APPENDIX. THE MELLIN TRANSFORM AND RELATED ANALYTIC TECHNIQUES. D. Zagier. 1. The generalized Mellin transformation

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1 APPENDIX. THE MELLIN TRANSFORM AND RELATED ANALYTIC TECHNIQUES D. Zagier. The generalized Mellin transformation The Mellin transformation is a basic tool for analyzing the behavior of many important functions in mathematics and mathematical physics, such as the zeta functions occurring in number theory and in connection with various spectral problems. We describe it first in its simplest form and then explain how this basic definition can be extended to a much wider class of functions, important for many applications. Let ϕt be a function on the positive real axis t > which is reasonably smooth actually, continuous or even piecewise continuous would be enough and decays rapidly at both and, i.e., the function t A ϕt is bounded on R + for any A R. Then the integral ϕs = ϕtt s dt converges for any complex value of s and defines a holomorphic function of s called the Mellin transform of ϕs. The following small table, in which α denotes a complex number and λ a positive real number, shows how ϕs changes when ϕt is modified in various simple ways: ϕλt t α ϕt ϕt λ ϕt ϕ t λ s ϕs ϕs+α λ s ϕλ s ϕ s s ϕs. 2 We also mention, although we will not use it in the sequel, that the function ϕt can be recovered from its Mellin transform by the inverse Mellin transformation formula ϕt = C+i 2πi C i ϕst s ds, where C is any real number. That this is independent of C follows from Cauchy s formula. However, most functions which we encounter in practise are not very small at both zero and infinity. If we assume that ϕt is of rapid decay at infinity but grows like t A for some real number A as t, then the integral converges and defines a holomorphic function only in the right half-plane Rs > A. Similarly, if ϕt is of rapid decay at zero but grows like t B at infinity for some real number B, then ϕs makes sense and is holomorphic only in the left half-plane Rs < B, while if ϕt has polynomial growth at both ends, say like t A at and like t B at with A < B, then ϕs is holomorphic only in the strip A < Rs < B. But it turns out that in many cases the function ϕt has a meromorphic extension to a larger half-plane or strip than the one in which the original integral converges, or even to the whole complex plane. Moreover, this extended Mellin transform can sometimes be defined even in cases where A > B, in which case the integral does not converge for any value of s at all.

2 Let us start with the frequently occurring case where ϕt is of rapid decay at infinity and is C at zero, i.e., it has an asymptotic expansion ϕt n= a nt n as t. Recall that this means that the difference ϕt N n= a nt n is Ot N as t for any integer N ; it is not required that the series a n t n be convergent for any positive t. Then for s with Rs > and any positive integer N the integral converges and can be decomposed as follows: ϕs = = ϕtt s dt + N ϕt n= ϕtt s dt a n t n t s dt + N n= a n n+s + ϕtt s dt. The first integral on the right converges in the larger half-plane Rs > N and the second for all s C, so we deduce that ϕs has a meromorphic continuation to Rs > N with simple poles of residue a n at s = n n =,...,N and no other singularities. Since this holds for every n, it follows that the Mellin transform ϕs in fact has a meromorphic continuation to all of C with simple poles of residue a n at s = n n =,,2,... and no other poles. The same argument shows that, more generally, if ϕt is of rapid decay at infinity and has an asymptotic expansion ϕt a j t α j t 3 j= as t tends to zero, where the α j are real numbers tending to + as j or complex numbers with real parts tending to infinity, then the function ϕs defined by the integral for Rs > min j Rα j has a meromorphic extension to all of C with simple poles of residue a j at s = α j j =,2,... and no other poles. Yet more generally, we can allow terms of the form t α logt m with λ C and m Z in the asymptotic expansion of ϕt at t = and each such term contributes a pole with principal part m m!/s + α m+ at s = α, because tα+s logt m dt = / α m tα+s dt = m m!/α+s m+ for Rs+α >. By exactly the same considerations, or by replacing ϕt by ϕt, we find that if ϕt is of rapid decay faster than any power of t as t but has an asymptotic expansion of the form ϕt b k t β k t 4 k= at infinity, where now the exponents β k are complex numbers whose real parts tend to, then the function ϕs, originally defined by in a left half-plane Rs < max k Rβ k, extends meromorphically to the whole complex s-plane with simple poles of residue b k at s = β k and no other poles. More generally, again as before, we can allow terms b k t β k logt n k in 3 which then produce poles with principal parts n k+ n k!b k /s+β k n k+ at s = β k. Now we can use these ideas to define ϕs for functions which are not small either at or at, even when the integral does not converge for any value of s. We simply assume that ϕt is a smooth or continuous function on, which has asymptotic expansions of the forms 3 and 4 at zero and infinity, respectively. Again, we could allow more general terms with powers of log t in the expansions, as already explained, but the corresponding modifications are easy and for simplicity of expression we will assume expansions purely in powers of t. For convenience we assume that the numbering s such that Rα Rα 2 and Rβ Rβ 2. Then, for 2

