The Riemann Zeta Function

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1 Physics 6A Witer 6 The Riema Zeta Fuctio I this ote, I will sketch some of the mai properties of the Riema zeta fuctio, ζ(x). For x >, we defie ζ(x) =, x >. () x = For x, this sum diverges. However, we shall see that it is possible to exted the defiitio of ζ(x) so that it is valid for all real x (more geerally, all complex x). There is a beautiful relatio betwee the zeta fuctio ad the Gamma fuctio. I derivig this relatio, we shall establish aother possible defiitio of ζ(x) [for x > ] i terms of a itegral. The trick here is to ote the followig formula: e t t x dt = Γ(x). x To prove this result, defie a ew variable u = t. The resultig itegral is recogized as the defiitio of the Gamma fuctio up to a overall factor of x. Now, we ca do somethig clever ad use the above result to write = x Γ(x) e t t x dt. If we ow sum both sides of the equatio from = to =, we obtai: ζ(x) = e t t x dt Γ(x) = = Γ(x) e t t x dt. () where the iterchage of the sum ad the itegral is justified by the uiform covergece of the sum. The sum over is just a ifiite geometrical series with the = term missig. Thus, usig r = r = r r = = r with r = e t, it follows that ζ(x) = Γ(x) = t x dt, x >. (3) e t

2 Note that this itegral form for ζ(x) coverges for x > ad diverges for x just as i eq. (). I particular, ζ() =. But, it is useful to determie the behavior of ζ(x) as x. Here we shall simply quote the relevat result: lim x ( ζ(x) x where γ is Euler s costat. Equivaletly, ) = γ, ζ(x) = + γ + O(x ), as x. (4) x The proof of this result ca be foud i the article o the Riema zeta fuctio that is posted o the class web site. We shall ow defie aother fuctio that is closely related to the Riema zeta fuctio: ( ) + η(x) =, x >. (5) x = Note that this series coverges for all x > by the alteratig series test. We ca also derive a itegral expressio for this fuctio by exactly the same procedure as above. This time, the geometric series that we ecouter is give by ( ) + r = + r = r + r = = r + Thus, we ed up with η(x) = Γ(x) t x dt, x >. (6) e t + I ecourage the reader to fill i the missig steps of this derivatio. Eqs. (3) ad (6) play a cetral role i the statistical mechaics of a free ideal Bose ad Fermi gas, respectively. The fuctio η(x) is related i a simple way to the Riema zeta fuctio as follows η(x) = x + 3 x 4 x = = + x + 3 x + 4 x + [ x + 4 x + 6 x + ] = ζ(x) x ( + x + 3 x + ) = ζ(x) x ζ(x) = ( x )ζ(x).

3 However, we must be careful. Although the series defiitio for η(x) [eq. (5)] coverges for x >, this covergece is coditioal for < x ad absolute oly for x >. Whereas the rearragemet of a absolutely coverget series do ot chage its sum, this is o loger true for a coditioally coverget series. Hece, the maipulatios just performed are mathematically soud oly if x >. Hece, we coclude that η(x) = ( x )ζ(x), x >. (7) If we solve eq. (7) for ζ(x), we ca write: ζ(x) = x ( ) + = x. (8) This expressio is valid for x >. If we try to set x =, we fid that ζ() = as expected. But, we also otice somethig remarkable. Eq. (8) seems to make perfect sese for < x <. We kow that the sum coverges coditioally if < x <. Moreover, there are o sigularities i sight. Thus, we shall exted the defiitio of the Riema zeta fuctio by defiig ζ(x) by eq. (8) for x >. We ca use eq. (8) to study the behavior of ζ(x) as x. This is a very istructive exercise i expasios, so I preset the details here. Let x = + ɛ, where ɛ is a very small parameter. We shall expad eq. (8) about ɛ =. The first step is to expad ( x ). Puttig x = ɛ, = x = ɛ e ɛ l. Expadig out the expoetial i the deomiator of the last expressio, x ( ɛ l + ɛ l ) ( ) ɛ l ɛ l ( + ɛ l ɛ l ). (9) The secod step is to expad out the factor x = +ɛ that appears i the deomiator of the summad i eq. (8). = +ɛ = ɛ e. ɛ l Workig to first order i ɛ, we expad the expoetial ad the expad the resultig expressio as follows: e ɛ l ( + ɛ l ) ( ɛ l ). 3

4 It the follows that ( ) + +ɛ = ( ) + [ ɛ l ] = ( ) + = l ɛ l + O(ɛ ), () = where we have idetified the well kow expasio, l = = ( )+ /. Applyig the results of eqs. (9) ad () to eq. (8), we fid [ ] ζ( + ɛ) = [ + ( ) + ɛ l ] l ɛ l + O(ɛ ) ɛ l = ɛ + l l = ( ) + l + O(ɛ). = Puttig back x = + ɛ, ad writig ( ) + = ( ), we arrive at our fial result: ζ(x) = x + l + l ( ) l + O(x ), as x. () = Comparig eq. () with eq. (4), we lear that: γ = l + l ( ) l, = which also provides us with the sum of the alteratig series: ( ) l = l [ γ l ]. = So far, we have maaged to exted the defiitio of ζ(x) to iclude all x >. As i the case of the Gamma fuctio, we would like to exted the defiitio further to iclude all real x (ad evetually all complex x). This ca be doe, but it requires methods beyod the scope of this ote. Oe of the key steps makes use of the followig remarkable fuctioal relatio: ζ(x) = x π x si ( πx) Γ( x) ζ( x). () Oe ca prove that this fuctioal relatio holds for < x <, where the zeta fuctio o both sides of the equatio are defied by eq. (8). The geeral proof 4

