Math212a1406 The Fourier Transform The Laplace transform The spectral theorem for bounded self-adjoint operators, functional calc

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1 Math212a1406 The Fourier Transform The Laplace transform The spectral theorem for bounded self-adjoint operators, functional calculus form The Mellin Transform September 18, 2014

2 1 Conventions, especially about. 2 Basic facts about the Fourier transform acting on S. 3 The Fourier transform on L 2. 4 Sampling. 5 The Heisenberg Uncertainty Principle. 6 Tempered distributions. Examples of Fourier transforms of elements of S. 7 The Laplace transform. 8 The spectral theorem for bounded self-adjoint operators, functional calculus form. 9 The Mellin trransform Dirichlet series and their special values

3 The space S. The space S consists of all functions on which are infinitely differentiable and vanish at infinity rapidly with all their derivatives in the sense that f m,n := sup{ x m f (n) (x) } <. x The m,n give a family of semi-norms on S making S into a Frechet space - that is, a vector space space whose topology is determined by a countable family of semi-norms.

4 The measure on. We use the measure 1 dx on and so define the Fourier transform of an element of S by ˆf (ξ) := 1 f (x)e ixξ dx and the convolution of two elements of S by (f g)(x) := 1 f (x t)g(t)dt.

5 1 We are allowed to differentiate f (x)e ixξ dx with respect to ξ under the integral sign since f (x) vanishes so rapidly at. We get ( ) d 1 f (x)e ixξ dx = 1 ( ix)f (x)e ixξ dx. dξ

6 1 We are allowed to differentiate f (x)e ixξ dx with respect to ξ under the integral sign since f (x) vanishes so rapidly at. We get ( ) d 1 f (x)e ixξ dx = 1 ( ix)f (x)e ixξ dx. dξ So the Fourier transform of ( ix)f (x) is d dξ ˆf (ξ).

7 1 We are allowed to differentiate f (x)e ixξ dx with respect to ξ under the integral sign since f (x) vanishes so rapidly at. We get ( ) d 1 f (x)e ixξ dx = 1 ( ix)f (x)e ixξ dx. dξ So the Fourier transform of ( ix)f (x) is d dξ ˆf (ξ). Integration by parts (with vanishing values at the end points) gives 1 f (x)e ixξ 1 dx = (iξ) f (x)e ixξ dx. So the Fourier transform of f is (iξ)ˆf (ξ).

8 The Fourier transform maps S to S. Putting these two facts together gives The Fourier transform is well defined on S and [( ) d m ( ) d n (( ix) n f )] ˆ= (iξ) m ˆf, dx dξ as follows by differentiation under the integral sign and by integration by parts. This shows that the Fourier transform maps S to S.

9 Convolution goes to multiplication. so (f g)ˆ(ξ) = 1 = 1 = 1 f (x t)g(t)dte ixξ dx f (u)g(t)e i(u+t)ξ dudt f (u)e iuξ du 1 (f g)ˆ= ˆf ĝ. g(t)e itξ dt

10 Scaling. For any f S and a > 0 define S a f by (S a )f (x) := f (ax). Then setting u = ax so dx = (1/a)du we have 1 (S a f )ˆ(ξ) = f (ax)e ixξ dx 1 = (1/a)f (u)e iu(ξ/a) du so (S a f )ˆ= (1/a)S 1/aˆf.

11 Fourier transform of a Gaussian is a Gaussian. The polar coordinate trick evaluates 1 e x2 /2 dx = 1. The integral 1 e x2 /2 xη dx converges for all complex values of η, uniformly in any compact region. Hence it defines an analytic function of η that can be evaluated by taking η to be real and then using analytic continuation.

12 The Fourier transform of the unit Gaussian is the unit Gaussian. For real η we complete the square and make a change of variables: 1 e x2 /2 xη 1 dx = e (x+η)2 /2+η 2 /2 dx = e η2 /2 1 e (x+η)2 /2 dx Setting η = iξ gives = e η2 /2. ˆn = n if n(x) := e x2 /2. We will make much use of this equation over the next few slides.

13 Scaling the unit Gaussian. ˆn = n if n(x) := e x2 /2. If we set a = ɛ in our scaling equation and define ρ ɛ := S ɛ n so ρ ɛ (x) = e ɛ2 x 2 /2, then (ρ ɛ )ˆ(x) = 1 ɛ e x2 /2ɛ 2.

14 (ρ ɛ )ˆ(x) = 1 ɛ e x2 /2ɛ 2. Notice that for any g S we have (by a change of varialbes) (1/a)(S 1/a g)(ξ)dξ = g(ξ)dξ so setting a = ɛ we conclude that 1 (ρ ɛ )ˆ(ξ)dξ = 1 for all ɛ.

15 Let and Then so 1 (ψ ɛ g)(ξ) g(ξ) = 1 = 1 (ρ ɛ )ˆ(ξ)dξ = 1 ψ := ψ 1 := (ρ 1 )ˆ ψ ɛ := (ρ ɛ )ˆ. ψ ɛ (η) = 1 ɛ ψ ( η ɛ ) [g(ξ η) g(ξ)] 1 ɛ ψ ( η ɛ [g(ξ ɛζ) g(ξ)]ψ(ζ)dζ. ) dη =

16 (ψ ɛ g)(ξ) g(ξ) = 1 = 1 [g(ξ η) g(ξ)] 1 ɛ ψ ( η ɛ [g(ξ ɛζ) g(ξ)]ψ(ζ)dζ. ) dη = Since g S it is uniformly continuous on, so that for any δ > 0 we can find ɛ 0 so that the above integral is less than δ in absolute value for all 0 < ɛ < ɛ 0. In short, ψ ɛ g g 0, as ɛ 0.

17 The multiplication formula. This says that ˆf (x)g(x)dx = f (x)ĝ(x)dx for any f, g S. Indeed the left hand side equals 1 f (y)e ixy dyg(x)dx. We can write this integral as a double integral and then interchange the order of integration which gives the right hand side.

