aa r X i v : . [ m a t h . HO ] A ug Basel problem: a physicist’s solution
Zurab K. Silagadze
To paraphrase renowned American physicist Richard Feynman’s quote “ev-ery theoretical physicist who is any good knows six or seven different theoreticalrepresentations for exactly the same physics” [1], every mathematician who isany good knows a dozen of solutions of the Basel problem, which asks for anevaluation of the infinite series ∞ P n =1 1 n . For example, Moreno in the arXiv versionof [2] gives more than 80 references related to various proofs of Euler’s famousformula, and some new ones have appeared since then [3, 4, 5, 6, 7, 8, 9]. EvenEuler himself gave at least four proofs [10, 11, 12] that ζ (2) ≡ ∞ X n =1 n = π . (1)In [13] W¨astlund reformulated the Basel problem in terms of a physical systemusing the proportionality of the apparent brightness of a star to the inversesquare of its distance. Inspired by this approach, here we give another physicalinterpretation which, in our opinion, is simpler, natural enough, and leads to aproof of (1) which is very Eulerian in its spirit. A physicist’s solution of the Basel problem
First of all, let’s notice [13] that, because of ∞ X n =1 n = ∞ X n =1 n − + 14 ∞ X n =1 n , we have ∞ X n =1 n = 43 ∞ X n =1 n − = 13 ∞ X n =1 n − ) . (2)A physicist can interpret (2) as representing a Coulomb force exerted on a unitcharge located at a point x = 1 / x = 1, x = 2, . . . The corresponding electrostatic potentialis U ( x ) = − ∞ X n =1 n − x , (3)1nd we can write ∞ X n =1 n = 13 F ( x ) (cid:12)(cid:12)(cid:12)(cid:12) x = = 13 (cid:18) − dU ( x ) dx (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) x = . (4)Unfortunately (3) diverges and hence we are returning here (temporarily) to thestandards of mathematical rigor of Eulerian times. However physicists are usedto infinities, and thus let’s regularize the potential (3): U ( x ) → U R ( x ) = U ( x ) − U (0) = − ∞ X n =1 (cid:18) n − x − n (cid:19) . (5)Note that our regularization procedure does not affect at all the force and hencewe have F ( x ) = ddx ∞ X n =1 (cid:18) n − x − n (cid:19) . (6)In (6) we recognize immediately the presence of the digamma function ψ ( x ) =Γ ′ ( x ) / Γ( x ) because ψ (1 − x ) = − γ − ∞ X n =1 (cid:18) n − x − n (cid:19) , (7) γ being the Euler constant (this relation will be discussed below). Therefore F ( x ) = ddx [ − ψ (1 − x ) − γ ] = ψ (1 − x ) , where ψ ( x ) = ψ ′ ( x ) is the trigamma function. Fortunately (4) indicates thatwe need the trigamma function at x = 1 / ∞ X n =1 n = 13 ψ (cid:18) (cid:19) , (8)and this is just the value at which the trigamma function can be simply calcu-lated thanks to the Euler’s reflection formula: ψ (1 − x ) + ψ ( x ) = π sin πx , (9)which gives ψ (cid:0) (cid:1) = π , and hence from (8) we immediately get Euler’s famousformula (1). Several remarks
Above we presented a physics-motivated approach to the Basel problem. Ofcourse the connection with physics is tenuous at best. However the interpre-tation of inverse squares as representing Coulomb forces was a crucial insight2n defining a connection with polygamma functions and the reflection formula.The resulting formalism is, in fact, quite elementary, in the sense that its basicpillars (7) and (9) can be obtained by elementary means.For example, (7) follows from Newman’s infinite product formula (used byWeierstrass as his definition of the gamma function)1Γ(1 + z ) = e γz ∞ Y n =1 (cid:16) zn (cid:17) e − z/n , (10)which by itself is just another version of Euler’s definition of the gamma functionas the limit [14] Γ( z ) = lim n →∞ n z n ! z ( z + 1) · · · ( z + n ) . (11)A simple consequence of (11) is the following interesting identity [15] ∞ Y k =0 ( k + α ) · · · ( k + α n )( k + β ) · · · ( k + β n ) = Γ( β ) · · · Γ( β n )Γ( α ) · · · Γ( α n ) , (12)where n ≥ α , . . . , α n , β , . . . , β n are nonzero complex numbers, none ofwhich are negative integers, such that α + . . . + α n = β + · · · + β n . In particular,when α = 1 + z , α = 1 − z , β = β = 1, we get ∞ Y n =1 (cid:18) − z n (cid:19) = 1Γ(1 + z ) Γ(1 − z ) , which, in combination with Γ(1 + z ) = z Γ( z ) and Euler’s celebrated formula ∞ Y n =1 (cid:18) − z n (cid:19) = sin πzπz , (13)implies the validity of the reflection formulaΓ( z ) Γ(1 − z ) = π sin πz , (14)from which other reflection formulas, like (9), do follow.It is tempting to consider the infinite product formula (13) to be a real back-bone of the presented approach, as in the Euler’s original first proof. Althoughseveral elementary proofs of Euler’s infinite product for the sine exist in theliterature (see, for example, [16, 17, 18, 19]), they do not seem to be signifi-cantly simpler than the original proof by Euler. Therefore it may appear thatthe reflection formula is fairly nontrivial to derive and its proof is as hard aproblem as the one we seek to solve. However this is actually not the case. Itis possible to avoid the use of the Euler’s infinite product in the derivation ofthe reflection formula. Below we provide one such proof which is simple andelementary enough and doesn’t rely on the infinite product formula for the sinefunction. 