11 Transforms of Fourier Series Representations of $U(x)$, $U'(x)$, and $U”(x)$

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Quite some time ago I noticed what appeared to be discrepancies between the Fourier transforms of the distributional and Fourier series representations of prime counting functions. I was initially trying to analyze these discrepancies in the context of the Fourier series representations of the second Chebyshev function $\psi(x)$ and it’s first and second-order derivatives, but I decided to shift my focus to the considerably simpler context of the Fourier series representations of $U(x)$ and it’s first and second-order derivatives defined below where $\theta(x)$ is the Heaviside step function (also referred to as the Unit step function) and $\delta(x)$ is the Dirac delta function.

(1) $\quad U(x)=-1+\theta(x+1)+\theta(x-1)$

(2) $\quad U'(x)=\delta(x+1)+\delta(x-1)$

(3) $\quad U”(x)=\delta'(x+1)+\delta'(x-1)$

The Fourier series representations of $U(x)$, $U'(x)$, and $U”(x)$,  are illustrated on pages 4.1, 5.1, and 6.1 respectively.


  • Laplace Transforms

The Laplace transforms of the distributional representations of $U(x)$, $U'(x)$, and $U”(x)$ defined in (1), (2), and (3) above are as follows.

(4) $\quad\mathcal{L}_x[U(x)](y)=\frac{e^{-y}}{y}\,,\quad\Re(y)>0$

(5) $\quad\mathcal{L}_x[U'(x)](y)=e^{-y}$

(6) $\quad\mathcal{L}_x[U”(x)](y)=y\,e^{-y}$

The Laplace transforms of the Fourier series representations of $U(x)$, $U'(x)$, and $U”(x)$ converge to the Laplace transforms of the distributional representations of $U(x)$, $U'(x)$, and $U”(x)$ for $Re(y)>0$ as illustrated on child page 11.1 below.


  • Fourier Transforms

The distributional representations of $U(x)$, $U'(x)$, and $U”(x)$ defined in (1), (2), and (3) above all have special relationships between their Fourier transforms and inverse Fourier transforms which are illustrated below where I believe the sign is related to whether each is an odd or even function. The Fourier transforms below and all subsequent Fourier transforms assume the Fourier parameters $\{0,\,-2\pi\}$.

(7) $\quad\mathcal{FT}_x[U(x)](z)=-\mathcal{FT}_x^{-1}[U(x)](z)=-\frac{i\,\cos(2\,\pi\,z)}{\pi\,z}$

(8) $\quad\mathcal{FT}_x[U'(x)](z)=\mathcal{FT}_x^{-1}[U'(x)](z)=2\,\cos(2\,\pi\,z)$

(9) $\quad\mathcal{FT}_x[U”(x)](z)=-\mathcal{FT}_x^{-1}[U”(x)](z)=4\,i\,\pi\,z\,\cos(2\,\pi\,z)$


The $sin$ and $cos$ terms associated with the Fourier series representations of $U(x)$, $U'(x)$, and $U”(x)$ also have special relationships between their Fourier transforms and inverse Fourier transforms which are illustrated below.

(10) $\quad\mathcal{FT}_x[\sin(a\,x)](z)=-\mathcal{FT}_x^{-1}[\sin(a\,x)](z)=i\,\pi\,\delta (a+2\,\pi\,z)-i\,\pi\,\delta(a-2\,\pi\,z)$

(11) $\quad\mathcal{FT}_x[\cos(a\,x)](z)=\mathcal{FT}_x^{-1}[\cos(a\,x)](z)=\pi\,\delta(a+2\,\pi\,z)+\pi\,\delta (a-2\,\pi\,z)$


Term-wise integration of the Fourier series representation of $U(x)$, $U'(x)$, and $U”(x)$ implies the Fourier transforms for the Fourier series representation of each of these functions consists of an infinite number of Dirac delta functions which is obviously inconsistent with the Fourier transforms of the distributional representation of $U(x)$, $U'(x)$, and $U”(x)$ defined in (7), (8), and (9) above.

The reason for the seeming discrepancies between the Fourier transforms of the distributional and Fourier series representations for $U(x)$, $U'(x)$, and $U”(x)$ is the idealization of the Fourier transforms of $sin$ and $cos$ functions as Dirac delta ($\delta$) functions is somewhat inaccurate and misleading as this idealization doesn’t hold up to reality. Note this discrepancy exists even for the Fourier transforms of the distributional and Fourier series representations of the Dirac comb. In the case of the distributional and Fourier series representations of the Dirac comb each representation transforms into the other representation which I suppose makes the discrepancy a bit easier to accept, but the case being explored here is considerably different.


The bilateral Laplace transforms of the Fourier series representations of $U(x)$, $U'(x)$, and $U”(x)$ don’t converge for $Re(y)=0$, and consequently they cannot be evaluated at $y=2\,\pi\,i\, z$ to obtain their corresponding Fourier transforms. However, the contribution of the left and right-half planes can be evaluated at $y=2\,\pi\,i\,z-\epsilon$ and $y=2\,i\,\pi\,z+\epsilon$ respectively and summed together to approximate their corresponding Fourier transforms for $z\in\mathbb{R}$.

The approximations to the Fourier transforms of the Fourier series representations of $U(x)$, $U'(x)$, and $U”(x)$ converge to the Fourier transforms of the distributional representations of $U(x)$, $U'(x)$, and $U”(x)$ defined in (7), (8), and (9) above for $z\in\mathbb{R}$ as $\epsilon\to 0^+$ as illustrated on child page 11.2 below.


  • Mellin Transforms

The Mellin transforms of the distributional representations of $U(x)$, $U'(x)$, and $U”(x)$ defined in (1), (2), and (3) above are as follows.

(12) $\quad\mathcal{M}_x[U(x)](s)=-\frac{1}{s}\,,\quad\Re(s)<0$

(13) $\quad\mathcal{M}_x[U'(x)](s)=1$

(14) $\quad\mathcal{M}_x[U”(x)](s)=1-s$


I don’t believe the Mellin transforms of the Fourier series representations of $U(x)$, $U'(x)$, and $U”(x)$ converge when evaluated at finite upper integration bounds, but I believe at infinite upper integration bounds it can be proven these transforms are equivalent to the Mellin transforms of the distributional representations defined in (12), (13), and (14) above using the definitions of the Riemann zeta function and the Riemann functional equation which analytically extends the Riemann zeta function to the entire complex plane.