3 any T > formerly we took T =, but the extra freedom of being able to choose any value of T will be very useful later we define two half-mellin transforms ϕ T s and ϕ T s by ϕ T s = ϕ T s = T T ϕtt s dt ϕtt s dt Rs > Rα, Rs < Rβ. Just as before, we see that for each integer J the function ϕ T s extends by the formula T J J ϕ T s = ϕt a j t α j t s a j dt + T s+α j s+α j j= to the half-plane Rs > Rα J+ and hence, letting J, that ϕ T s is a meromorphic function of s with simple poles of residue a j at s = α j j =,2,... and no other poles. Similarly, ϕ T s extends to a meromorphic function whose only poles are simple ones of residue b k at s = β k. We now define j= ϕs = ϕ T s + ϕ T s. 4 2 This is a meromorphic function of s and is independent of the choice of T, since the effect of changing T to T is simply to add the everywhere holomorphic function T T ϕtts dt to ϕ T s and subtract the same function from ϕ T s, not affecting the sum of their analytic continuations. In summary, if ϕt is a function of t with asymptotic expansions as a sum of powers of t or of powers of t multiplied by integral powers of logt at both zero and infinity, then we can define in a canonical way a Mellin transform ϕs which is meromorphic in the entire s-plane and whose poles reflect directly the coefficients in the asymptotic expansions of ϕt. This definition is consistent with and has the same properties 2 as the original definition. We end this section by giving two simple examples, while Sections 2 and 3 will give further applications of the method. Example. Letϕt = t α, whereαisacomplexnumber. Thenϕhasanasymptoticexpansion3 at with a single term α = α, a =, and an asymptotic expansion 4 at with a single term β = α, b =. We immediately find that ϕ T s = T s+α /s + α for Rs + α > and ϕ T s = T s+α /s + α for Rs + α <, so that, although the original Mellin transform integral does not converge for any value of s, the function ϕs defined as the sum of the meromorphic continuations of ϕ T s and ϕ T s makes sense, is independent of T, and in fact is identically zero. More generally, we find that ϕs whenever ϕt is a finite linear combination of functions of the form t α log m t with α C, m Z. These are exactly the functions whose images ϕ λ t = ϕλt under the action of the multiplicative group R + span a finite-dimensional space. In particular, we see that the generalized Mellin transformation is no longer injective. Example 2. Let ϕt = e t. Here the integral converges for Rs > and defines Euler s gamma-function Γs. From the fact that ϕt is of rapid decay at infinity and has the asymptotic here even convergent expansion n= tn /n! at zero, we deduce that Γs = ϕs has a meromorphic continuation to all s with a simple pole of residue n /n! at s = n n =,,... and no other poles. Of course, in this special case these well-known properties can also be deduced from the functional equation Γs + = sγs proved for Rs > by integration by parts in the integral defining Γs, N applications of which gives the meromorphic extension Γs = s s+ s+n Γs+N of Γs to the half-plane Rs > N. From the first of the properties listed in 2, we find the following formula, which we will use many times: ϕt = e λt ϕs = Γsλ s λ >. 5 3

4 2. Dirichlet series and their special values In this section we look at functions ϕt for which the Mellin transform defined in Section is related to a Dirichlet series. The key formula is 5, because it allows us to convert Dirichlet series into exponential series, which are much simpler. Example 3. Define ϕt for t > by ϕt = /e t. This function is of rapid decay at infinity and has an asymptotic expansion actually convergent for t < 2π e t = = t + t2 2 + t3 6 + r= B r r! tr 6 with certain rational coefficients B =, B = 2, B 2 = 6,... called Bernoulli numbers. From the results of Section we know that the Mellin transform ϕs, originally defined for Rs > by the integral, has a meromorphic continuation to all s with simple poles of residue B r /r! at s = r r =,,2,... On the other hand, since e t > for t >, we can expand ϕt as a geometric series e t + e 2t + e 3t +, so 5 gives first in the region of convergence ϕs = Γsζs, where ζs = m= m s Rs > 7 is the Riemann zeta function. Since Γs, as we have seen is also meromorphic, with simple poles of residue n /n! at non-positive integral arguments s = n and no other poles, and since Γs as is well-known and easily proved never vanishes, we deduce that ζs has a meromorphic continuation to all s with a unique simple pole of residue /Γ = at s = and that its values at non-positive integral arguments are rational numbers expressible in terms of the Bernoulli numbers: ζ n = n B n+ n+ n =,, 2,... 8 Example 4. To approach ζs in another way, we choose for ϕt the theta function ϑt = n= e πn2 t t >. 9 The factor π in the exponent has been included for later convenience. We can write this out as ϑt = + 2e πt + 2e 4πt +, and since the generalized Mellin transform of the function is identically by Example, we deduce from 5 that ϕs = 2ζ 2s, where ζ s = π s/2 Γs/2ζs. To obtain the analytic properties of ζs from the results of Section, we need the asymptotics of ϑt at zero and infinity. They follow immediately from the following famous result, due to Jacobi: 4