18 The inversion formula. This says that for any f S f (x) = 1 ˆf (ξ)e ixξ dξ. To prove this, we first observe that for any h S the Fourier transform of x e iηx h(x) is just ξ ĥ(ξ η) as follows directly from the definition. Taking g(x) = e itx e ɛ2 x 2 /2 in the multiplication formula gives 1 ˆf (t)e itx e ɛ2 t 2 /2 dt = 1 f (t)ψ ɛ (t x)dt = (f ψ ɛ )(x).

19 1 ˆf (t)e itx e ɛ2 t 2 /2 dt = 1 f (t)ψ ɛ (t x)dt = (f ψ ɛ )(x). We know that the right hand side approaches f (x) as ɛ 0. Also, e ɛ2 t 2 /2 1 for each fixed t, and in fact uniformly on any bounded t interval. Furthermore, 0 < e ɛ2 t 2 /2 1 for all t. So choosing the interval of integration large enough, we can take the 1 left hand side as close as we like to ˆf (x)e ixt dt by then choosing ɛ sufficiently small.

20 Plancherel s theorem. Let f (x) := f ( x). Then the Fourier transform of f is given by 1 f ( x)e ixξ dx = 1 f (u)e iuξ du = ˆf (ξ) so Thus ( f )ˆ= ˆf. (f f )ˆ= ˆf 2. The inversion formula applied to f f and evaluated at 0 gives (f f )(0) = 1 ˆf 2 dx.

21 (f f )(0) = 1 ˆf 2 dx. The left hand side of this equation is 1 f (x) f (0 x)dx = 1 Thus we have proved Plancherel s formula 1 f (x) 2 dx = 1 f (x) 2 dx. ˆf (x) 2 dx.

22 Extending the Fourier transform to L 2. 1 f (x) 2 dx = 1 ˆf (x) 2 dx. Define L 2 () to be the completion of S with respect to the L 2 norm given by the left hand side of the above equation. Since S is dense in L 2 () we conclude that the Fourier transform extends to unitary isomorphism of L 2 () onto itself.

23 The Poisson summation formula. This says that for any g S we have k Z g(k) = 1 m Z ĝ(m).

24 Proof. Let h(x) := k g(x + k) so h is a smooth function, periodic of period and h(0) = k g(k). Expand h into a Fourier series a m = 1 0 h(x)e imx dx = 1 Setting x = 0 in the Fourier expansion h(x) = m a me imx where h(x) = 1 ĝ(m)e imx g(x)e imx dx = 1 ĝ(m). gives h(0) = 1 ĝ(m). m

25 The Shannon sampling theorem. Let f S be such that its Fourier transform is supported in the interval [ π, π]. Then a knowledge of f (n) for all n Z determines f. This theorem is the basis for all digital sampling used in information technology. More explicitly, f (t) = 1 π n= f (n) sin π(n t). (1) n t Proof. Let g be the periodic function (of period ) which extends ˆf, the Fourier transform of f. So and is periodic. g(τ) = ˆf (τ), τ [ π, π]

26 Expand g into a Fourier series: g = n Z c n e inτ, where c n = 1 π g(τ)e inτ dτ = 1 ˆf (τ)e inτ dτ, π or, by the Fourier inversion formula, c n = 1 f ( n). () 1 2 But f (t) = 1 () () 1 2 π π π ˆf (τ)e itτ dτ = 1 g(τ)e itτ dτ = () 1 2 π 1 f ( n)e i(n+t)τ dτ. () 1 2

27 f (t) = 1 () 1 2 π ˆf (τ)e itτ dτ = 1 n () 1 2 π π g(τ)e itτ dτ = 1 1 f ( n)e i(n+t)τ dτ. () 1 2 π () 1 2 eplacing n by n in the sum, and interchanging summation and integration, which is legitimate since the f (n) decrease very fast, this becomes f (t) = 1 π f (n) e i(t n)τ dτ. π But π π e i(t n)τ dτ = ei(t n)τ i(t n) π π = ei(t n)π e i(t n)π i(t n) = 2 sin π(n t). n t

28 escaling the Shannon sampling theorem. It is useful to reformulate this via rescaling so that the interval [ π, π] is replaced by an arbitrary interval symmetric about the origin: In the engineering literature the frequency λ is defined by ξ = λ. Suppose we want to apply (1) to g = S a f. We know that the Fourier transform of g is (1/a)S 1/aˆf and supp S 1/aˆf = a supp ˆf. So if supp ˆf [ λ c, λ c ] we want to choose a so that aλ c π or

29 For a in this range (1) says that or setting t = ax, f (ax) = 1 π f (t) = n= a 1 2λ c. (2) sin π(x n) f (na), x n f (na) sin( π a (t na) π a (t na). (3)

30 The Nyquist rate. This holds in L 2 under the assumption that f satisfies supp ˆf [ λ c, λ c ]. We say that f has finite bandwidth or is bandlimited with bandlimit λ c. The critical value a c = 1/2λ c is known as the Nyquist sampling interval and (1/a) = 2λ c is known as the Nyquist sampling rate. Thus the Shannon sampling theorem says that a band-limited signal can be recovered completely from a set of samples taken at a rate the Nyquist sampling rate.

31 The Heisenberg Uncertainty Principle. Let f S() with f (x) 2 dx = 1. We can think of x f (x) 2 as a probability density on the line. The mean of this probability density is x m := x f (x) 2 dx. If we take the Fourier transform, then Plancherel says that ˆf (ξ) 2 dξ = 1 as well, so it defines a probability density with mean ξ m := ξ ˆf (ξ) 2 dξ.

32 The Heisenberg Uncertainty Principle. Suppose for the moment that these means both vanish. The Heisenberg Uncertainty Principle says that ( ) ( ) xf (x) 2 dx ξˆf (ξ) 2 dξ 1 4. In other words, if Var(f ) denotes the variance of the probability density f 2 with similar notation for ˆf then Var(f ) Var(ˆf ) 1 4.

33 Proof. Write iξf (ξ) as the Fourier transform of f and use Plancherel to write the second integral as f (x) 2 dx. Then the Cauchy - Schwarz inequality says that the left hand side is the square of xf (x)f (x) dx e(xf (x)f (x))dx = 1 2 x(f (x)f (x) + f (x)f (x))dx = 1 2 x d dx f 2 dx = 1 2 f 2 dx = 1 2.