3his proof of the reflection formula was inspired by Richard Dedekind’s 1852proof [20] of (14) which seems to be not as well known as it deserves to be. Itappears as an exercise in [21] and was popularized in [22]. We prove not (14),but the reflection formula for the digamma function ψ ( x ) − ψ (1 − x ) = − π cot πx, (15)from which (9) follows by differentiation.During the proof, which seems to be much simpler than the Dedekind’soriginal one, we freely interchange the order of integrals and differentiate underthe integral signs, as physicist are generally accustomed to doing. A genuinemathematician, of course, will resort in these cases to Fubini’s theorem and toLebesgue’s dominated convergence theorem to justify these operations [22].We have the following well known integral representation for the digammafunction: ψ (1 − x ) = − γ + Z − t − x − t dt. (16)Indeed, expanding 1 / (1 − t ) in geometric series, interchanging the order of sum-mation and integration and thus integrating term by term, we get (7).Therefore, φ ( x ) ≡ ψ ( x ) − ψ (1 − x ) = Z t − x − t x − − t dt = 12 Z ∞ t − x − t x − − t dt. (17)The validity of the last step can be checked by breaking the correspondingintegral into two integrals, over (0 ,
1) and (1 , ∞ ), and putting y = 1 /t in thesecond integral.Naively, the two parts of the last integral in (17) Z ∞ t − x − t dt and Z ∞ − t x − − t dt, (18)appear to be the same, because the second transforms into the first under thechange of integration variable y = 1 /t . However the separate integrals in (18) areill-defined because of a singularity at t = 1. Nevertheless these integrals becomewell-defined and equal in the sense of Cauchy principal value. Therefore, φ ( x ) = P Z ∞ t − x − t dt. (19)To get rid of inconvenient principal value, we use quantum physicists’ favoriteformula (Sokhotski-Plemelj formula [23, 24])1 z ± iǫ = P z ∓ iπδ ( z ) , (20) Sokhotski-Plemelj formula is a relation between the generalized functions, that is it isassumed that both sides of (20) are multiplied by a smooth function, which is non-singularin a neighborhood of the origin, then integrated over a range of z containing the origin, andfinally a limit ǫ → φ ( x ) ± iπ = lim ǫ → Z ∞ t − x − t ∓ iǫ dt. (21)Multiplying these two representations of φ ( x ), we get φ ( x ) + π = lim ǫ → Z ∞ t − x − t + iǫ dt Z ∞ s − x − s − iǫ ds. (22)We can substitute s = y/t in the second integral, interchange the order ofintegrations, solve the resulting simple integral in t ,lim ǫ → Z ∞ − t + iǫ )( t − y − iǫt ) dt = − ln y − y , (23)and end up with φ ( x ) + π = − Z ∞ y − x ln y − y dy. (24)On the other hand, if we differentiate (19) by x , we get φ ′ ( x ) ≡ dφ ( x ) dx = − Z ∞ t − x ln t − t dt. (25)(there is no longer a need for the principal value after differentiation, becausethe singularity softens and becomes integrable). Comparing (24) and (25), wesee that the function φ ( x ) satisfies differential equation φ ′ ( x ) = π + φ ( x ) . (26)Note that this differential equation is much simpler than the differential equation F ( x ) F ′′ ( x ) = ( F ′ ( x )) + F ( x ) obtained by Dedekind in [20] for the function F ( x ) = Γ( x )Γ(1 − x ).Under the initial condition φ (1 /
2) = 0, which follows from the definition of φ ( x ), (26) can be solved immediately: x −
12 = Z φ dφπ + φ = 1 π arctan φπ . (27)Therefore φ ( x ) = π tan (cid:16) πx − π (cid:17) = − π cot πx, (28)and this completes the proof of (15). Zeta function values at positive even integers
The above approach can be easily generalized (this time without any physicsinput) to enable a calculation of all ζ (2 k ). Because of ∞ X n =1 n k = ∞ X n =1 n − k + 12 k ∞ X n =1 n k ,