5 Proposition. The function ϑt satisfies the functional equation ϑt = ϑ t >. 2 t t Proof. Formula 2 is a special case of the Poisson summation formula, which says that = n Zfn fn 3 n Z for any sufficiently well-behaved i.e., smooth and small at infinity function f : R C, where fy = fxe2πixy dx is the Fourier transform of f. To prove this, note that the function Fx = n Z fn+xisperiodicwithperiod, sohasafourierexpansionfx = m Z c me 2πimx with c m = Fxe 2πimx dx = f m. Now set x =. Now consider the function f t x = e πtx2. Its Fourier transform is given by f t y = e πtx2 +2πixy dx = e πy2 /t e πtx+iy/t2 dx = c t f /t y, where c > is the constant c = e πx2 dx. Applying 3 with f = f t therefore gives ϑt = ct /2 ϑ/t, and taking t = in this formula gives c = and proves equation 2. Now we find from that ϑt has the asymptotic expansions ϑt = +Ot N as t and ϑt = t /2 +Ot N as t, where N > is arbitrary. It follows from the results of Section that its Mellin transform ϑs has a meromorphic extension to all s with simple poles of residue and at s = /2 and s =, respectively, and no other poles. From the formula ζ s = 2 ϑs/2 we deduce that the function ζ s defined in is meromorphic having simple poles of residue and at s = and s = and no other poles and hence using once again that Γs has simple poles at non-positive integers and never vanishes that ζs itself is holomorphic except for a single pole of residue at s = and vanishes at negative even arguments s = 2, 4,... This is weaker than 8, which gives a formula for ζs at all non-positive arguments and also shows the vanishing at negative even integers because it is an exercise to deduce from the definition 6 that B r vanishes for odd r >. The advantage of the second approach to ζs is that from equation 2 and the properties of Mellin transforms listed in 2 we immediately deduce the famous functional equation ζ s = ζ s 4 of the Riemann zeta-function which was discovered for integer values, of s by Euler in 749 and proved for all complex values, of s by Riemann in 859 by just this argument. We next generalize the method of Example 3. Consider a generalized Dirichlet series Ls = c m λm s 5 m= where the λ m are real numbers satisfying < λ < λ 2 < and growing at least as fast as some positive power of m. This is an ordinary Dirichlet series if λ m = m for all m. Assume that the series converges for at least one value s of s. Then it automatically converges in a half-plane for instance, if λ m = m then the fact that c m = Om s implies convergence in the half-plane Rs > Rs + and the associated exponential series ϕt = c m e λ mt t >, 6 m= converges for all positive values of t. We then have: 5