34 The general case. If f has norm one but the mean of the probability density f 2 is not necessarily zero (and similarly for for its Fourier transform) the Heisenberg uncertainty principle says that ( ) ( ) (x x m )f (x) 2 dx (ξ ξ m )ˆf (ξ) 2 dξ 1 4. The general case is reduced to the special case by replacing f (x) by f (x + x m )e iξmx.

35 The topology on S. The space S was defined to be the collection of all smooth functions on such that f m,n := sup{ x m f (n) (x) } <. x The collection of these norms define a topology on S which is much finer that the L 2 topology: We declare that a sequence of functions {f k } approaches g S if and only if f k g m,n 0 for every m and n. A linear function on S which is continuous with respect to this topology is called a tempered distribution.

36 The space of tempered distributions is denoted by S. For example, every element f S defines a linear function on S by φ φ, f = 1 φ(x)f (x)dx. But this last expression makes sense for any element f L 2 (), or for any piecewise continuous function f which grows at infinity no faster than any polynomial. For example, if f 1, the linear function associated to f assigns to φ the value 1 φ(x)dx. This is clearly continuous with respect to the topology of S but this function of φ does not make sense for a general element φ of L 2 ().

37 The Dirac delta function. Another example of an element of S is the Dirac δ-function which assigns to φ S its value at 0. This is an element of S but makes no sense when evaluated on a general element of L 2 ().

38 Defining the Fourier transform of a tempered distribution. If f S, then the Plancherel formula formula implies that its Fourier transform F(f ) = ˆf satisfies (φ, f ) = (F(φ), F(f )). But we can now use this equation to define the Fourier transform of an arbitrary element of S : If l S we define F(l) to be the linear function F(l)(ψ) := l(f 1 (ψ)).

39 Examples of Fourier transforms of elements of S. The Fourier transform of the constant 1. If l corresponds to the function f 1, then F(l)(ψ) = 1 (F 1 ψ)(ξ)dξ = F ( F 1 ψ ) (0) = ψ(0). So the Fourier transform of the function which is identically one is the Dirac δ-function.

40 Examples of Fourier transforms of elements of S. The Fourier transform of the δ function. If δ denotes the Dirac δ-function, then (F(δ)(ψ) = δ(f 1 (ψ)) = ( (F 1 (ψ) ) (0) = 1 ψ(x)dx. So the Fourier transform of the Dirac δ function is the function which is identically one.

41 Examples of Fourier transforms of elements of S. The Fourier transform of the δ function. If δ denotes the Dirac δ-function, then (F(δ)(ψ) = δ(f 1 (ψ)) = ( (F 1 (ψ) ) (0) = 1 ψ(x)dx. So the Fourier transform of the Dirac δ function is the function which is identically one.in fact, this last example follows from the preceding one: If m = F(l) then But (F(m)(φ) = m(f 1 (φ)) = l(f 1 (F 1 (φ)). F 2 (φ)(x) = φ( x). So if m = F(l) then F(m) = l where l(φ) := l(φ( )).

42 Examples of Fourier transforms of elements of S. The Fourier transform of the function x This assigns to every ψ S the value 1 ψ(ξ)e ixξ xdξdx = 1 1 dψ(ξ) i e ixξ dξdx = i dx ( dψ(ξ) = iδ dx d ( e ixξ) dξdx = dξ ψ(ξ) 1 i ( ( ( dψ(ξ) F (F 1 dx ). ))) (0)

43 Examples of Fourier transforms of elements of S. For an element of S we have dφ dx f dx = 1 φ df dx dx. So we define the derivative of an l S by ( dl dx (φ) = l dφ ). dx Thus the Fourier transform of x is i dδ dx.

44 Definition of the (one sided) Laplace transform The inversion problem. Let f be a (possibly vector valued) bounded piecewise differentiable function, so that the integral F (z) = 0 e zt f (t)dt converges for z with z > 0. F is called the Laplace transform of f. The inversion problem is to reconstruct f from F.

45 The Laplace transform as a Fourier transform. Let f be a bounded piecewise differentiable function defined on [0, ). Let c > 0, z = c + iξ and h the function given by { () 1 2 e h(t) = ct f (t) t 0 0 t < 0 Then h is integrable and ĥ(ξ) = 0 e zt f (t)dt = F (z), z = c + iξ.

46 ĥ(ξ) = 0 e zt f (t)dt = F (z), z = c + iξ. If the function F were integrable over the line Γ given by z = c, then the Fourier inversion formula would say that 1 e zx F (z)dz = 1 i ecx e ixξ ĥ(ξ)dξ = f (x) for x 0. Γ The condition that the function F be integrable over Γ, which is the same as the condition that ĥ be integrable over, would imply that h is continuous. But h will have a jump at 0 (if f (0) 0). So we need to be careful about the above formula expressing f in terms of F.

47 The philosophy we have been pushing up until now in today s lecture has been to pass to tempered distributions. But for applications that I have in mind (to the theory of semi-groups of operators) later on in this course, I need to go back to 19th century mathematics - more precisely to the analogue for the Fourier transform of Dirichlet s theorem about Fourier series that we proved in Lecture 2. ecall that Dirichlet proved the convergence of the symmetric Fourier sum n n a ke ikx to 1 2 (f (x +) + f (x )) under the assumptions that the periodic function f is piecewise differentiable. The analogue of the limit of the symmetric sum for the case of an integral is the Cauchy principal value : lim.

48 Fourier inversion à la Dirichlet. Theorem Let h L() be bounded and such that there is a finite number of real numbers a 1,... a r such that h is differentiable on (, a 1 ), (a 1, a 2 ),... (a r, ) with bounded derivative (and right and left handed derivatives at the endpoints). Then for any x we have 1 2 [h(x + ) + h(x )] = lim 1 () 1 2 e ixξ ĥ(ξ)dξ.

