Fourier, Laplace and Dirichlet relationship

1. The Function Space as an Infinite-Dimensional Vector Space

We consider a space of functions of a real variable (allowing generalized functions such as the Dirac delta distribution $δ (x)$ ) as an infinite-dimensional vector space $F$ , where an abstract element (a "vector") is denoted by $v$ .

We highlight two fundamental continuous bases for this space:

Position / Dirac Basis: ${δ_{x}}$
Frequency / Exponential Basis: ${e_{ω}}$ (where $e_{ω} (x) = e^{- i ω x}$ )

Using continuous index Einstein-style notation—where matching upper and lower continuous indices imply integration over the domain rather than summation —the same abstract vector $v$ can be represented in both bases:

v = f^{x} δ_{x}

v = {\hat{f}}^{ω} e_{ω}

$f^{x}$ represents the components of vector $v$ in the continuous Dirac delta basis (which corresponds to the standard function values $f (x)$ ).
${\hat{f}}^{ω}$ represents the components of vector $v$ in the continuous exponential basis (which corresponds to the Fourier transformed function $\hat{f} (ω)$ ).

2. The Fourier Transform as a Basis Change Matrix

The Fourier transform acts as a continuous, infinite-dimensional transition matrix $T_{x}^{ω}$ that changes components from the position basis to the frequency basis:

{\hat{f}}^{ω} = T_{x}^{ω} f^{x}

Component Examples:

Dirac Delta Component: A vector with components zero everywhere except a single spike at $x_{0}$ transforms into a uniform constant component vector (1 everywhere) across frequencies.
Sine Component: A function $\sin (x)$ decomposes into two exponential components (two discrete active components in frequency space).

3. The Laplace Transform & Holomorphic Extension

While the Fourier transform maps real position variables to components indexed by real frequencies $ω$ , the Laplace transform extends this framework to construct a full complex/holomorphic function $F (s)$ over $s = c + i ω \in C$ .

Here we use the unilateral Laplace transform — the natural choice since the Dirac combs we consider are supported on $x \geq 0$ ( $\ln 1 = 0$ ). The ROC is a right half-plane $Re (s) > σ_{c}$ .

Operationally, for each fixed real parameter $c > σ_{c}$ , we damp the position components $f^{x}$ by $e^{- c x}$ and apply the Fourier basis transform matrix:

Laplace transform: F (c + i ω) = {\hat{f}}_{c}^{ω} = T_{x}^{ω} (f^{x} e^{- c x})

Each value of $c$ corresponds to a 1D vertical line in the complex plane.
Together, these lines form a single holomorphic function $F (s)$ .
Via analytic continuation, knowing the Fourier transform along one line determines $F (s)$ on the entire half-plane $Re (s) > σ_{c}$ .

4. Fourier Series as Uniformly Spaced Impulses

In this unified picture, a Fourier Series is not a distinct entity. It emerges when the continuous position component vector $f^{x}$ consists of discrete unit impulses (Dirac deltas) spaced uniformly at integer steps:

f^{x} = \sum_{n \in Z} a_{n} δ (x - n)

Applying the basis change $T_{x}^{ω}$ transforms these uniformly spaced impulses into a linear combination of exponentials:

{\hat{f}}^{ω} = \sum_{n \in Z} a_{n} e^{- i n ω}

This yields a periodic function, which is precisely a standard Fourier Series.

5. Dirichlet Series from Logarithmic Impulses

A Dirichlet Series arises from the exact same process, but with a Dirac comb whose impulses are spaced logarithmically rather than uniformly:

f^{x} = \sum_{n \geq 1} a_{n} δ (x - \ln n)

Fourier Transform Step: Applying $T_{x}^{ω}$ to logarithmically spaced impulses converts continuous translations into logarithmic powers, producing an almost-periodic function (a particular subfamily of almost-periodic functions, namely those with logarithmically spaced frequencies $ω_{n} = \ln n$ ):
${\hat{f}}^{ω} = \sum_{n \geq 1} a_{n} n^{- i ω}$
Laplace Extension Step: Applying the full Laplace transform package ( $s = c + i ω$ ) holomorphically extends this almost-periodic function:
$F (s) = \sum_{n \geq 1} a_{n} n^{- s}$

Important examples: L-functions.

6. The Z-Transform: Completing the Analogy

The same "Laplace extension" applied to the Fourier series (Section 4) yields the Z-transform — the discrete-time analogue of the Laplace transform:

Starting from the uniformly-spaced Dirac comb $f^{x} = \sum_{n \in Z} a_{n} δ (x - n)$ , apply the Laplace package (damp by $e^{- c x}$ , then Fourier $T_{x}^{ω}$ with kernel $e^{- i ω x}$ ):

F (c + i ω) = T_{x}^{ω} (f^{x} e^{- c x}) = \sum_{n \in Z} a_{n} e^{- c n} e^{- i ω n} = \sum_{n \in Z} a_{n} e^{- (c + i ω) n}

Define $z := e^{s} = e^{c + i ω}$ . Then:

F (s) = \sum_{n \in Z} a_{n} e^{- s n} = \sum_{n \in Z} a_{n} z^{- n} with z = e^{s}

This is precisely the (bilateral) Z-transform. When the sequence is one-sided ( $a_{n} = 0$ for $n < 0$ ), it reduces to the unilateral Z-transform $\sum_{n = 0}^{\infty} a_{n} z^{- n}$ , with ROC $| z | > R$ (the exterior of a disk, corresponding to the half-plane $Re (s) > \ln R$ ).