6 Proposition 2. Let Ls be a generalized Dirichlet series as in 5, convergent somewhere, and assume that the function ϕt defined by 6 has an asymptotic expansion of the form ϕt a n t n t. 7 n= as t. Then Ls has a meromorphic continuation to all s, with a simple pole of residue a at s = and no other singularities, and its values at non-positive integers are given by L n = n n!a n n =,, 2,... 8 Proof. The function ϕt is of rapid decay at infinity and has the asymptotic expansion 7 at zero, so by the results of Section we know that its Mellin transform ϕs extends meromorphically to all s, with simple poles of residue a n at s = n n =,,,... On the other hand, ϕs is equal to ΓsLs by formula 5, and we know that Γs has simple poles of residue n n! at s = n n =,,..., has no other zeros or poles, and equals at s =. The result follows. Example 5. Consider the Dirichlet series defined by L 4 s = s 3 s + 5s Rs >. 9 Here the function defined by 6 is given by ϕt = e t e 3t + e 5t = e t +e t = 2 cosht geometric series. The asymptotic expansion of this function at t = has the form ϕt = /2 + t 2 /2! + t 4 /4! + = 2 n= E n t n n! where the coefficients E =, E =, E 2 =,... are certain integers called the Euler numbers. It follows from the proposition that the function L 4 s has a holomorphic continuation to all s and that L 4 n = E n /2 for all n. We can omit the factor n because E n = for n odd. The same method works for any Dirichlet series of the form m= χmm s with coefficients χm which are periodic of some period here 4, the most important case being that of Dirichlet L-series, where the coefficients χm also satisfy χm m 2 = χm χm 2 for all m and m 2. Example 6. As a final example, consider the Hurwitz zeta function, defined by ζs,a = n+a s a >, Rs >. 2 Here ϕt = n= n= e n+at = e at e t. But for any x we have the expansion e xt e t n= B n x n! r= t n t, 2 where the B n x are the Bernoulli polynomials n n B n x = B r x n r 22 r B x =, B x = x 2, B 2x = x 2 x+ 6,.... We deduce that ζs,a has a meromorphic continuation in s with a simple pole of residue independent of a at s =, and, generalizing 8, that its values at non-positive integers are given by ζs, n = B n+a n =,, 2, n+ 6

7 3. Application: the Casimir effect In the study of the Casimir effect, described in Chapter 6 of this book, one encounters the function defined by the series Fλ = 2π l,m,n Z l2 +m 2 +λ 2 n 2, 24 where λ is a positive real variable; the factor 2π here has been included for later convenience. Of course the series is divergent. There are several natural questions: A. How can Fλ be defined rigorously? B. How can Fλ be computed effectively for a given λ >? C. How does Fλ behave asymptotically as λ and as λ? For the analysis of the Casimir effect, it is the asymptotics at λ which are important. Using the ideas explained in the previous two sections, we will show that the answers are as follows: A. For complex s with Rs > 3 2, define Zλ,s = l,m,n Z l 2 +m 2 +λ 2 n 2 s λ >, s C, Rs > the prime on the summation sign means that the term l,m,n =,, is to be omitted, a so-called Epstein zeta function. Then Zλ, s has a meromorphic continuation to all s, with a simple pole of residue 2π/λ at s = 3/2 and no other poles, and satisfies the functional equation where Z λ,s = λ Z λ, 3 2 s, 26 Z λ,s := π s ΓsZλ,s. 27 Then one makes sense of 24 by setting Fλ = Z λ, 2. Note that π/2 Γ 2 = 2π. B. The value of the function Fλ is given for any positive real number T by Fλ = T l,m,n γ 2 πtl 2 +m 2 +λ 2 n 2 + π λt 2 γ 2 T l2 +m 2 +λ 2 n 2, 28 l,m,n where the sums are taken over all triples l,m,n Z 3 and the functions γ x and γ 2x are 2 defined by the formulas γ x = e xt dt x 29 2 t 3/2 a variant of the error function and γ 2 x = x + x 2 e x if x >, 2 if x =, respectively. Since both γ 2 x and γ 2x are Oe x as x, formula 28 makes Fλ rapidly computable. More precisely, if we choose T = λ 2/3 then there are uniformly in λ only OM 3/2 7 3

8 terms in the two sums in 28 for which the arguments of γ 2 or γ 2 are M, so that a relatively small number of terms suffices to compute Fλ to high precision. Here are some sample values: t Ft Each of these numbers was computed independently using 28 for several different values of T. The fact that the answers agreed to the precision given, even though the individual terms of the sums are completely different, gives a high degree of confidence in the correctness of the theoretical and numerical calculations. Of course we could also use the functional equation 26 to give a much simpler convergent series expansion for Fλ than 28, namely Fλ = λ Z λ,2 = λ3 π 2 l,m,n λ 2 l 2 +λ 2 m 2 +n 2 2, 32 but this formula would be useful for practical purposes because of the slow convergence: summing over l, m, n N involves ON 3 terms and gives an error of the order of /N, so that we would need some terms to achieve even three digits of precision. C. The value of Fλ is given for small λ by and for large λ by Fλ = 2C 3 λ + π 3 λ + Oλe π/λ λ 33 Fλ = π2 45 λ3 + C + O λe πλ λ 34 where C = = L 42 with L 4 as in 9 is Catalan s constant and C = 2 π ζ3 2 L The values obtained for λ =. and λ = by retaining only the first two terms in equations 33 or 34, respectively, are F , F , extremely close to the exact values for these two numbers given above. Let us now prove each of these assertions. A. From 5 and the fact that the generalized Mellin transform of the constant function vanishes, we deduce that the function Z λ,s defined by 27 equals the Mellin transform ϕ λ s of the function ϕ λ t = e πl2 +m 2 +λ 2 n 2 t = ϑt 2 ϑλ 2 t λ, t >, l,m,n Z where ϑt is the theta series defined in 9. The functional equation 2 of ϑt implies the functional equation ϕ λ t = λ t 3/2 ϕ /λ t of ϕ λ t, and the meromorphic continuation, description of poles and funcctional equation 26 of Z s then follow as in Example 4 above. 8

