Approximation of $\arctan(\cot(\pi x))$ That is Integrable

Question

As per the title, are there any approximations of $\arctan(\cot(\pi x))$, with arbitrary error (that hopefully can be controlled by some parameter $k$ ) whose integral can be expressed with elementary functions. I know there's a Fourier series, but it requires more terms to get more accurate, and it can become computationally expensive to calculate.

Answer 1

Write $\{x\} = x - \lfloor x\rfloor$. Since $$\arctan(\cot(\pi x)) = \arctan(\cot(\pi\{x\})) = \arctan(\tan(\frac{\pi}{2}-\pi\{x\})) \\ -\frac{\pi}{2} < \frac{\pi}{2} - \pi\{x\} \leq \frac{\pi}{2}$$ it follows that $$\arctan(\cot(\pi x)) = \frac{\pi}{2} - \pi\{x\}, \qquad x \in \mathbb{R}\setminus\mathbb{Z}.$$

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Arguably $\arctan$ and $\cot$ are both "elementary" functions, but this gives an exact way to write $\arctan(\cot(\pi x))$ without appealing to anything more than linear functions and the operation of "rounding down to the nearest integer".

It has as an antiderivative the function $$\int_0^x \left(\frac{\pi}{2} - \pi\{t\}\right)dt = \frac{\pi}{2}\{x\}(1-\{x\})$$

Answer 2

We have

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

so it is continuous on intervals $[n,n+1)$ for $n$ an integer.