5 of this relatio ca be foud i the article o the Riema zeta fuctio that is posted o the class web site. Eq. () provides the way to exted the defiitio of ζ(x). For values of x <, the right had side of eq. () is perfectly well behaved. Thus, we ca use eq. () to defie ζ(x) for all egative values of x. Usig eq. (), we ca work out the value of ζ(). First, we multiply eq. () by ( x) ad use ( x)γ( x) = Γ( x). The we take the limit as x : lim( x)ζ(x) = ζ(). x But, accordig to eq. (4), lim x ( x)ζ(x) =. We coclude that ζ() =. This remarkable result plays a sigificat role i the Casimir effect, which is the pheomeo (predicted by quatum field theory) that two ucharged coductig plates i the vacuum actually attract each other. I a aive versio of the theoretical computatio, oe obtais a result that ivolves lim x = (/x ) =. I the more sophisticated aalysis, it turs out that the actual result of the computatio ivolves lim x ζ(x) =. It seems that ature is tellig us that: =. While we are at it, we ca also evaluate η() = ( )ζ() =. If we iterpret this as the x limit of eq. (5), we would coclude that =, a result we have see before whe substitutig x = ito the geometric series ( x) = = x. Of course, you should ot take these last two equatios literally! From eq. (8), it is clear that ζ(x) for ay value of x >. For egative values of x, we ca examie eq. (). I this case, the right had side of eq. () ca vaish oly whe si(πx/) =. For x <, we have si(πx/) = whe πx = π ( =,, 3,...). Hece, we coclude that: ζ( ) =, =,, 3,... These are the oly zeros of the zeta fuctio for real values of x. Whe the defiitio of the zeta fuctio is exteded to complex values of x, Riema cojectured that the oly other zeros of the Riema zeta fuctio occur for x = + iy. A ifiite umber of zeros appear for various discrete values of y. This cojecture is called the Riema zeta hypothesis ad is the most famous outstadig usolved problem i mathematics (ow that Fermat s last theorem has bee prove). The For the case of x =, we ote that lim x si(πx/)ζ( x) = π/, which implies agai that ζ() =. You should be able to derive this limit from the results preseted above. 5

6 proof of this hypothesis would have very profoud results for the behavior of prime umbers. Remarkably, there seems to be a amazig coectio with physics as well. The distributio of the discrete values of y correspodig to the zeros of ζ(x) for x = + y seems to mirror precisely the distributio of eergy levels of certai complex quatum systems. It would be remiss ot to metio the value of the zeta fuctio at other iteger values. There is a deep coectio betwee the zeta fuctio ad the Beroulli umbers. As a result, the followig relatio holds: ζ( ) = B, =,, 3,..., which provides the value of the zeta fuctio evaluated at egative odd itegers i terms of the Beroulli umbers B. The geeral proof of this relatio ca be foud i the article o the Riema zeta fuctio that is posted o the class web site. Usig this result ad eq. (), oe ca ow easily derive the value of the zeta fuctio evaluated at the positive eve itegers: ζ() = ( )+ (π) B ()!, =,, 3,... (3) I derivig this result, I iserted x = i eq. () ad used Γ() = Γ( + ) = ()! ad si( π( )) = ( ). Note that ( ) + B = B, ad we recover eq. () o p. 53 of McQuarrie. As a test, you should put = i eq. (3) ad fid ζ() = 4 (π) B = π /6. Similarly, ζ(4) = π 4 /9, etc. The oly itegers we have ot yet cosidered are the positive odd itegers. You might thik that we could get a formula for ζ( + ) usig eq. (). However, give that ζ( ) =, eq. () with x = (for positive itegers ) simply reads =, ad we fail to extract a value for ζ( + ). I fact, there are o kow formulae for ζ( + ) i terms of powers of π or ay other kow costats. Whether such formulae could ever be discovered is a ope questio, although there is strog evidece that o such formulae exist. McQuarrie itroduces two other series o page 5: λ(x) β(x) = = ( ( + ) = ) ζ(x), x >, x x ( ) ( + ) x, x >. (4) You will otice that I have ot give a relatio betwee β(x) ad ζ(x). As far as I kow, o such relatio exists. The values of β(x) evaluated at the eve positive itegers do ot appear to be related to ay other kow costats. I fact β() A ice itroductio to the Beroulli umbers ca be foud o the class web site. 6

7 is called Catala s costat (usually deoted by G), ad appears from time to time i mathematical aalysis. The values of β(x) evaluated at the odd positive itegers are related to powers of π as follows: β( + ) = ( ) π + E 4 + ()!, (5) where the E are the Euler umbers (ot to be cofused with Euler s costat). The Euler umbers are relatives of the Beroulli umbers ad ca be defied as coefficiets of a particular power series: sech x cosh x = e x + e x = = E x! = x + 5x4 4 6x6 7 +, x < π. Sice cosh( x) = cosh x, oly eve powers of x appear i the Taylor series, so E + =. If you compute the Taylor series of sech x by had, you ca evaluate the o-zero Euler umbers. The first few are: E =, E =, E 4 = 5, E 6 = 6, E 8 = 385,.... Note that ( ) E = E, so that β( + ) give i eq. (5) is positive as required. Usig eq. (5), we ca quickly compute β() = π/4 ad β(3) = π 3 /3. The former result cofirms a summatio that we have already established i class: β() = = ( ) + = π 4. 7

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