The Fourier transform
Bill Casselman
University of British Columbia [email protected]
This essay is a brief introduction to the Fourier transform on Rn. Special features of the transform in dimension one are covered elsewhere.
Contents
1. Motivation . . . 1
2. The Fourier transform . . . 2
3. The Plancherel theorem . . . 5
4. Schwartz’ Paley-Wiener theorem . . . 6
5. Tempered distributions . . . 7
6. Examples . . . 9
7. More about the Dirac delta . . . 12
8. Poisson summation . . . 13
9. Riemann’s zeta function . . . 14
10. References . . . 15
1. Motivation
The theory of Fourier series tells us that if f (x) is a smooth function of period 1 on R then f (x) =X n fne2πinx, where fn= Z 1 0 f (x)e−2πinxdx = Z 1/2 −1/2 f (x)e−2πinxdx .
The Fourier coefficients fn are of rapid decrease in the variable n ∈ Z, so the series certainly converges absolutely and uniformly.
We can generalize this to deal with functions of arbitrary period. If f (x) has period T > 0 then ϕ(x) = f (xT ) has period 1, so ϕ(x) =X n ϕne2πinx where ϕn= Z 1/2 −1/2 ϕ(x)e−2πinxdx = 1 T Z T /2 −T /2 f (y)e−2πiny/Tdy (y = xT ) .
f (x) = 1 T X n fn/Te2πi(n/T )x where fn/T = Z T /2 −T /2 f (y)e−2πi(n/T )ydy , or f (x) = 1 T X λ∈Z/T fλe2πiλx where fλ= Z T /2 −T /2 f (y)e−2πiλydy .
Suppose now that f (x) is a smooth function on R of support on the interval (−C/2, C/2). Let fT(x) be the function one gets by extending it outside the interval (−C/2, C/2) to be of period T > C:
fT(x) = X
n
f (x + nT ) .
Thus fThas period T , and it agrees with f on (−T/2, T/2). What we have just seen is that inside the interval (−T/2, T/2) f (x) = fT(x) = 1 T X λ∈Z/T fλe2πiλx where fλ= Z T /2 −T /2 f (x)e−2πiλxdx .
As T → ∞ the coefficient fλhas limit b f (λ) =
Z ∞ −∞
f (x)e−2πiλxdx .
This makes sense for any λ ∈ R, and the formula for f(x) (which is a Riemann sum with interval 1/T ) becomes f (x) = Z ∞ −∞ b f (λ)e2πiλxdλ .
This formula expresses the Fourier inversion formula for f in Cc∞(R). In the next section we’ll derive this formula, and in fact a somewhat more general one, in a different way.
2. The Fourier transform
For x in Rnlet
kxk = q
x2
1+ · · · + x2n.
Define theSchwartz spaceS(Rn) to be the space of all smooth functions f on Rnsuch that all the semi-norms 1 + kxkk ∂mf ∂xm
are bounded. In other words, f and all its derivatives decrease more rapidly at infinity than any negative power ofkxk. This definition is invariant under invertible linear transformations, hence allows us to define the Schwartz spaceS(V ) of any finite-dimensional real vector space V , but I’ll work only with Rn.
The point of working with Rnis that we can assign Rn the usual Euclidean inner product and measure, so that it may be identified with its own linear dual. For y in Rnthe function x7→ e2πix•yis a unitary character of Rn–that is to say, a continuous homomorphism from Rnto the multiplicative group of complex numbers of absolute value 1. For f inS(Rn), define its Fourier transform to be the function on Rn:
b f (y) =
Z Rn
f (x)e−2πix•ydx .
The integral is finite, and more precisely for k≫ 0 bf (y) ≤ Z Rn f(x) dx = Z Rn 1 + kxk k f(x) 1 1 + kxkk dx ≤ sup Rn 1 + kxk k f(x) Z Rn 1 1 + kxkk dx .
2.1. Lemma.Suppose f to be inS(Rn). The Fourier transform of ∂f/∂x
mis 2πiymf (y), and ∂ bb f /∂ymis the Fourier transform of−2πixmf (x).
This is the whole point of the Fourier transform—it transforms problems in analysis (differentiation) into problems of algebra (multiplication).
Proof. The second claim is straightforward, the first follows from integration by parts.
2.2. Corollary.The Fourier transform takesS(Rn) to itself.
In other words, because f is smooth, its Fourier transform is rapidly decreasing, and because f is rapidly decreasing, its Fourier transform is smooth.
