Find range of the function $f : \mathbb{R} \to \mathbb{R}$
$$f(x)=\frac{\left\{x\right\}}{1+(\lfloor x\rfloor)^2}$$
My try:
Obviously range contains zero, since for integers $\left\{x\right\}=0$
If $x$ is not an integer we have $x=n+f$ Then
$$f(n+f)=\frac{f}{1+n^2} \lt 1$$
So range is $\left[0, \:\: 1\right)$
is this right approach?
For negative number $-n-f$, $$f(-n-f)=\frac{-f}{1+(n+1)^2}$$
$$0\le f<1\Leftrightarrow 0\ge f(-n-f)>\frac{-1}{1+(n+1)^2}\ge-\frac12$$
Therefore, the range is $(-\frac12,1)$.
Bonus: $$\int_0^\infty f(x)dx=\frac{\pi \coth \pi +1}4$$