9 B. Since the function ϕt = ϕ λ t equals to all orders in t as t and by virtue of its functionalequationequalsλ t 3/2 toallordersintast, thetwopiecesofthedecompositionr 2 of its Mellin transform are given by ϕ λ, T s = T ϕλ t t s dt Ts s 35 and ϕ λ, T s = T ϕλ t λ t 3/2 t s dt + λ T s 3 2 s 3 2 respectively. Using the functional equation of ϕ λ t, we see that these are exchanged when we replace λ, s and T by λ, 3 2 s and T, respectively again making the functional equation of Z λ,s = ϕ λ s evident, so we only have to study ϕ λ, T s. Substituting the definition of ϕ λ t into 35, we find ϕ λ, T s = T s γ s πtl 2 +m 2 +λ 2 n 2 Ts s, l,m,n where γ s x, essentially the incomplete gamma function, is defined for x > by γ s x = t s e xt dt x >. The extra term T s /s in this formula can be omitted if we drop the prime from the summation signs and define the value of γ s as /s which is indeed the limiting value of γ s x as x if Rs <. The final result, with this convention for γ s, is therefore Z λ, s = T s l,m,n γ s πtl 2 +m 2 +λ 2 n 2 + [ s 3 2 s, λ λ, T T ]. The special case s = 2 gives the expansion 28. C. The leading term in the expansion of Fλ when λ is very big or very small can be obtained from equation 32: if λ is small then the dominating terms in 32 are those with n =, so Fλ is asymptotically equal to π 2 λ l,m l2 + m 2 2 = 4π 2 ζ2l 4 2λ here we have used that 4 l,m l2 + m 2 s is the Dedekind zeta function of the field Qi, which factors as ζs times L 4 s, while for large λ the dominating terms are those with l = m = and we get Fλ 2π 2 ζ4λ 3. To get a more precise estimate, we use equation 28. Consider first the case λ, and choose T in 28 so that T, λ 2 T. The best choice will turn out to be T = λ. Then in view of the exponential decay γ s x = O s x e πx of γ s x when x, we find that the only terms in 28 which are not exponentially small are those with l,m =, in the first sum and those with n = in the second one. Hence Fλ = [f λ 2 T + O T T + [ λt 2 f 2 + O T l,m l,m ] e πtl2 +m 2 e πλ2 Tn 2 e πl2 +m 2 /T λ 2 T = f λ 2 T + T λt 2 f 2 + O T 9 n n λt 2 e πt + λe π/λ ] e πn2 /λ 2 T 2 T, 36

10 where f ε = n Z γ /2 πεn 2, f 2 ε = l,m Z γ 2 πεl 2 +m 2. From the definition 29 and the functional equation of ϑx we find for ε small since f ε = = ε + ε t 3/2 ϑεtdt = ε /ε ϑx dx = 2 ε ϑt dt t 3/2 = ε /ε ϑx dx ϑx dx = ε + ε π 3 + O e π/ε n= e πn2x dx = 2 π n= n 2 = π 3. Similarly, noting the exceptional case x = in the definition 3 of γ 2 x, we find f 2 ε + 2 = t ϑεt 2 dt = ε 2 xϑx 2 dx ε = ε ε 2 xϑx 2 dx x+oe π/x dx where this time we have used x ϑx 2 dx = = 2C 3 ε 2 ε Oe π/ε, l,m Inserting these formulas into 36, we find Fλ = T [ + λt 2 xe πl2 +m 2 x dx = π 2 λ T + π 3 λ T + Oλ T e π/λ [ ] 2C 3 T2 T + Oe πt l,m ] 2T = 2C 3 λ 2 + π 3 λ + O λt 2 e πt + λe π/λ 2 T, l 2 +m 2 2 = 4 π 2 ζ2l O λt 2 e πt + λe π/λ. Note how the two non-exponentially small terms which depend on T cancel, as they have to. Taking T = /λ gives 33. The proof of 34 is completely analogous. From 28 and the exponential decay of γ s x we get Fλ = f 3 T + T λt 2 f 4 λ 2 T + O λ 2 e πλ T5/2 2T + e π/t T/2 as λ, where T with λ 2 T and where f 3 ε are f 4 ε are defined by f 3 ε = /2 πεl l,m Zγ 2 +m 2, f 4 ε = n Z γ 2 πεn 2. 2 T

