It is an error (see my comment). You can graph
on Desmos.com by looking at the complement of
The Unshaded region below is the graph of your equation.
Alternatively, taking the complement of
Floor function error in Desmos? - Mathematics Stack Exchange
how do you type y=[[x]] in desmos?
Continuous approximations of the Floor and Ceiling functions
integration - Integral concerning the floor function - Mathematics Stack Exchange
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It is an error (see my comment). You can graph
on Desmos.com by looking at the complement of
The Unshaded region below is the graph of your equation.
Alternatively, taking the complement of
It is not difficult to show that 
if and only if 
.
Try graphing in Desmos
Desmos does not show it accurately on the boundaries, however.
It is also the case that 
if and only if 
, but Desmos errs once again on the boundary.
i am looking to graph the integer-only line, but whenever i try to type [[]], it tells me that you cannot store a list of numbers in a list.
Inspired by the "absolute value of x without using absolute value" I thought I'd try to do the floor and ceiling function. Not an exact solution but it was fun!
Verification:
$$\begin{align} \int_{1}^{\infty}\frac{\sin\left(\frac{\pi}{2}\{x\}\right)^{[x]}}{[x]}\cos\left(\frac{\pi}{2}\{x\}\right)\,dx &=\sum_{k=1}^\infty\int_0^1\frac{\sin\left(\frac{\pi}{2}x\right)^k}k\cos\left(\tfrac{\pi}{2}x\right)\,dx \\&=\frac2\pi\sum_{k=1}^\infty\left.\frac{\sin\left(\frac{\pi}{2}x\right)^{k+1}}{k(k+1)}\right|_0^1\\&=\frac2\pi\sum_{k=1}^\infty\frac{1}{k(k+1)}. \end{align}$$
Given the tortuous shape of the function and its angular points, plus the slow decay, it is more than likely that numerical evaluation of this improper integral is challenged.
You might ask Desmos to evaluate on a few $[k,k+1]$ intervals. (WA replies exact values.)
Let $$f(x)=\frac{\sin\left(\frac{\pi}{2}\{x\}\right)^{[x]}}{[x]}\cos\left(\frac{\pi}{2}\{x\}\right)$$ and use Mathematica with the following syntax
R[p_]:=NIntegrate[f[x], {x, 1, Infinity}, WorkingPrecision -> p]
There is no problem or warning. Nice but the result depends very much on the value of the working precision $$\left( \begin{array}{cc} p & R(p) \\ 10 & 0.6317221476 \\ 20 & 0.6300419923 \\ 30 & 0.6319526217 \\ 40 & 0.6321773132 \\ 50 & 0.6327733152 \\ 60 & 0.6319059416 \\ 70 & 0.6320912143 \\ 80 & 0.6317180264 \\ 90 & 0.6318379131 \\ 100 & 0.6329749100 \\ 200 & 0.6326666679 \\ 300 & 0.6322638105 \\ 400 & 0.6321032001 \\ 500 & 0.6321174712 \\ 600 & 0.6322364980 \\ 700 & 0.6322364980 \\ \end{array} \right)$$
Even when the result stabilizes, we are still far away from $\frac 2 \pi$