49 In the proof of this theorem we may take x = 0 by a shift of variables. So we want to evaluate the limit of 1 = 1 ĥ(ξ)dξ = 1 h(t) h(t) eit e it dt = 1 it π = 1 π 0 [h(t) + h( t)] e iξt dξdt h(t) sin t dt t sin t dt t where the interchange of the order of integration in the first equation is justified by the assumption that h is absolutely integrable.

50 The next fact that we will use is the evaluation of the Dirichlet integral sin t dt = π t 2. 0 There are many ways of establishing this classical result. For a proof using integration by parts see Wikipedia under Dirichlet integral. An alternative proof can be given via a contour integral. In fact, this evaluation will also be a consequence of what follows: it is clear that the integral converges, so if we let k denote the value and carry the k throughout the proof, we will find that k = π 2 since the above formula is a special case of our Laplace inversion formula for the Heaviside function.

51 The proof of the theorem will proceed by integration by parts: Let so s(y) := y sin t dt t s (y) = sin y y. Let H(x) := h(x) + h( x) so we are interested in evaluating the limit of v() := 0 H(u)s (u)du as. The function H is piecewise differentiable with a finite number of points of non-differentiablity where the right and left handed derivatives exist as in the theorem. We break the integral on the right up into the sum of the integrals over intervals of differentiability.

52 For example, the last integral will contribute p Integration by parts gives q p H(u)s (u)du := lim q q p H(u)s (u)du. H(u)s (u)du = [H(q)s(q) H(p)s(p)]+ q p H (u)s(u)du. We are assuming that H and H are bounded. Since s(u) 0 as and s(q) 0 as q, the contribution of these terms tend to zero.

53 The same integration by parts argument applies to any interval of the form (a, b) where a 0. For the interval (0, c) we have c 0 H(u)s (u)du = H(c)s(c)+H(0)s(0)+ c The first and third terms tend to zero as before, and since s(0) = π 2 we are left with π 2 H(0) proving the theorem. 0 H (u)s(u)du.

54 Comment. The hypotheses on h in the theorem are far too strong. In fact, for for a scalar valued function h, all that need be assumed is that h is in L and is locally of bounded variation. Even weaker hypotheses work. See for example Widder, The Laplace Transform.

55 The Mellin inversion formula. In any event, we have proved the Mellin inversion formula f (x) = 1 e zx F (z)dz i where the left hand side is interpreted to mean 1 2 (f (x + ) + f (x )) and the contour integral on the right is interpreted as a Cauchy principal value. Γ

56 The exponential series for a bounded operator Suppose that B is a bounded operator on a Banach space B. For example, any linear operator on a finite dimensional space. Then the series e tb t k = k! Bk converges for any t. (We will concentrate on t real, and eventually on t 0 when we get to more general cases.) Convergence is guaranteed as a result of the convergence of the usual exponential series in one variable. (There are serious problems with this definition from the point of view of numerical implementation which we will not discuss here.) 0

57 The standard proof using the binomial formula shows that e (s+t)b = e sb e tb. Also, the standard proof for the usual exponential series shows that the operator valued function t e tb is differentiable (in the uniform topology) and that d ( e tb) = B e tb = e tb B. dt

58 The unitary group associated with a bounded self-adjoint operator In particular, let A be a bounded self-adjoint operator on a Hilbert space H and take B = ia. So let U(t) := e ita. Then t U(t) defines a one parameter group of bounded transformations on H. Furthermore, d dt (U(t)U(t) ) = U(t)(iA + (ia) )U(t) = 0, and since U(0)U(0) = I we conclude that U(t)U(t) I. In words: the operators U(t) are unitary.

59 The functional calculus for functions in S. ecall the Fourier inversion formula for functions f S which says that f (x) = 1 ˆf (t)e itx dt. If we replace x by A and write U(t) instead of e ita this suggests that we define f (A) := 1 ˆf (t)u(t)dt. (4) We want to check that this assignment f f (A) has the properties that we would expect from a functional calculus. This map is clearly linear in f since the Fourier transform is. We now check that it is multiplicative:

60 Checking that (fg)(a) = f (A)g(A). To check this we use fact that the Fourier transform takes multiplication into convolution, i.e. that (fg)ˆ= ˆf ĝ so (fg)(a) = 1 ˆf (t s)ĝ(s)u(t)dsdt = 1 ˆf (r)ĝ(s)u(r + s)drds = 1 ˆf (r)ĝ(s)u(r)u(s)drds = f (A)g(A).

61 Checking that the map f f (A) sends f (f (A)). For the standard Fourier transform we know that the Fourier transform of f is given by ˆ f (ξ) = ˆf ( ξ). Substituting this into the right hand side of (4) gives 1 ˆf ( t)u(t)dt = 1 ˆf ( t)u ( t)dt = ( 1 ) ˆf ( t)u( t)dt = (f (A)) by making the change of variables s = t.

62 Checking that f (A) f. Let f denote the sup norm of f, and let c > f. Define g by g(s) := c c 2 f (s) 2. So g is a real element of S and g 2 = c 2 2c so f f 2cg + g 2 = 0. So by our previous results, = 2cg f f c 2 f 2 + c 2 f 2 f (A) f (A) cg(a) cg(a) + g(a) g(a) = 0.

63 f (A) f (A) cg(a) cg(a) + g(a) g(a) = 0 i.e. f (A) f (A) + (c g(a)) (c g(a) = c 2. So for any v H we have f (A)v 2 f (A)v 2 + (c g(a))v 2 = c 2 v 2 proving that f (A) f. (5)

64 Enlarging the functional calculus to continuous functions vanishing at infinity. The inequality f (A) f (5) allows us to extend the functional calculus to all continuous functions vanishing at infinity. Indeed if ˆf is an element of L 1 so that its inverse Fourier transform f is continuous and vanishes at infinity (by iemann-lebesgue) we can approximate f in the norm by elements of S and so the formula (4) applies to f. We will denote the space of continuous functions vanishing at infinity by C 0 ().