The fundamental theorem of the modern theory of the Fourier transform is this:
2.3. Theorem.The map f 7→ bf is an isomorphism of S(Rn) with S(Rn). Explicitly, I claim that the inverse map takes F (y) to
F∨(x) = Z
Rn
F (y)e2πix•ydy
In other words, we want to prove that
f (x) = Z
R b
f (y) e2πix•ydy .
Proof. I do this in several steps.
Step 1.I first show that the inversion formula is valid for one particular function.
2.4. Lemma.The Fourier transform of e−πkxk2is itself.
Proof. Complete the square in the Fourier integral and make a simple contour shift.
Step 3.It remains to show that f (0) = Z R b f (y) dy for all f inS(Rn). But e−πkxk2
lies in the Schwartz space, Fourier inversion is valid for it, and f (x)− f (0)e−πkxk2
vanishes at 0. So it suffices to show that the inversion formula is valid for x = 0 when f (0) = 0.
Step 4.It must be shown that if f (0) = 0 then Z
R b
f (y) dy = 0 . This will follow in the next step from:
2.5. Lemma.If f lies inS(Rn) then f (0) = 0 if and only if there exist functions fkinS(Rn) with
f =X
k xkfk.
This is the only hard part of the entire proof of Fourier inversion.
Proof of the Lemma. One way is clear. For the other, I’ll prove something stronger.
2.6. Lemma.If f lies inS(Rn) and f (0, . . . , 0, xm+1, . . . , xn) = 0 for all xi(i > m) then there exist functions fkinS(Rn) with f = m X k=1 xkfk.
The proof of this will be induction on m, starting with m = 2. For x in Rnlet x∗= (x2, . . . , xn) in Rn−1. So suppose f (0, x∗) = 0 for all x∗in Rn−1.
Define ϕx(t) = f (tx1, x∗) for each x in Rnand t∈ [0, 1]. By the fundamental theorem of calculus f (x) = f (x) − f(0, x∗) = ϕx(1) − ϕx(0) = Z 1 0 ϕ′ x(t) dt = Z 1 0 ∂f ∂x1 (tx1,x∗) d tx1 dt dt = x1 Z 1 0 ∂f ∂x1 (tx1,x∗) dt = (say) x1f1(x) . I leave it as an exercise to prove that f1lies inS(Rn). Now for the induction step. Suppose m≥ 3. Let
ϕ(x) = f (x) − f(0, . . . , xm−1, xm, . . . , xn)e−(x 2
1+···+x2m−2). On the one hand, the function f (0, . . . , xm−1, xm, . . . , xn)e−(x
2
Step 5. Why does the Lemma imply the integral formula we want? It suffices to see that Fourier inversion holds for a function of the form xmfm(x) with fminS(Rn).
The Fourier inversion theorem may be restated:
2.7. Corollary.If bf (y) = ϕ(y) then bϕ(x) = f (−x).
It will be useful to know how the Fourier transform behaves with respect to certain simple transformations.
2.8. Proposition.Suppose f to be inS(Rn).
(a) If ϕ(x) = f (ax) with a6= 0 then bϕ(y) = (1/|a|) bf (y/a); (b) if ϕ(x) = f (x + a) then bϕ(y) = e2πiayf (y).b
3. The Plancherel theorem
The Fourier transform extends to a map from L2(Rn) to itself, and induces an isometry of Hilbert spaces.
3.1. Proposition.Suppose f , g inS(Rn). Then Z Rnf (x)bg(x) dx = Z Rn b f (y)g(y) dy .
Proof. Both are equal to Z
Rn Z Rn f (x)g(y)e−2πix•ydx dy . 3.2. Corollary.For f , g inS(Rn) Z Rn f (x)g(x) dx = Z Rn b f (x)bg(y) dy .
3.3. Corollary.(The Plancherel Theorem)The map fromS(Rn) to itself extends to an isometry of L2(Rn) with itself.
4. Schwartz’ Paley-Wiener theorem
A very general question one might naturally ask is: if f satisfies such and such conditions, what properties does its Fourier transform possess? Or: if U is a vector space of functions in Rn, can we characterize all functions in the image of U with respect to the Fourier transform? We have seen this last question answered forS(Rn) and L2(Rn), but there are other spaces of interest. Many of them involve conditions of support. Questions of this kind were first answered in joint work of R. E. A. C. Paley and Norbert Wiener, particularly in the book [Paley-Wiener:1934], and a certain category of such results (those involving Fourier transforms that are analytic functions) are often calledPaley-Wiener theorems.
