}}(z) lying inside the strip {\mathscr{S}} are simple and purely imaginary.*

It follows from Equation (17) and from Stirling’s formula that, in the strip *all* the zeros” with “*all but finitely many* zeros”, can be proved using only Equations (28) and (29); a proof of the full result needs more.

Consider the branch of *smallest* positive value of *not* hold by

Denote by

H_n denote the change in\Im\log{\mathfrak{G}}(z) whenz moves in a straight line from1+iy_n to-1+iy_n ;N_n denote the number of*purely imaginary*zeros of{\mathfrak{G}}(z) lying insideR_n , counted*without*multiplicity;N_n+N_n^* denote the number of*all*zeros of{\mathfrak{G}}(z) lying insideR_n , counted*with*multiplicity.

So

The total change of

By definition,

Then

- For
y=y_n in Equation (28), the leading term in Equation (28) dominates over the error term{\mathcal{O}} in all of the segment-1\leqslant x\leqslant 1 , and so{\mathfrak{G}}(z) has no zeros on the horizontal sides ofR_n , as we had previously claimed. - Moving along an arc in this segment, the
e^{i\pi x/2} factor dominates in{\mathfrak{G}}(z) , or, more precisely, for arbitrary\varepsilon we haveH_n < -\pi + \varepsilon \tag{33} forn sufficiently large. Then Equations (32) and (33) give2\pi N_n^* < 2\varepsilon.

Thus

Now, (IV), (V), and (VI) form a complete partition of the plane, and show that *all the zeros of {\mathfrak{G}}(z) are purely imaginary and simple*.
The above observation also shows that the number of zeros with ordinate between

# 4

In order to be able to prove all of the claims in the introduction, we will need two simple, general lemmas.

Let

*Proof*. Since

If we take

Let ^{4}

*Proof*. The hypotheses on

The fact that

By Equations (8), (17), and (19), we also see that, in the half-plane ^{5}

B. Riemann, Werke (1876), S. 138.↩︎

This was casually mentioned by Prof. Landau in a conversation in 1913.↩︎

See G. Pólya and G. Szegö, Aufgaben und Lehrsätze aus der Analysis (Berlin 1925), Bd. 1, Aufgaben III 333, III 339.↩︎

The corresponding result for polynomials is a special case of a theorem of Ch. Biehler. See e.g. G. Pólya and G. Szegö, op. cit., Aufgabe III 25.↩︎

*Note added during editing (5th February, 1926).*By suitably extending Lemma II, we can prove that the function\xi^{**}(z) = 4\pi\int_0^\infty\left[2\pi\left(e^{\frac{9u}{2}} + e^{-\frac{9u}{2}}\right)-3\left(e^{\frac{5u}{2}} + e^{-\frac{5u}{2}}\right)\right]e^{-\pi(e^{2u}+e^{-2u})} \cos zu \operatorname{d}\!u, which “better” approximates the true\xi -function, also has only real zeros.↩︎