65 Preview of coming attractions. We will be devoting eight or nine lectures to generalizing this result in two different ways: We will extend the result from bounded to unbounded self-adjoint operators. Of course this will require us to define unbounded self-adjoint operators. We will also greatly extend the class of functions to which the functional calculus is defined. For example, suppose that we have extended the calculus so as to include functions of the form 1 I where I is an interval on the real line. Since 1 I is real valued, we conclude that 1 I (A) = 1 I (A), i.e. it is self-adjoint. Since 1 2 I = 1 I, we conclude that 1 I (A) 2 = 1 I (A). In other words, 1 I (A) is a projection. We will examine the meaning of the image of this projection.

66 But our first item of business will be to try to understand more deeply the meaning of e tb and its relation to the resolvent. This will take at least three lectures.

67 Temporary change in notation The rest of this lecture is a bonus: A look at some applications of Fourier analysis to analytic number theory, specifically to the iemann zeta function. I will follow verbatim an article by Zagier which appeared in the book Quantum field theory by Zeidler. Many of the slides are photographic copies from the book. The number theorists use a different convention for the Fourier transform than the one that we have been using and will continue to use after this interruption. They define the Fourier transform as ˆf (y) = F(f )(y) = f (x)e ixy dx so that the Poisson summation formula has the more symmetrical looking form f (n) = ˆf (n). n Z n Z As we are following Zagier, we will use this convention for the rest

68 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 including the Casimir effect. 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. The Mellin transform Let ϕ :]0, [ C be a function on the positive real axis which is reasonably smooth (actually, continuous or even piecewise continuous would be enough) and decays rapidly at both 0 and, i.e.,thefunctiont φ(s) = φ(t)t s 1 A ϕ(t) isboundedon]0, [ for any A. Then the integral dt. (1) Outline Conventions, especially about. Basic facts about the Fourier transform acting on S. The Fourier transform on L 2. Sam 0Z ϕ(s) = ϕ(t)t s 1 dt (6.31) 0 This is the Mellin transform. If φ is a function on the positive real axis which is piecewise continuous and decays rapidly at both 0 converges and, for any thiscomplex integral value converges s and defines for any a holomorphic complex value function of sofand s called the Mellin transform of ϕ(t). The following small table, which α denotes a complex defines a holomorphic function of s. The following small table, in number and λ apositiverealnumbershows how ϕ(s) changeswhenϕ(t) ismodified in various whichsimple α denotes ways: a complex number and λ a positive real number, shows how φ changes when φ is modified in various simple ways: ϕ(λt) t α ϕ(t) ϕ(t λ ) ϕ(t 1 ) ϕ (t) λ s ϕ(s) ϕ(s + α) λ s ϕ(λ 1 s) ϕ( s) (1 s) ϕ(s 1) (6.32) (2) We extend the definition of the Mellin transform:

69 Extending the definition of the Mellin transform Let us start with the frequently occurring case where ϕ(t) is of rapid decay at infinity and is C at zero, i.e., it has an asymptotic expansion ϕ(t) n=0 a nt n as t 0. (ecall that this means that the difference ϕ(t) N 1 n=0 a nt n is O(t N ) as t 0 for any integer N 0; it is not required that the series a n t n be convergent for any positive t.) Then for s with (s) > 0 and any positive integer N the integral (1) converges and can be decomposed as follows: ϕ(s) = = ϕ(t) t s 1 dt + 1 ϕ(t) t s 1 dt ( N 1 ) ϕ(t) a n t n t s 1 dt + n=0 N 1 n=0 a n n + s + 1 ϕ(t) t s 1 dt. The first integral on the right converges in the larger half-plane (s) > N and the second for all s C, so we deduce that ϕ(s) has a meromorphic continuation to (s) > N with simple poles of residue a n at s = n (n =0,...,N 1) and no other singularities. Since this holds for every n, it follows that the Mellin transform ϕ(s) in fact has a meromorphic continuation to all of C with simple poles of residue a n at s = n (n =0, 1, 2,...) and no other poles. The same argument shows that, more generally, if ϕ(t) is of rapid decay at infinity and has an asymptotic expansion simple poles of residue a n at s = n, n = 0, 1, 2,... and no other poles. Math212a1406 The Fourier Transform The Laplace ϕ(t) transform The aj spectral t αj theorem (t for 0) bounded self-adjoint operators, functional (3) calc

70 1 Outline Conventions, especially ϕ(s) = about. ϕ(t) Basic t s 1 facts dt + about the ϕ(t) Fourier t s 1 transform dt acting on S. The Fourier transform on L 2. Sam ( N 1 ) N 1 = ϕ(t) a n t n t s 1 a n dt + 0 n + s + ϕ(t) t s 1 dt. n=0 n=0 1 Extending the definition of the Mellin transform, 2 The first integral on the right converges in the larger half-plane (s) > N and the second for all s C, so we deduce that ϕ(s) has a meromorphic continuation to (s) > N with simple poles of residue a n at s = n (n =0,...,N 1) and no other singularities. Since this holds for every n, The same argument shows that, more generally, if φ is of rapid decay at infinity and has an asymptotic expansion it follows that the Mellin transform ϕ(s) in fact has a meromorphic continuation to all of C with simple poles of residue a n at s = n (n =0, 1, 2,...) and no other poles. The same argument shows that, more generally, if ϕ(t) is of rapid decay at infinity and has an asymptotic expansion ϕ(t) a j t αj (t 0) (3) j=1 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 ϕ(s) defined by the integral (1) for (s) > min j (α j ) has a meromorphic extension to all of C with simple poles of residue a j at s = α j (j =1, 2,...) and no other poles. Yet more generally, we can allow terms of the form t α (log t) m with λ C and m Z 0 in the asymptotic expansion of ϕ(t) at t = 0 and each such term contributes a pole with principal part ( 1) m m!/(s + α) m+1 at s = α, because 1 0 tα+s 1 (log t) m dt =( / α) m 1 0 tα+s 1 dt =( 1) m m!/(α + s) m+1 for (s + α) > 0. By exactly the same considerations, or by replacing ϕ(t) byϕ(t 1 ), we find that if ϕ(t) is of rapid decay (faster than any power of t) as t 0 but has an asymptotic expansion of the form ϕ(t) b k t βk (t ) (4) Math212a1406 The Fourier Transform The Laplace transformk=1 The spectral theorem for bounded self-adjoint operators, functional calc