One of the most useful of these characterizations is for Cc∞(Rn), and is due to Laurent Schwartz rather than Paley and Wiener. If f is integrable and has support in the compact subset Ω of Rn, then the integral
b f (s) = Z Rn f (x)e−2πis•xdx = Z Ω f (x)e−2πis•xdx
is absolutely and uniformly convergent for all s in Cn. For R > 0 let BRbe the Euclidean ballkxk ≤ R.
4.1. Proposition.If f is a smooth function on Rnwith support in BR, then its Fourier transform F (s) satisfies these conditions: PW(a) F (s) is holomorphic on Cn; PW(b) for all N > 0 F (s) ≪N eR|IM(s)| (1 + ksk)N
Proof. Since all derivatives of f still satisfy the conditions of the Proposition, this is straightforward.
4.2. Proposition.Conversely, if F (s) satisfies the conditions PW of the previous proposition, then it is the Fourier transform of a function in Cc∞(Rn) with support in BR.
Proof. Just to simplify things slightly, I’ll prove this only for n = 1. It must be shown that the inverse transform of F is well defined, that it is smooth, and that it has support in [−R, R]. So suppose F to satisfy conditions PW. I first claim that t7→ F (σ + it) lies in S(R) for each σ. It has to be verified that its derivatives are rapidly decreasing as|t| → ∞. But this follows from the Cauchy integral formula expressing F(n)(s) as a path integral around small circles surrounding s.
The inverse Fourier transform of F is the integral
f (x) = Z ∞
−∞
F (s)e2πisxds .
The Fourier inversion theorem for Schwartz functions implies that this lies inS(R), and it must be shown that f (x) = 0 if|x| > R. By assumption, there exists C2> 0 such that
F (σ + it) ≤ C2e2πR|t| 1 + |t|2 for all real σ, t. We may shift the integration path for f (x) to get
f (x) = Z ∞+it
−∞+it
But then f(x) ≤ C2 Z ∞+it −∞+it e2π(R−x)t 1 + |s|2 ds f(x) ≤ C2e2π(R−x)t Z ∞+it −∞+it 1 1 + |s|2ds
But if x > R then letting t→ ∞ we see that f(x) must be 0. If we let t → −∞ we deduce that f(x) = 0 for x < −R.
5. Tempered distributions
The vector spaceS(Rn) has a topology. That is to say, we can measure how close one function in S(Rn) is to another. Actually, there are several ways to measure this. Define
kfkN,n= sup
x∈Rn 1 + kxk N
f(n)(x) .
These aresemi-normsthat measure how close f is to 0. Atempered distributionis a linear function Φ: S(Rn) → C
that’s continuous in this topology— if f approaches 0 inS(Rn) then so should hΦ, fi in C. In other words,
we require that
hΦ, fi ≪N,nkfkN,n
for some finite set of N , n. For example, if F (x) is a locally integrable function on Rnof moderate growth in
the sense that
F (x)/kxkN
∈ L1(Rn) for some N , then
f 7−→ Z
Rn
ϕ(x)f (x) dx defines a tempered distribution.
In particular, if F is an integrable function on Rn, then integrating against F defines a tempered distribution, for which one might reasonably expect to have an integral formula:
5.1. Proposition.If F lies in L2(Rn) then bF is a bounded continuous function on Rn, and b
F (y) = Z
Rn
F (x)e−2πix•ydx .
The Riemann-Lebesgue Lemma asserts that bF (y) → 0 as kyk → ∞, but I won’t prove or need that here. Proof. The integral
G(y) = Z
Rn
F (x)e−2πix•ydx
is a bounded function of y. It remains to show that integration against G defines the distribution bF . Suppose f to lie in S(R). By definition
and there are no obstacles to expressing this a double integral hF, bf i == Z Rn F (y) bf (y) dy = Z Rn Z Rn
F (y)f (x)e−2πix•ydx dy = hG, fi .
So we can construct lots of tempered distributions from functions with various growth conditions, but as we shall see in the next section there are others of a more exotic kind. The point is that one can define the derivatives of an arbitrary tempered distribution, which is another tempered distribution. Thus every L1 function has a derivative in this sense.O brave new world!
How to define the derivative of a tempered distribution? If f is in the Schwartz space and ϕ is a smooth function of moderate growth whose derivatives are of moderate growth, then integration by parts gives us
hϕ, ∂f/∂xmi = Z Rn ϕ(x)∂f ∂xmdx = − Z Rn ∂ϕ ∂xmf (x) dx = −h∂ϕ/∂x m, f i . This suggests defining the partial derivative of any tempered distribution Φ by the formula
h∂Φ/∂xm, f i = −hΦ, ∂f/∂xmi .
