Using Riemann's methods to derive the floor function of x from the zeta function

Question

I thought it would be interesting to try and use Riemann's methods to derive (rather than the quantity of primes less than $x$) the quantity of natural numbers less than $x$. This function is, of course, the floor function $\lfloor x \rfloor$.
First we write the definition for the Zeta function: \[\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}\] \[\frac{1}{n^s}=s\int_n^\infty x^{-s-1}dx\] \[\implies\frac{\zeta(s)}{s}=\sum_{n=1}^{\infty}\int_n^\infty x^{-s-1}dx\] \[\frac{\zeta(s)}{s}=\int_0^\infty F(x)x^{-s-1}dx\] See that the function $F(x)$ has jumps at each whole number, which means that $F(x)=\begin{cases}0&x\leq0\\\lfloor x\rfloor&0<x\notin\mathbb{N}\\\lfloor x \rfloor-0.5&x\in\mathbb{N}\end{cases}$

Now by substitution of $x=e^{t}$ \[\frac{\zeta(s)}{s}=\int_{-\infty}^\infty F(e^{t})e^{-st}dt\] Which is a Laplace transform, meaning it can be inverted by \[F(e^{t})=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\zeta(s)}{s}e^{st}ds\] Now by reverting the substitution $t=ln(x)$ \[F(x)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\zeta(s)}{s}x^sds\] Now taking the residue at the two poles of the function $s=1$ and $s=0$ we find that $Res(\frac{\zeta(s)}{s}x^s,1)=x$ and $Res(\frac{\zeta(s)}{s}x^s,0)=-\frac{1}{2}$ which means that $F(x)=x-\frac{1}{2}$ which is a good approximation, but is not the exact value of this function, which is what i was expecting. Can anybody tell me what went wrong in these steps.

Answer 1

The relationship between the floor function representation

$$\lfloor x\rfloor=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \theta(x-n)\right),\quad x\ge 0\tag{1}$$

(where $\theta(x)$ is the Heavide step function) and the Dirichlet series

$$\zeta(s)=\underset{N\to\infty}{\text{lim}} \left(\sum_{n=1}^N \frac{1}{n^s}\right),\quad\Re(s)>1\tag{2}$$

for the Riemann zeta function is the term-wise Mellin and inverse Mellin transforms

$$s\, \mathcal{M}_x[\theta(x-n)](-s)=s \int\limits_0^\infty \theta(x-n)\, x^{-s}\, dx=s \int\limits_n^\infty x^{-s}\, dx=\frac{1}{n^s}\tag{3}$$

and

$$\mathcal{M}_s^{-1}\left[\frac{1}{n^s}\, \frac{1}{s}\right]\left(\frac{1}{x}\right)=\int\limits_{\beta-i \infty}^{\beta+i \infty} \frac{1}{n^s}\, \frac{x^s}{s}\, ds=\theta(x-n)\tag{4}$$

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Also, the Fourier series representation

$$f(x)=x-\frac{1}{2}+\underset{K\to\infty}{\text{lim}} \left(\frac{1}{\pi} \sum\limits_{k=1}^K \frac{\sin(2 k \pi x)}{k}\right)\tag{5}$$

for the function

$$f(x)=\underset{\epsilon\to 0}{\text{lim}} \left(\frac{\lfloor x-\epsilon \rfloor +\lfloor x+\epsilon \rfloor}{2}\right)\tag{6}$$

which converges to $\lfloor x \rfloor$ except at integer values of $x$ can be derived from the inverse Mellin transform

$$\mathcal{M}_s^{-1}\left[\frac{\zeta(s)}{s}\right]\left(\frac{1}{x}\right)=\int\limits_{\alpha -i \infty}^{\alpha+i \infty} \zeta(s)\, \frac{x^s}{s}\, ds\,,\quad \alpha<0\tag{7}$$

where the inverse Mellin transform integral in formula (7) above is evaluated term-wise from the representation

$$\zeta(s)=\underset{K\to\infty}{\text{lim}} \left(2^s\, \pi^{s-1}\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(1-s)\, \sum\limits_{k=1}^K \frac{1}{k^{1-s}}\right),\quad\Re(s)<0\tag{8}.$$

The result of the inverse Mellin transform integral in formula (7) above corresponds to the sum over $k$ in formula (5) above, but one must also account for the residues of $\zeta(s)\, \frac{x^s}{s}$ at $s=1$ and $s=0$ which correspond to the $x$ and $-\frac{1}{2}$ terms in formula (5) above.