11 This time we find the expansions f 3 ε = 2 3 ε + C ε /2 + Oe π/ε, f 4 ε = π2 45 ε ε /2 + O εe π/ε forεsmall, thenon-exponentially smalltermswhich dependont againcancel, andtakingt = /λ to make the error as small as possible we obtain 34. The details are left to the reader. As a final comment, we mention that the constants C and C occurring in 33 and 34 can be evaluated rapidly by the same method as in B, with ϑt 2 instead of ϑt 2 ϑλt, using the fact that the Dirichlet series 4ζsL 4 s is the Mellin transform of ϑt 2. Their numeical values are C = , C = Asymptotics of sums of the form fnt In this section we describe an extremely useful, and not sufficiently well known, asymptotic formula for functions given by expansions of the form gt = ft + f2t + f3t +, 37 where f :, C is a smooth function of sufficiently rapid decay to ensure the convergence of the series say ft = Ot ε as t and having a known asymptotic expansion at t =. In the simplest situation, we assume that f has a power series expansion which may be only asymptotic rather than convergent, i.e. f need only be differentiable rather than analytic at the origin ft b n t n t. 38 n= First, let us try to guess what the answer should be. On the one hand we can argue à la Euler, simply substituting the expansion 38 into 37 and interchanging the order of summation, without worrying about convergence problems. This gives gt b n mt n = m=n= b n m t n n = n= m= b n ζ nt n. 39 n= Of course this calculation is meaningless, since not only is the interchange of summation not permitted, but each of the interior sums m mn is divergent. Nevertheless, we know from Section 2, and Euler knew non-rigorously in 749, that the numbers ζ n do make sense and are certain rational numbers given by equation 8, so that at least the final expression in 39 makes sense as a formal power series. Alternatively, we can proceed à la Riemann and consider 37 for t small as /t times an approximation to the integral I f = ftdt. 4 This suggests instead that the correct asymptotic expansion of gt near should be given by gt I f t t, 4

12 and indeed this formula is true, by the very definition of Riemann integrals as limits of sums, if we interpret the symbol of asymptotic equality in its weak sense, as saying simply that the ratio of the expressions on its left and right tends to as t. However, we are using in its strong sense, where gt λ a λt λ means that the difference between gt and the finite sum λ<c a λt λ is Ot C as t for any value of C, no matter how large. In this stronger sense, neither 39 nor 4 gives the correct asymptotic development of g. Remarkably enough, however, their sum does give the right answer: Proposition 3. Let f be a C function on the positive real line which has the asymptotic development 38 at the origin and, rogether with all its derivatives, is of rapid decay at infinity. Then the function gt defined by 37 has the asymptotic development gt I f t as t, where I f is defined by equation 4. + n= b n B n+ n+ tn 42 Remark. The relationship of Proposition 3 to the discussion of the Mellin transform in 3 is very easy to describe. Suppose for simplicity that ft is of rapid decay as t. Then 38 implies that the Mellin transform fs, defined for Rs > as ftt s dt, has a meromorphic continuation to all s with simple poles of reside b n at s = n n =,, 2,... and with no other singularities. But the transformation rules 2 of the Mellin transform imply that the Mellin transform gs of gt equals ζs fs, so it has at most simple poles of residue f = I f at s = and residue ζ nb n at s = n n =,, 2,... Since gt is small at infinity, it follows that, if gt has an asymptotic expansion as t of the form j a jt α j or even j a jt α j logt m j, then this expansion is necessarily given by eq. 42. Proof. We begin by describing the Euler-Maclaurin summation formula. To state it, we need the Bernoulli polynomials B n x, which can be described by the generating function 22, by the explicit formula 23 in terms of Bernoulli numbers, or, most beautiful, by the property that a+ a B n xdx = a n for every a. It is easy to see that there is only one polynomial B n x with this property for each n. From any of these definitions we can deduce without difficulty that B nx = nb n x and that B n x+ B n x = nx n. Now let f be a smooth function on the positive reals. Integration by parts and the fact that B = 2 = B but B n+ = B n+ = B n+ for n 2 gives { 2 [f+f] if n =, f n x B nx n! dx = and hence, by induction on N, fxdx = N 2 [f+f] + f n+ x B n+x n+! dx + n= n B n+ n+! B n+ n+! [fn f n ] if n [f n f n ] + N f N x B Nx N! for every integer N. Replacing fx by fx+m and summing over m =,2,...,M gives M fxdx = f 2 + M m= fm+ fm 2 + N n= n B n+ [f n M f n ] n+! M + N f N x B N x dx 43 N! 2 dx