This definition can be extended to any differential operator with constant coefficients.
According to Proposition 3.1, if f and g are both inS(Rn) then hf, bgi = h bf , gi, so here again it is consistent to define the Fourier transform of an arbitrary tempered distribution by duality:
hbΦ, f i = hΦ, bf i .
Since the Fourier transform is an isomorphism ofS(Rn) with itself, it induces also an isomorphism of the space of tempered distributions with itself. We shall see examples in the next section.
Certian operations and formulas involving the Fourier transform of functions inS(Rn) are valid also for tempered distributions. (a) If F (x) is a smooth function on Rn all of whose derivatives are of moderate growth, and f lies inS(Rn), then F (x)f (x) is again in S(Rn). If Φ is a tempered distribution on Rnthen the product F· Φ is defined by the formula
hF · Φ, fi = hΦ, F · fi . (b) If g is in GLn(R) then the transformation µgf of f by g is defined by
[µgf ](x) = f (g−1x) . If Φ is a tempered distribution then the scale of Φ by g is defined by
hµgΦ, f i = hΦ, µg−1f i . (c) The translation λvf of f by v in Rnis defined by
[λvf ](x) = f (x − v) . The translation of Φ by v is defined by
hλvΦ, f i = hΦ, λ−vf i .
In comparing these to the analogous properties of Schwartz functions, you must keep in mind that it is not really a function f (x) that defines a distribution, but the measure f (x)dx.
How do these interact with the Fourier transformation?
5.2. Proposition.Suppose Φ to be a tempered distribution. (a) The Fourier transformation of xk· Φ is (−1/2πi) ∂ bΦ/∂yk; (b) the Fourier transform of µgΦ is | det(g)|−1µtg−1Φ;b (c) the Fourier transform of λvΦ is e2πiv•yΦ;b
(d) the Fourier transform of ∂Φ/∂xkis (2πiyk) · bΦ.
6. Examples
Example. The step function. Let
H(x) = 0 if x < 0 1 if x ≥ 0. Thus hH, fi = Z ∞ 0 f (x) dx and hH′, f i = −hH, f′i = − Z ∞ 0 f′(x) dx = f (0) .
Thus the derivative of H is the distribution that takes f to f (0). This is called theDirac deltaδ0. (I’ll say something in the next section about an intuitive approach to these matters.)
Example. The Dirac delta. If Φ = δ0then
hΦ′, f i = hΦ, −f′i = −f′(0) , and continuing on one obtains
hΦ(n), f i = (−1)nf(n)(0) . What is the Fourier transform of δ0?
hbΦ, f i = hδ0, bf i = bf (0) = Z ∞
−∞f (x) dx = h1, fi so bΦ is the constant function 1. And conversely, the Fourier transform of 1 is δ0.
6.1. Lemma.A distribution Φ satisfies xkΦ = 0 for all k if and only if it is a scalar multiple of δ0.
Proof. It follows from what we have just done (calling on Fourier inversion), but it can also be proved directly from Lemma 2.5.
Example. The principal value of 1/x . The function log|x| is integrable near 0 and defines a tempered distribution. What is its derivative? Its derivative in the naive sense is the function 1/x. What relationship does it have with the distributional derivative of log|x|?
since f (ε)− f(−ε) ∼ 2εf′(0) as ε → 0.
Theprincipal valuedistributionP(1/x) associated to the function 1/x is the limit that appears here:
hP, fi = lim ε→0 Z −ε −∞ f (x) x dx + Z ∞ ε f (x) x dx .
Roughly speaking, the limit exists because 1/x takes equal and negative values on opposite sides of 0. The distributionP(1/x) is determined without trouble on the subspace m of f in S(R) such that f(0) = 0, since then f (x) = xϕ(x) for some ϕ inS(R), and
hP(1/x), fi = lim ε→0 Z |x|≥ε f (x) x dx = Z R ϕ(x) dx .
The codimension of m inS(R) is one, and if we are given one extension Φ of P(1/x) to all of S(R) then we obtain all such extensions as Φ + cδ0. How do we distinguishP(1/x)?