The Full Symmetry

\begin{aligned} Real frequency (s = i ω) : \int f (x) e^{- i ω x} d x & \overset{x = n}{\to} \sum_{n \in Z} a_{n} e^{- i ω n} \overset{n \mapsto \ln n}{\to} \sum_{n \geq 1} a_{n} n^{- i ω} \end{aligned}

\begin{aligned} ↓ damp by e^{- c x} \end{aligned}

\begin{aligned} Complex frequency (s = c + i ω) : \int f (x) e^{- s x} d x & \overset{x = n}{\to} \sum_{n \in Z} a_{n} z^{- n} \overset{n \mapsto \ln n}{\to} \sum_{n \geq 1} a_{n} n^{- s} \end{aligned}

\begin{aligned} Laplace & Z-transform Dirichlet series \end{aligned}

The bottom row is unified by a single complex variable $s$ : the Z-transform is $\sum_{n \in Z} a_{n} e^{- s n}$ and the Dirichlet series is $\sum_{n \geq 1} a_{n} e^{- s \ln n} = \sum_{n \geq 1} a_{n} n^{- s}$ . The only difference is the spacing of the underlying Dirac impulses — and the index range: the Z-transform comb is two-sided ( $n \in Z$ ), while the map $n \mapsto \ln n$ forces the Dirichlet comb to be one-sided ( $n \geq 1$ , since $\ln n$ is real only for $n > 0$ ).

7. The Mellin Transform: the Laplace Package Applied After the Relabeling $L$

For starter, think of a change of coordinates in the space of function $F$ as a linear map that reorders the basis. Discretely: for a basis ${e_{1}, e_{2}, e_{3}, \dots}$ in a vector space $V$ , the map sending $e_{1} \to e_{2}$ , $e_{2} \to e_{3}$ , and so on, is just a change of indices $j = i + 1$ — a permutation whose matrix is filled with 1s and 0s. The same basis vectors are merely relabeled.

For a continuous basis it is the same. The relabeling $x \mapsto g (x)$ acts on the Dirac basis ${δ_{x}}$ by $δ_{x} \to δ_{g (x)}$ , permuting the continuum of basis elements. Its "matrix" is still 1s and 0s, now a collection of deltas $δ (y - g (x))$ . But in this case, the Jacobian of the transformation enters the scene.

The logarithmic/exponential change of coordinates $u = e^{x}$ (i.e. $x = \ln u$ ) is one map of this class — call it $L$ . It is applied on the top layer, before any transform: it reorders the basis by $x = \ln u$ and acts on inputs as

L : δ_{u} \mapsto δ_{\ln u}, (L f) (x) = f (e^{x}) .

(On basis elements the relabeling carries the Jacobian $1 / u$ , which is exactly the factor that turns $d x$ into $d u / u$ .)

Mellin = the Laplace package applied to the image of $L$ . The Laplace transform package — damp by $e^{- c x}$ , Fourier $T_{x}^{ω}$ , glue into a holomorphic function — is run on $L f$ , not on $f$ :

M {f} (s) = \int_{0}^{\infty} f (u) u^{- s} \frac{d u}{u} = \int_{- \infty}^{\infty} (L f) (x) e^{- s x} d x = L {L f} (s)

The diagram is commutative. The exponential basis after Laplace and the power basis after Mellin are the same seen through $L$ — $e^{- s x} |_{x = \ln u} = u^{- s}$ .

Zeta and Dirichlet series as Mellin transforms of equispaced combs

A comb at the positive integers is equispaced in $u$ : $\sum_{n \geq 1} a_{n} δ (u - n)$ . Under $L$ it is precisely the log-spaced comb $\sum_{n \geq 1} a_{n} δ (x - \ln n)$ of Section 5 — the same comb in two coordinate systems. Its Mellin transform is immediate:

M {\sum_{n \geq 1} a_{n} δ (u - n)} (s) = \sum_{n \geq 1} a_{n} n^{- s}

(The index runs over $n \geq 1$ only: the Mellin integral lives on $(0, \infty)$ , so spikes at negative integers lie outside the domain, and the spike at $n = 0$ sits at the boundary, where $0^{- s}$ is undefined. A naive two-sided sum $\sum_{n \in Z} n^{- s}$ would give $(1 + e^{- i π s}) ζ (s)$ , not $ζ (s)$ .)

Two conclusions:

Every Dirichlet series is the Mellin transform of a general equispaced comb $\sum_{n \geq 1} a_{n} δ (u - n)$ .
Zeta is the Mellin transform of the unit equispaced comb ( $a_{n} = 1$ ), $ζ (s) = M {\sum_{n \geq 1} δ (u - n)} (s)$ for $Re (s) > 1$ and by meromorphic continuation beyond.