13 where B N x = B N x [x]. This is the Euler-Maclaurin summation formula. Nowassumethatf andeachofitsderivativesisofrapiddecayatinfinity, sothat f N x dx converges. Since B N x is periodic and hence bounded, we can let M in 43 to get fm = m= fxdx + N n= n B n+ n+! f n N Replacing fx by ftx and then x by x/t with t > changes this formula to fmt = t m= fxdx + N n= n B n+ n+! f n t n + t N f N x B N x N! dx. f N x B N x/t N! The last integral is bounded as t with N fixed for the same reason as before, so the final term is O N t N. Substituting f n = n!b n from 38, we get the desired asymptotic formula 42. Before giving examples of Proposition 3, we mention three extensions to more general sums.. First, instead of 37 we can look at shifted sums of the form gt = m=fm+at, where a >. Here the Riemann Ansatz and the Euler Ansatz predict I f t and n= b nζ n,at n for the asymptotic expansion of g, and again the correct answer is the sum of these two: m= fm+at I f t + n= dx. b n B n+ a n+ tn 44 cf. equation 23. The proof is similar to that of Proposition 3 and will be omitted. By taking rational values of a in 44 and forming suitable linear combinations, we can use this formula to give the asymptotic development of m χmfmt as t for any periodic function χm, such as a Dirichlet character. 2. Next, we can allow non-integral exponents of t in 38. If the expansion of ft at the origin contains terms b λ t λ with arbitrary real numbers λ > or complex numbers with real part greater than, then the formula for g need only be modified by adding the corresponding terms b λ ζ λt λ. Terms with λ < are not interesting since they can simply be subtracted from ft, since the sum m b mmtλ then converges absolutely. The limiting case λ = is of interest because it occurs in various applications. Here the answer proved most easily by taking one function, like t e t, which has a /t singularity at the origin and for which fmt can be computed exactly is ft b λ t λ fmt b log t t + I f + b λ ζ λ t λ, 45 λ m= where I f = ft b e t /tdt. 3. Finally, we can also allow terms of the form t λ logt n in the expansion of ft, the corresponding contribution to gt being simply the nth derivative with respect to λ of ζ λt λ, e.g. a term t λ log t in the expansion of ft at leads to a term tλ ζ λlog t +ζ λ in the expansion λ> at of gt. In particular, using the known value ζ = 2 log2π we find ft b log t + b n t n n= fmt I f t 3 m= b 2 log 2π t + b n ζ nt n. 46 n=

14 We end by giving four examples two easy and two harder to illustrate how these asymptotic formulas work. Example. Take ft = e λt with λ >. This function is smooth, small at infinity, and has an expansion 38 at t = with b n = λ n /n!. The integral I f equals /λ. Hence 42 gives gt λt + B n+ n= n+! λtn as the asymptotic expansion of gt = fmt = m= e λt, in accordance with the definition 6 of the Bernoulli numbers. Example 2. Now take ft = e λt2 with λ >. This function is again smooth and small at infinity, and has an expansion 38 at t = with b 2n = λ n /n! and b n = for n odd. Since all Bernoulli numbers with odd indices > vanish, the asymptotic expansion in 42 breaks off after two terms, and we find gt = I f /t /2 + Ot N for all N, with I f = π/4λ. In this case, of course, we know much more, because gt is simply 2 ϑλt 2 /π with ϑt as in eq. 9, and therefore equation 2 gives the much more precise statement gt = I f /t /2+O t e π2 /λt 2. The same applies to any function ft whose expansion at t = has only even powers of t. For such a function, the expansion 42 collapses to gt I f /t b /2, but this is always just a weakening of the Poisson summation formula 3, because we can extend ft by ft = f t to a smooth even function on the real line and use 4 to get the exact formula 2gt+f = n Z fnt = t n ζ fnt = 2I f t + 2 t fnt, n= and the smoothness of f implies that the function fy decays at infinity more rapidly than any negative power of y. The right way to think of Proposition 3 is therefore as a replacement for the Poisson summation formula when one is confronted with a sum over only positive integers rather than a sum over all of Z. Such sums are very much harder to study than sums over all integers just think of the special values of the Riemann zeta function, where the numbers ζ2k can be obtained in closed form because they can be written as 2 n Z n k, while the numbers ζ2k +, which cannot be reduced to sums over all of Z in this way, are not known exactly. Example 3. As our next example, we consider the function g k q = σ k nq n < q <, n= where k is an integer greater than and σ k n denotes the sum of the k st powers of the divisors of a natural number n. In the theory of modular forms, it is shown that if k is even and larger than 2, then g k satisfies the functional equation B k 2k + g k e 2πt = k/2 t k B k 2k + g k e 2π/t k = 4, 6, 8,...; t >. In particular, g k e 2πt has the terminating asymptotic expansion g k e 2πt = k/2b k 2k t k + B k 2k + OtN N > 47 as t in these cases. Let us see how Proposition 3 permits us to recover this asymptotic formula using without knowing the modularity, and at the same time tells us why 47 fails for k = 2 or k odd and what replaces it in those cases. 4