6.2. Proposition.The distributionP(1/x) is, up to scalar multiple, the only distribution Φ such that µaΦ = sgn(a)Φ.
Here
sgn(a) =n 1 if a > 0 −1 if a < 0. Proof. Let Φ =P(1/x). Let a = σ|a| with σ = ±1. We have
hµaΦ, f i = hΦ, µ1/af i = lim ε→0 Z −ε −∞ f (ax) x dx + Z ∞ ε f (ax) x dx = lim ε→0 Z −aε −∞ f (σx) x dx + Z ∞ aε f (σx) x dx = lim ε→0 Z −ε −∞ f (σx) x dx + Z ∞ ε f (σx) x dx = σ · lim ε→0 Z −ε −∞ f (x) x dx + Z ∞ ε f (x) x dx .
But it is trivial to verify that µaδ0 = δ0, soP(1/x) has exactly one extension which is an eigenvector with eigencharacter sgn.
Example. The step function, again. What is the Fourier transform of H? If Φ is any tempered distribution, then the Fourier transform of Φ′is (2πiy)bΦ. Since H′= δ
0we have 2πiy · bH = 1 .
The formal deduction from this is that bH = P(1/y), but muliplication by y has δ0in its kernel, so at most we may deduce that bH = P(1/y) + c δ0for some c. What is c?
There is a clever approach to this problem. Let F = H(x)− 1/2, so
F (x) =
One can easily show that µaF = aF = sgn(a)|a|F (as distribution). But then bF must be by Proposition 5.2 equal to some of cP(1/x). The value of c can be decided by applying it to the odd function f(x) = xe−πx2
. By Proposition 5.2, bf = −if(x). So on the one hand
h bF , f i = chP(1/x), fi = c Z ∞ −∞ e−πx2dx = c
and on the other
h bF , f i = hF, bf i =−i 2 − Z 0 −∞ xe−πx2 dx + Z ∞ 0 xe−πx2 dx = −i Z ∞ 0 xe−πx2 dx =−i 2π Z ∞ 0 (2πx)e−πx2dx = 1 2πi Z ∞ 0 e−xdx = 1 2πi. Therefore b H = 1 2πiP(1/x) + 1 2δ0.
Example. Parties finies. It might be a bit more surprising that a distribution can be associated to every function 1/xn+1(n≥ 0).
For ε > 0 and f inS we have Z ∞ ε f (x) xn+1dx = Z 1 ε f (x) xn+1dx + Z ∞ 1 f (x) xn+1dx = Z 1 ε f0+ f1x + f2x2+ · · · + fnxn+ something divisible by xn+1 xn+1 dx + something finite = nf0ε−n+ (n − 1)f1ε−(n−1)+ · · · + fnlog ε + something finite where f(k)(0)/k! = fk. The limit
lim ε→0 Z ∞ ε f (x) xn+1dx − nf (0)ε−n+ (n − 1)f′(0)ε−(n−1)+ · · · +f(n)(0) log ε n!
therefore exists, and defines a distribution called thepartie finieof 1/xn+1on (0,∞), or Pf(1/xn+1) x>0. Similarly, the limit
lim ε→0 Z −ε −∞ f (x) xn+1dx + nf (0)(−ε)−n+ (n − 1)f′(0)(−ε)−(n−1)+ · · · + f (n)(0) log ε n!
exists, and defines a distribution called thepartie finie of 1/xn+1on (−∞, 0), or Pf(1/xn+1)
7. More about the Dirac delta
I attempt in this section to justify the physicists’ notion that δ0is a function that vanishes everywhere except x = 0, but has total area 1 underneath its graph—it is the limit of functions that come closer and closer to that ideal.
The step function H(x) can be well approximated by smooth functions.
What happens to it derivative as this convergence takes place?
First of all, I recall an explicit construction of functions in Cc∞(R). All derivatives of e−1/xvanish at x = 0,as do all of e1/(1−x)at x = 1. Therefore the function
f (x) = 0 for x < 0 e−1/xe−1/(1−x) 0 ≤ x ≤ 1 0 1 < x
is smooth on all of R. If we scale it and translate it, we can find a non-negative function ϕ1(x) which vanishes outside [−1/2, 1/2] and has total area 1 underneath its graph. Now if we define
ϕc(x) = (1/c) ϕ(x/c) then the support of ϕcis contained in [−c/2, c/2] but retains area 1.
As c→ 0, the function ϕchas limit δ0in the sense that
hϕc, f i = Z c/2
−c/2
ϕc(x)f (x) dx → f(0) .
and the function Z x
−∞
ϕc(t) dt , whose derivative is ϕc, has as limit the function H(x).