71 Outline Conventions, especially about. Basic facts about the Fourier transform acting on S. The Fourier transform on L 2. Sam ϕ(t) a j t αj (t 0) (3) Extending the definition of the Mellin transform, 3 j=1 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 ϕ(s) defined by the integral (1) for (s) > min j (α j ) has a meromorphic extension to all of C with simple poles of residue a j at s = α j (j =1, 2,...) and no other poles. Yet more generally, we can allow terms of the form t α (log t) m with λ C and m Z 0 in the asymptotic expansion of ϕ(t) at t = 0 and each such term contributes a pole with principal part ( 1) m m!/(s + α) m+1 at s = α, because 1 0 tα+s 1 (log t) m dt =( / α) m 1 0 tα+s 1 dt =( 1) m m!/(α + s) m+1 for (s + α) > 0. By exactly the same considerations, or by replacing ϕ(t) byϕ(t 1 ), we find that if ϕ(t) is of rapid decay (faster than any power of t) as t 0 but has an asymptotic expansion of the form ϕ(t) b k t βk (t ) (4) k=1 at infinity, where now the exponents β k are complex numbers whose real parts tend to, then the function ϕ(s), originally defined by (1) in a left half-plane (s) < max k (β 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 termsb k t βk (log t) nk in (3) which then produce poles with principal parts ( 1) nk+1 n k! b k /(s + β k ) nk+1 at s = β k.) Now we can use these ideas to define ϕ(s) for functions which are not small either at 0 or at, even when the integral (1) does not converge for any value of s. We simply assume that ϕ(t) is a smooth (or continuous) function on (0, ) 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 Math212a1406 assume The that Fourier thetransform numbering The slaplace such that transform (α The ) (α spectral ) theorem for and bounded (β ) self-adjoint (β ) operators,. Then, functional for calc

72 ϕ(t) b k t (t ) (4) Outline Conventions, especially about. Basic facts k=1 about the Fourier transform acting on S. The Fourier transform on L 2. Sam at infinity, where now the exponents β k are complex numbers whose real parts tend to, then the function ϕ(s), originally defined by (1) in a left half-plane (s) < max k (β 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 termsb k t βk (log t) nk in (3) which then produce poles with principal parts ( 1) nk+1 n k! b k /(s + β k ) nk+1 at s = β k.) Extending the definition of the Mellin transform, 4 Now we can use these ideas to define ϕ(s) for functions which are not small either at 0 or at, even when the integral (1) does not converge for any value of s. We simply assume that ϕ(t) is a smooth (or continuous) function on (0, ) 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 (α 1 ) (α 2 ) and (β 1 ) (β 2 ). Then, for 2 any T>0 (formerly we took T = 1, 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 0 T ϕ(t) t s 1 dt ϕ(t) t s 1 dt ( (s) > (α1 ) ), ( (s) < (β1 ) ). Just as before, we see that for each integer J 1 the function ϕ T (s) extends by the formula T ( J ) J ϕ T (s) = ϕ(t) a j t αj t s 1 a j dt + T s+αj s + α j 0 j=1 to the half-plane (s) > (α J+1 ) 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 = 1, 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.wenowdefine j=1

73 ϕ T (s) = ϕ(t) t dt (s) > (α 1 ), Outline Conventions, especially about. Basic 0facts about the Fourier transform acting on S. The Fourier transform on L 2. Sam ϕ T (s) = ϕ(t) t s 1 ( dt (s) < (β1 ) ). T Just as before, we see that for each integer J 1 the function ϕ Extending the definition of the Mellin T (s) extends by the formula T ( transform, 5 J ) J ϕ T (s) = ϕ(t) a j t αj t s 1 a j dt + T s+αj 0 s + α j=1 j=1 j to the half-plane (s) > (α J+1 ) 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 = 1, 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.wenowdefine ϕ(s) = ϕ T (s) + ϕ T (s). (4 1 2 ) This is a meromorphic function of s and is independent of the choice of T,sincetheeffectof changing T to T is simply to add the everywhere holomorphic function T T ϕ(t) ts 1 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 ϕ(t) 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 log t) at both zero and infinity, then we can define in a canonical way a Mellin transform ϕ(s) which is meromorphic in the entire s-plane and whose poles reflect directly the coefficients in the asymptotic expansions of ϕ(t). This definition is consistent with and has the same properties (2) as the original definition (1). We end this section by giving two simple examples, while Sections 2 and 3 will give further applications of the method. Example 1. Let ϕ(t) =t α, where α is a complex number. Then ϕ has an asymptotic expansion (3) at 0 with a single term α 1 = α, a 1 = 1, and an asymptotic expansion (4) at with a single term β 1 = α, b 1 = 1. We immediately find that ϕ T (s) = T s+α /(s + α) for (s + α) > 0 and ϕ T (s) = T s+α /(s + α) for (s + α) < 0, so that, although the original Mellin transform integral (1) does not converge for any value of s, the function ϕ(s) 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 ϕ(s) 0wheneverϕ(t) is a finite linear combination Math212a1406 of functions The Fourier of the Transform t α The log m Laplace t withtransform α C, The m spectral Z. theorem (These for arebounded exactlyself-adjoint the functions operators, whose functional calc