15 We first note that g k q = m q k n = n= m n m k q m +q 2m + = m= m= m k q m q m and hence that g k can be written after a change of variables in the form g k e t = t k f k mt, m= f k t = tk e t. The function f = f k satisfies the hypotheses of Proposition 3, with integral I f = t k e t dt = and with Taylor expansion ft = r= t k e t +e 2t + dt = k!ζk B r r! tr+k 2 at zero. Proposition 3 therefore gives g k e t k!ζk t k + r+k B r r! r= B r+k r +k tr t. 48 If k is even and 4, then all products B r B r+k with r vanish, since r and r +k have opposite parity and all odd-index Bernoulli numbers except B are zero. Thereofre only two terms of 48 survive, and replacing t by 2πt and using the well-known formula for ζk in terms of B k, we recover 47. If k = 2, then the argument is the same except that now the r = term also gives a non-zero contribution, so that we find g 2 e 2πt = 24t 2 4πt OtN N >, instead of 47, in accordance with the known near-modularity property of g 2. Finally, if k is odd then we still get an explicit asymptotic formula with rational coefficients, but it now no longer terminates. Thus for k = 3 we find the expansion g 3 e t 2ζ3 t 3 2t + t 44 + t t t as t, even though g k has no modularity properties in this case. We leave it as an exercise to the reader to calculate the corresponding expansion when k =, where one has to use equation 45 instead of equation 42. Example 4. As our final example, consider the function Pq = m= q m q <, the generating power series of the partition function. To study the behavior of the partition function, we need to know how Pq blows up as q approaches from below or, more generally, any root of unity from within the unit circle. Here again the known modularity properties of the function Pq imply the non-trivial functional equation e πt/2 P e 2πt = te π/2t P e 2π/t t >, 5

16 from which one immediately obtains the asymptotic expansion logp e 2πt = π 2t + 2 logt π 2 t + OtN N >. 49 To obtain this formula without knowing anything about the modularity of P, we observe that logp e t has the form m= fmt with ft = log e t. This function is small at infinity, has integral I f = ζ2 as one sees by integrating by parts once and then calculating as in Example 2, and has an asymptotic development ft log t n= B n n n! tn, as one sees easily by differentiating once. Hence equation 47 applies and gives logp e t ζ2 t + 2 log t 2π n B n n n! n= B n+ n+ tn. Again all terms except for the first in the sum on the right vanish because n and n+ have opposite parity, and replacing t by 2πt and using ζ2 = π 2 /6 we recover 49. The same method using 44 with rational values of a lets us compute the exact asymptotics of Pq as q approaches any root of unity, recovering precisely the same result as that given by the modularity. Moreover, just as in the case of Example 2 for k odd, the method applies even when modularity fails. For example, if we define P 2 q = q m m q <, m= a generating function that occurs in connection with the theory of plane partitions, then an analysis like the one just given for Pq, but now with ft = tlog e t, produces the complete asymptotic expansion P 2 e t = ct /2 e ζ3/t2 t t t with c = e ζ = , and using 46 one can get the expansion of P 2 αe t for any root of unity α. Furthermore, for reasons similar to those which applied to g k with k even, one finds that the corresponding expansions for the logarithm of P 3 q = q n n2 when q tends to a root of unity are terminating, even though there are no modularity properties in this case. Warning. Equations numbering have to be checked, e.g., two eqs. 5, one now changed to