3.1 $U[x]=-1+\theta[x+1]+\theta[x-1]$

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$U[x]$ – A function which takes unit steps of at $|x|=1$.

The $U[x]$ function isn’t really a prime counting function, but the Fourier series representation of the $U[x]$ function is closely related to the Fourier series representations of the rational and log-step staircase functions $S[x]$ and $T[x]$ and the six prime counting functions illustrated on this website. The $U[x]$ function is defined as follows.

$\quad U[x]=-1+\theta[x+1]+\theta[x-1]$


The following plot illustrates the $U[x]$ function takes unit steps at $|x|=1$.

$U[x]=-1+\theta[x+1]+\theta[x-1]$

The first and second-order derivatives of $U[x]$ are as follows

$\quad U'[x]=\delta[x+1]+\delta[x-1]$

$\quad U”[x]=\delta'[x+1]+\delta'[x-1]$


Note that $U'[x]$ evaluates to $\delta[x-1]$ in the right-half plane and the Laplace transform of $U'[x]$ is $e^{-y}$. Consequently the Laplace transform of the Fourier series representation of $U'[x]$ provides a new formula for $e^{-y}$ which is illustrated on page 12.1. This new formula can also be used to derive new formulas for functions which are defined by integrals containing an $e^{-y}$ term such as $\frac{1}{x}$, $E_i[x]$, $E_n[x]$, and $\Gamma[a,\,x]$.


Also note that the Mellin transform of $\delta'[x-1]$ is one, and the Fourier series representation of $U'[x]$ can be substituted in Mellin convolutions such as the following to derive new formulas for a wide variety of functions. This topic is addressed further on page 13.

$\quad g(y)=\delta(x-1)\,*_\mathcal{M_1}\,g(x)=\int_0^\infty \frac{\delta(x-1)}{x}\,g\left(\frac{y}{x}\right)\,dx$

$\quad g(y)=\delta(x-1)\,*_\mathcal{M_2}\,g(x)=\int_0^\infty\delta(x-1)\,g(y\,x)\,\,dx$