This closes the square of Section 6: the Z-transform is the Laplace transform of the two-sided equispaced comb $\sum_{n \in Z} a_{n} δ (x - n)$ , while a Dirichlet series is the Mellin transform of the one-sided equispaced comb $\sum_{n \geq 1} a_{n} δ (u - n)$ — same comb, same transform package, different relabeling $L$ and different index range.

Still not understood

The other brick is $Γ (s) = M {e^{- u}} (s)$ , and Perron's formula is the Mellin inversion that recovers the partial sums of a Dirichlet series. The functional equation of $ζ$ is the Poisson self-duality of the integer comb, expressed through the modularity of the theta function $θ (1 / x) = \sqrt{x} θ (x)$ . The analytic-continuation mechanism itself is now understood: see the callout below.

Analytic continuation of

ζ

Chart 1 — the comb itself ( $Re (s) > 1$ ). Take the one-sided integer comb and apply the recipe with no further transformation:

f (u) = \sum_{n \geq 1} δ (u - n) \overset{L}{⟶} \sum_{n \geq 1} δ (x - \ln n) \overset{Laplace}{⟶} M {f} (s) = \sum_{n \geq 1} n^{- s} = ζ (s) .

Chart 2 — the mirrored comb ( $Re (s) < 0$ ). Transform the same comb first, then run the identical recipe. The transformation is a composition of three linear maps on $F$ :

g = M_{1 / u} \circ R \circ T_{- 1} (f), {\begin{cases} T_{- 1} : f \mapsto f - 1 & subtract the constant (the k = 0 Poisson mode) \\ R : u \mapsto 1 / u & relabeling — the modular involution, x \mapsto - x in log coords \\ M_{1 / u} : \times \frac{1}{u} & the Jacobian weight, e^{- x} in log coords \end{cases}

In deltas, $R$ pulls the spike $δ_{n}$ back to $\frac{1}{n^{2}} δ_{1 / n}$ (the pullback Jacobian), and the weight supplies the factor $n$ , leaving unit spikes at reciprocal positions:

g (u) = \frac{1}{u} (Ш_{+} (\frac{1}{u}) - 1) = \sum_{n \geq 1} \frac{1}{n} δ (u - \frac{1}{n}) - \frac{1}{u}

Then $L$ sends this to the reflected log-comb $\sum_{n \geq 1} δ (x + \ln n) - e^{- x}$ , and the Laplace package (Abel-regularized, via the Poisson expansion $Ш_{+} (u) - 1 = 2 \sum_{n \geq 1} \cos (2 π n u)$ on $(0, \infty)$ and $\int_{0}^{\infty} \cos (u) u^{- s} d u = Γ (1 - s) \sin (π s / 2)$ ) yields:

M {g} (s) = 2 (2 π)^{s - 1} Γ (1 - s) \sin (\frac{π s}{2}) ζ (1 - s) = χ (s) ζ (1 - s) = ζ (s), Re (s) < 0

with $χ (s) = 2^{s} π^{s - 1} \sin (\frac{π s}{2}) Γ (1 - s)$ — the functional equation factor.

Why the mirror works. In the logarithmic coordinate, the inversion $u \mapsto 1 / u$ is the reflection $x \mapsto - x$ — a pure basis permutation with unit Jacobian, the same class of relabeling as $L$ itself. The reflection acts on the transform variable as $s \mapsto - s$ ; the weight $e^{- x}$ acts as $s \mapsto s + 1$ ; the renormalization of the removed constant mode completes the composite into exactly $s \mapsto 1 - s$ . The analytic continuation is therefore not a new transform: it is the same Laplace package, applied to the same comb relabeled by the second change of variables $u \mapsto 1 / u$ — the modular mirror of $L$ .

The critical strip $0 < Re (s) < 1$ is covered by neither chart. There the bridge is the theta function $θ (x) = \sum_{n \in Z} e^{- π n^{2} x}$ — the heat-kernel mollification of the same comb — whose Mellin transform gives the completed $ξ (s)$ and converges for all $s$ .

Exploration: Fourier duality as the driver of Chart 2

The three operations in Chart 2 ( $T_{- 1}, R, M_{1 / u}$ ) feel structurally different from the clean $L$ of Chart 1. But at heart Chart 2's mechanism relies on the Poisson summation formula — i.e., Fourier duality — not on geometric mirroring alone.

Conjecture. Define $\tilde{L} = L \circ F$ , where $F$ is the continuous basis-change matrix $T_{x}^{ω}$ from Section 2 (mapping position components to frequency components). Then Chart 2 is the same Laplace package applied to $\tilde{L} f$ — reading the comb in frequency space rather than position space. The factor $χ (s)$ would then be understood as the action of $F$ on the log-spaced comb.

At present this is notationally imprecise ( $L$ expects a position-domain input and $F$ outputs a frequency-domain object — the typing does not close), but the conceptual kernel is worth pursuing: the apparent asymmetry between Charts 1 and 2 may be not a bug but the signature of the position-frequency duality built into the Fourier transform from the very beginning.