Formula for the floor function

It is well-known that the cotangent function has period $\pi$, so that the cotangent of $\pi x$ has period $1$.

With the usual definition of the arc tangent, you get a number between $-\frac\pi2$ and $\frac\pi2$, or, after division by $\pi$, a number between $-\frac12$ and $\frac12$.

Hence,

$$\frac{\arctan\cot\pi x}\pi=\frac{\arctan\tan\left(\dfrac\pi2-\pi x\right)}\pi=\frac12-x\bmod1$$

Now,

$$-\frac12+\frac12+x-x\bmod1=x-\{x\}=\lfloor x\rfloor.$$

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Anyway, the formula is virtually useless, as it is undefined for integer $x$.

Hint:

Write down $x=n + \frac12 + q$ where $n$ is an integer and $q\in\left(-\frac12, \frac12\right)$.

Then use trigonometric addition theorems and the fact that $$\cot(n\pi+\frac\pi2) = \frac{\cos(n\pi + \frac\pi2)}{\sin(n\pi +\frac\pi2)} = 0 $$

Consider

$$f(x) = \frac{1}{\pi}\int_{0}^{\cot(\pi x)}\frac{dy}{1+y^2}=\frac{\arctan(\cot(\pi x))}{\pi},$$

with the latter equation obtained by substituting $y=\tan u.$ Since $x\to \cot(\pi x)$ is manifestly periodic of period $1,$ and $f$ integrates to $0$ over one period (since $\cot$ is an odd function), $f$ also is periodic.

When $x$ is not an integer $f$ is differentiable at $x$ because both $\cot(\pi x)$ and the integral (qua function of its upper limit) are. The Chain Rule and Fundamental Theorem of Calculus together imply

$$f^\prime(x) = \frac{1}{\pi}\left(\frac{1}{1 + (\cot(\pi x))^2}\right) \frac{d}{dx}\left(\cot(\pi x)\right)=\frac{-\pi\csc(\pi x)^2}{\pi\csc(\pi x)^2}=-1.$$

This is the key insight, because it shows $f$ has the basic properties needed to construct functions that are periodic and linear between their points of discontinuity. The rest is just algebra.

Since $f(1/2)$ is an integral from $0$ to $0=\cot(\pi/2),$

$$f(1/2) = 0.$$

This information completely determines $f.$ To summarize, at nonintegral values $f$ falls linearly with slope $-1,$ equals $0$ at $1/2,$ and repeats this pattern between each successive pair of integers. Consequently the function

$$\frac{1}{2} - f(x)$$

must rise linearly from $0$ at $x=0$ up to a limit of $1$ as $x\to 1.$ Because it is periodic, it drops back to $0$ when $x=1$ and repeats this pattern ad infinitum in both directions. Obviously that describes the remainder ("fractional part") function. That is,

$$x - \lfloor x \rfloor = \frac{1}{2} - f(x).$$

Solving for the floor,

$$\lfloor x \rfloor = x - \left( \frac{1}{2} - f(x)\right) = -\frac{1}{2} + x + \frac{\arctan(\cot(\pi x))}{\pi}.$$