5e have ζ (2 k ) ≡ ∞ X n =1 n k = 2 k k − ∞ X n =1 n − k = 12 k − ∞ X n =1 n − ) k . (29)On the other hand, differentiating (7) 2 k − ψ k − (1 − x ) = (2 k − ∞ X n =1 n − x ) k , (30)where ψ n ( x ) = d n dx n ψ ( x ) is the polygamma function. Therefore ζ (2 k ) = ψ k − (cid:0) (cid:1) (2 k − k − . (31)Since π sin πx = − π ddx cot πx, differentiating (9) 2 k − ) times, we get ψ k − (1 − x ) + ψ k − ( x ) = − π d k − dx k − cot πx. (32)It follows from this reflection formula that ψ k − (cid:18) (cid:19) = − π s k − , (33)where the s n numbers are defined through s n = d n dx n cot πx (cid:12)(cid:12)(cid:12)(cid:12) x = . (34)Thanks to identity tan π (cid:0) x − (cid:1) = − cot πx , we can express the s n numbersthrough more familiar tangent numbers [25] T n = d n dx n tan x (cid:12)(cid:12)(cid:12)(cid:12) x =0 (35)as s n = − π n T n and (31) takes the form ζ (2 k ) = π k T k − k − k − . (36)We can calculate tangent numbers recursively. To do so, note that d n dx n tan x = d n − dx n − x = d n − dx n − (cid:0) x (cid:1) = d n − dx n − (tan x tan x ) , n -th derivative of a product of two functions.We get the recurrence relation T n = n − X r =0 (cid:18) n − r (cid:19) T r T n − − r . (37)From the definition (35) we find T = 0 , T = 1 and it is not hard to prove byinduction that (37) implies the vanishing of tangent numbers if their index iseven. For an odd index, let’s take r = 2 m − T k − = k − X m =1 (cid:18) k − m − (cid:19) T m − T k − m ) − . (38)In principle (36) and (38) solve the generalized Basel problem and allow tocalculate ζ (2 k ) at least for small values of k . For example, we easily get T =2 , T = 16 , T = 272 , T = 7936 , T = 353792 which imply ζ (4) = π , ζ (6) = π , ζ (8) = π , ζ (10) = π , ζ (12) = 691 π . However for large values of k more efficient algorithms are needed to calculatetangent numbers. One of them can be found in [26].A connection with Bernoulli numbers is established by the well-known for-mula (see, for example, [26]), valid for n > B n = − n T n − (2 i ) n (2 n − . We also can use (31), (33) and s n = ( − n (2 π ) n − (2 n − n B n , proved in [27]. In either way we get the well known result ζ (2 k ) = ( − k +1 (2 π ) k B k k )! . Recurrence formula for ζ (2 k ) With some extra effort, it is possible to obtain a nice recurrence relation for ζ (2 k ) [28, 29] which allows a calculation of ζ (2 k ) recursively, and thus alsoprovides a solution of the generalized Basel problem.Let’s introduce another sequence of numbers related to the cotangent func-tion: S n = d n dx n ( x cot x ) | x =0 . (39)7n [28] the coefficients of the Taylor expansion of x cot x were related to thevalues of ζ (2 k ). As an alternative to that argument, we will study the numbers S n and, using the insights from the previous section, relate them to ζ (2 k ).Because of relation x tan x = x cot x − x cot 2 x, (40)which can be simply checked, the numbers introduced are related to the tangentnumbers. Namely, differentiating (40) n times and setting x = 0 produces theequality n T n − = S n − n S n . Therefore T k − = − k − k S k and ζ (2 k ) = − π k S k k )! . (41)Now we obtain a recurrence relation for the S n . By using cot ′ x = − (1 + cot x )it can be checked that [28] x ( x cot x ) ′ = x cot x − x cot x − x . (42)Differentiating both sides of this relation n > x = 0, we get nS n = S n − n X r =0 (cid:18) nr (cid:19) S r S n − r = S n − S n − n − X r =1 (cid:18) nr (cid:19) S r S n − r . Therefore the S n numbers obey the recurrence relation( n + 1) S n = − n − X r =1 (cid:18) nr (cid:19) S r S n − r . (43)As S = 1 , S = 0, it follows from this recurrence relation (by induction) that S n = 0, if n is odd (if n is odd, one of the numbers r , n − r is also odd).Therefore, taking r = 2 m , we can write the recurrence relation (43) in the form(2 k + 1) S k = − k − X m =1 (cid:18) k m (cid:19) S m S k − m . (44)If we substitute (44) into (41), we get a recurrence relation for the zeta-functionthat was called “highly elegant” in [29]: (cid:18) k + 12 (cid:19) ζ (2 k ) = k − X m =1 ζ (2 m ) ζ (2 k − m ) . (45)In fact (45) is equivalent to Euler’s recurrence relation for Bernoulli numbers(independently found by Ramanujan [30])(2 n + 1) B n = − n − X m =1 (cid:18) n m (cid:19) B m B n − m . (46)8oth (45) and (46) were rediscovered many times [31]. For example Williams[32] thought he was the first to explicitly state the recurrence relation in the form(45). Actually, this recurrence relation is given implicitly in Euler’s work [33](paper E130 at ), and is explicitlystated at least as early as 1906 in the book [34] (with the remark that thisrecurrence relation is well known). Nowadays the proof of the recurrence relation(45) is often given as an exercise in number theory courses (see, for example,[35]). Acknowledgments
The work is supported by the Ministry of Education and Science of the RussianFederation. The author thanks Professor Juan Arias de Reyna for indicatingseveral interesting references, as well as an anonymous referee for constructivecomments which helped to improve the presentation.
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