74 k k Outline Conventions, especially about. Basic facts about the Fourier transform acting on S. The Fourier transform on L 2. Sam ϕ(s) = ϕ T (s) + ϕ T (s). (4 1 2 ) This is a meromorphic function of s and is independent of the choice of T,sincetheeffectof changing T to T is simply to add the everywhere holomorphic function T ϕ(t) ts 1 dt to ϕ T (s) and subtract the same function from ϕ T (s), not affecting the sum of their analytic continuations. T Extending the definition of the Mellin transform, 6 In summary, if ϕ(t) 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 log t) at both zero and infinity, then we can define in a canonical way a Mellin transform ϕ(s) which is meromorphic in the entire s-plane and whose poles reflect directly the coefficients in the asymptotic expansions of ϕ(t). This definition is consistent with and has the same properties (2) as the original definition (1). We end this section by giving two simple examples, while Sections 2 and 3 will give further applications of the method. Example 1. Let ϕ(t) =t α, where α is a complex number. Then ϕ has an asymptotic expansion (3) at 0 with a single term α 1 = α, a 1 = 1, and an asymptotic expansion (4) at with a single term β 1 = α, b 1 = 1. We immediately find that ϕ T (s) = T s+α /(s + α) for (s + α) > 0 and ϕ T (s) = T s+α /(s + α) for (s + α) < 0, so that, although the original Mellin transform integral (1) does not converge for any value of s, the function ϕ(s) 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 ϕ(s) 0 whenever ϕ(t) is a finite linear combination of functions of the form t α log m t with α C, m Z 0. (These are exactly the functions whose images ϕ λ (t) =ϕ(λt) under the action of the multiplicative group + span a finite-dimensional space.) In particular, we see that the generalized Mellin transformation is no longer injective. Example 2. Let ϕ(t) =e t. Here the integral (1) converges for (s) > 0 and defines Euler s gamma-function Γ(s). From the fact that ϕ(t) is of rapid decay at infinity and has the asymptotic (here even convergent) expansion n=0 ( t)n /n! at zero, we deduce that Γ(s) = ϕ(s) has a meromorphic continuation to all s with a simple pole of residue ( 1) n /n! at s = n (n =0, 1,...) and no other poles. Of course, in this special case these well-known properties can also be deduced from the functional equation Γ(s + 1) = sγ(s) (proved for (s) > 0 by integration by parts in the integral defining Γ(s)), N applications of which gives the meromorphic extension Math212a1406 Γ(s) =s The 1 Fourier (s + 1) Transform 1 (s The + NLaplace 1) 1 transform Γ(s + The N) spectral of Γ(s) theorem to for half-plane bounded self-adjoint (s) > N. operators, functional calc

75 Outline Conventions, with and has especially the same aboutproperties. Basic facts (2) as about the the original Fourierdefinition transform acting (1). We on S. endthe this Fourier section transform by giving on L 2. Sam two simple examples, while Sections 2 and 3 will give further applications of the method. Example 1. Let ϕ(t) =t α, where α is a complex number. Then ϕ has an asymptotic expansion (3) at 0 with a single term α 1 = α, a 1 = 1, and an asymptotic expansion (4) at with a single term β 1 = α, b 1 = 1. We immediately find that ϕ T (s) = T s+α /(s + α) for (s + α) > 0 and Extending the definition of the Mellin transform, 7 - the Gamma function ϕ T (s) = T s+α /(s + α) for (s + α) < 0, so that, although the original Mellin transform integral (1) does not converge for any value of s, the function ϕ(s) 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 ϕ(s) 0wheneverϕ(t) is a finite linear combination of functions of the form t α log m t with α C, m Z 0. (These are exactly the functions whose images ϕ λ (t) =ϕ(λt) under the action of the multiplicative group + span a finite-dimensional space.) In particular, we see that the generalized Mellin transformation is no longer injective. Example 2. Let ϕ(t) =e t. Here the integral (1) converges for (s) > 0 and defines Euler s gamma-function Γ(s). From the fact that ϕ(t) is of rapid decay at infinity and has the asymptotic (here even convergent) expansion n=0 ( t)n /n! at zero, we deduce that Γ(s) = ϕ(s) has a meromorphic continuation to all s with a simple pole of residue ( 1) n /n! at s = n (n =0, 1,...) and no other poles. Of course, in this special case these well-known properties can also be deduced from the functional equation Γ(s + 1) = sγ(s) (proved for (s) > 0 by integration by parts in the integral defining Γ(s)), N applications of which gives the meromorphic extension Γ(s) =s 1 (s + 1) 1 (s + N 1) 1 Γ(s + N) of Γ(s) to the half-plane (s) > N. From the first of the properties listed in (2), we find the following formula, which we will use many times: ϕ(t) = e λt ϕ(s) = Γ(s) λ s (λ > 0). (5) 3

76 Dirichlet series and their special values We now look at functions φ for which the Mellin transform defined in Section 1 is related to a Dirichlet series. The key formula is (5), 2. Dirichlet series and their special values because it allows us to convert Dirichlet series into exponential series, which are much simpler. In this section we look at functions ϕ(t) for which the Mellin transform defined in Section 1 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 ϕ(t) fort>0byϕ(t) =1/(e t 1). This function is of rapid decay at infinity and has an asymptotic expansion (actually convergent for t<) 1 e t 1 = 1 = t + t2 2 + t3 6 + r=0 B r r! tr 1 (6) with certain rational coefficients B 0 = 1, B 1 = 1 2, B 2 = 1 6,... called Bernoulli numbers. From the results of Section 1 we know that the Mellin transform ϕ(s), originally defined for (s) > 1 by the integral (1), has a meromorphic continuation to all s with simple poles of residue B r /r! at s =1 r (r =0, 1, 2,...). On the other hand, since e t > 1 for t>0, we can expand ϕ(t) as a geometric series e t + e 2t + e 3t +, so (5) gives (first in the region of convergence) ϕ(s) = Γ(s)ζ(s), where 1 ζ(s) = m s ((s) > 1) (7) m=1 is the iemann zeta function. Since Γ(s), as we have seen is also meromorphic, with simple poles of residue ( 1) n /n! at non-positive integral arguments s = n and no other poles, and since

