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
The translator (Tim Hosgood) takes full responsibility for any errors introduced, and claims no rights to any of the mathematical content herein.
Version: 9bcba86
[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
With regards to the Riemann hypothesis, one could ask the following question2: 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
In what follows, I provide a proof of the fact that all the zeros of
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
We know, by Equation (11), that
II
Let
It follows, from Stirling’s formula, that there exists some constant
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
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
This has already been proven for
VI
All the zeros of
It follows from Equation (17) and from Stirling’s formula that, in the strip
Consider the branch of
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
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
Proof. The hypotheses on
The fact that
By Equations (8), (17), and (19), we also see that, in the half-plane
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.↩︎