8. Poisson summation Suppose f (x) to be inS(Rn). Then Θf(x) = X n f (x + n)
converges absolutely to a smooth function of period 1 on Rn. It may therefore be expanded in a Fourier series Θf(x) = X m Θme2πimx where Θm= Z 1 0 Θf(x)e−2πimxdx = Z 1 0 X n f (x + n)e−2πimxdx =X n Z 1 0 f (x + n)e−2πim(x+n)dx = Z ∞ −∞ f (x)e−2πimxdx = bf (m) .
8.1. Proposition.(Poisson summation formula)For f inS(Rn) X n f (x + n) =X m b f (m)e2πimx.
Suppose we apply this to f (x) = e−πtx2. It follows from Lemma 2.4 and Proposition 2.8 that its Fourier transform is (1/√t)e−πx2/t
. Poisson summation tells us that if ϑ(t) =X n e−πn2t then ϑt= 1 √ tϑ1/t
9. An application: Riemann’s functional equation
Riemann’s zeta function is defined to be
ζ(s) =X n>0
1 ns.
The series converges forRE(s) > 1, while for s = 1 we have the series
X n>0 1 n
which is well known not to converge. For reasons that will become clear in a moment, the more convenient function is ξ(s) = π− s 2Γ s 2 ζ(s)
The importance of these functions is that unique factorization of the positive integers allows us to deduce that ζ(s) may be expanded in the Euler product
ζ(s) =Y p 1 + 1 ps + 1 p2s + · · · =Y p 1 − p1s −1 .
This suggests that analytic properties of ζ(s) might imply properties of the prime numbers. As a simple example, that ζ(1) sums to∞ implies that there are an infinite number of primes, and even says something, albeit rather weak, about their distribution.
9.1. Theorem.The function ξ(s) extends meromorphically to all of C, is holomorphic everywhere except for
two simple poles at s = 0 and s = 1, and satisfies the functional equation ξ(s) = ξ(1 − s) .
Proof. It all comes down to the consequences of Poisson summation for the function ϑ(t) =X
n>0
e−πn2t.
Poisson summation tells us that
1 + 2ϑ(t) = √1
t 1 + 2ϑ(1/t)
.
The function ϑ(t) decreases exponentially at∞. As t → 0 it goes off to ∞, and this equation tells us exactly how: ϑ(t) = √1 t 1 + 2ϑ(1/t) −√t 2 ∼ 1 2√t. I recall that the Gamma function is defined to be
Γ(s) = Z ∞
0
e−ttsdt t . forℜ(s) > 0. By a change of variable t = πn2x this becomes
This and the asymptotic behaviour of ϑ at 0 and∞ tell us that forRE(s) > 1 ξ(s) = Z ∞ 0 X n>0 e−πn2x xs/2dx x = Z ∞ 0 ϑ(x)xs/2dx x . We can now write (still forRE(s) > 1)
ξ(s) = Z ∞ 0 ϑ(x)xs/2dx x = Z 1 0 ϑ(x)xs/2dx x + Z ∞ 1 ϑ(x)xs/2dx x = Z ∞ 1 ϑ(1/x)x−s/2dx x + Z ∞ 1 ϑ(x)xs/2dx x = Z ∞ 1 √ xϑ(x) + √x 2 − 1 2 x−s/2dx x + Z ∞ 1 ϑ(x)xs/2dx x = Z ∞ 1 ϑ(x)x(1−s)/2 dx x + Z ∞ 1 ϑ(x)xs/2dx x + 1 2 Z ∞ 1 x−(s−1)/2− x−s/2 dxx = Z ∞ 1 ϑ(x)x(1−s)/2 dx x + Z ∞ 1 ϑ(x)xs/2dx x − 1 s(1 − s).
Since ϑ(x) decreases exponentially fast, the integrals converge for all s. Since the expression is invariant if s and 1− s are swapped, we are through.
Furthermore, the pole of ξ(s) at s = 1 is simple, with residue 1.
10. References
1. Raymond Edward Alan Christopher Paley and Norbert Wiener,Fourier transforms in the complex domain, American Mathematical Society, 1934.
2. Michael Reed and Barry Simon, Functional analysis, volume I of the series Methods of Modern Mathematical Physics, Academic Press, 1972.
3. Michael Reed and Barry Simon,Fourier analysis, self-adjointness, volume II of the seriesMethods of Modern Mathematical Physics, Academic Press, 1975.
4. Laurent Schwartz,Th ´eorie des distributions, Hermann, 1966. This is the first edition of the original two volumes in one.