77 1 Outline Conventions, especially about. Basic facts about the Fourier transform acting on S. The Fourier transform on L 2. Sam e t 1 = 1 B r = t + t2 Dirichlet series and their special values 2 + t3 6 + r! tr 1 (6) r=0 with certain rational coefficients B Euler s calculation of the 0 = 1, B zeta 1 = function 1 2, B 2 = 1 6,... at called negative Bernoulli numbers. integers From the results of Section 1 we know that the Mellin transform ϕ(s), originally defined for (s) > 1 by the integral (1), has a meromorphic continuation to all s with simple poles of residue B r /r! at s =1 r (r =0, 1, 2,...). On the other hand, since e t > 1 for t>0, we can expand ϕ(t) as a geometric series e t + e 2t + e 3t +, so (5) gives (first in the region of convergence) ϕ(s) = Γ(s)ζ(s), where ζ(s) = 1 m s ((s) > 1) (7) m=1 is the iemann zeta function. Since Γ(s), as we have seen is also meromorphic, with simple poles of residue ( 1) n /n! at non-positive integral arguments s = n and no other poles, and since Γ(s) (as is well-known and easily proved) never vanishes, we deduce that ζ(s) has a meromorphic continuation to all s with a unique simple pole of residue 1/Γ(1) = 1 at s = 1 and that its values at non-positive integral arguments are rational numbers expressible in terms of the Bernoulli numbers: ζ( n) = ( 1) n B n+1 n +1 (n =0, 1, 2,...). (8) Example 4. To approach ζ(s) in another way, we choose for ϕ(t) thetheta function For n = 1 we find ζ( 1) = 1 which was our mysterious formula in Lecture ϑ(t) 1 = e πn2 = t 1 12 n= (t >0). (9) (The factor π in the exponent has been included for later convenience.) We can write this out as

78 is the iemann zeta function. Since Γ(s), as we have seen is also meromorphic, with simple poles Outline Conventions, of residue ( 1) especially n /n! about at non-positive. Basic facts integral about the arguments Fourier transforms = n actingand on S. no The other Fourier poles, transform and since on L 2. Sam Γ(s) (as is well-known and easily proved) never vanishes, we deduce that ζ(s) has a meromorphic Dirichlet series and their special values continuation to all s with a unique simple pole of residue 1/Γ(1) = 1 at s = 1 and that its values at non-positive integral arguments are rational numbers expressible in terms of the Bernoulli numbers: Enter the theta function ζ( n) = ( 1) n B n+1 n +1 (n =0, 1, 2,...). (8) Example 4. To approach ζ(s) in another way, we choose for ϕ(t) thetheta function ϑ(t) = n= e πn2 t (t >0). (9) (The factor π in the exponent has been included for later convenience.) We can write this out as ϑ(t) = 1 + 2e πt +2e 4πt +, (10) and since the generalized Mellin transform of the function 1 is identically 0 by Example 1, we deduce from (5) that ϕ(s) =2ζ (2s), where ζ (s) = π s/2 Γ(s/2) ζ(s). (11) To obtain the analytic properties of ζ(s) from the results of Section 1, we need the asymptotics of ϑ(t) at zero and infinity. They follow immediately from the following famous result, due to Jacobi: 4

79 Dirichlet series and their special values Jacobi s functional equation of the theta function Proposition 1. The function ϑ(t) satisfies the functional equation ϑ(t) = 1 ϑ (1 ) (t >0). (12) t t Proof. Formula (12) is a special case of the Poisson summation formula, which says that f(n) (13) n Z f(n) = n Z for any sufficiently well-behaved (i.e., smooth and small at infinity) function f : C, where f(y) = f(x) eixy dx is the Fourier transform of f. (To prove this, note that the function F (x) = n Z f(n+x) is periodic with period 1, so has a Fourier expansion F (x) = m Z c me imx with c m = 1 0 F (x)e imx dx = f( m). Now set x = 0.) Now consider the function f t (x) = e πtx2. Its Fourier transform is given by f t (y) = e πtx2 +ixy dx = e πy2 /t e πt(x+iy/t)2 dx = c t f 1/t (y), where c>0 is the constant c = dx. Applying (13) with f = f e πx2 t therefore gives ϑ(t) = ct 1/2 ϑ(1/t), and taking t = 1 in this formula gives c = 1 and proves equation (12). Now we find from (10) that ϑ(t) has the asymptotic expansions ϑ(t) =1+O(t N ) as t and Math212a1406 ϑ(t) The =t 1/2 Fourier +O(t Transform N ) asthe t Laplace 0, where transform N>0 Theis spectral arbitrary. theorem It follows for bounded fromself-adjoint the results operators, of Section functional 1 calc

80 f(y) = Outline Conventions, f(x) e dx is the Fourier transform of f. (To prove this, note that the function especially about. Basic facts about the Fourier transform acting on S. The Fourier transform on L 2. Sam F (x) = n Z f(n+x) is periodic with period 1, so has a Fourier expansion F (x) = m Z c me imx Dirichlet series withand c m their = 1 special values 0 F (x)e imx dx = f( m). Now set x = 0.) Now consider the function f t (x) = e The functional πtx2. Its Fourier transform is given by equation of the zeta function f t (y) = e πtx2 +ixy dx = e πy2 /t e πt(x+iy/t)2 dx = c f 1/t (y), t where c>0 is the constant c = e πx2 dx. Applying (13) with f = f t therefore gives ϑ(t) = ct 1/2 ϑ(1/t), and taking t = 1 in this formula gives c = 1 and proves equation (12). Now we find from (10) that ϑ(t) has the asymptotic expansions ϑ(t) =1+O(t N ) as t and ϑ(t) =t 1/2 +O(t N ) as t 0, where N>0 is arbitrary. It follows from the results of Section 1 that its Mellin transform ϑ(s) has a meromorphic extension to all s with simple poles of residue 1 and 1 at s =1/2 and s = 0, respectively, and no other poles. From the formula ζ (s) = 1 2 ϑ(s/2) we deduce that the function ζ (s) defined in (11) is meromorphic having simple poles of residue 1 and 0 at s = 1 and s = 0 and no other poles and hence (using once again that Γ(s) has simple poles at non-positive integers and never vanishes) that ζ(s) itself is holomorphic except for a single pole of residue 1 at s = 1 and vanishes at negative even arguments s = 2, 4,... This is weaker than (8), which gives a formula for ζ(s) 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>1). The advantage of the second approach to ζ(s) is that from equation (12) and the properties of Mellin transforms listed in (2) we immediately deduce the famous functional equation ζ (1 s) = ζ (1 s) (14) of the iemann zeta-function which was discovered (for integer values 0, 1 of s) by Euler in 1749 and proved (for all complex values 0, 1 of s) by iemann in 1859 by just this argument. We next generalize the method of Example 3. Consider a generalized Dirichlet series L(s) = c m λm s (15) m=1