The function $f(z)=\frac{\pi}{\sin \pi z}$ has simple poles of residue 1 at the integers. Hence, by the residue theorem, I consider the interesting idea of drawing a (perhaps rectangular, for example) contour around the real line. The value of the line integral would essentially count the amount of integers inside the contour, so there seems to be the possibility of writing the floor function as a contour integral and hence eventually a real-valued integral!
Could someone attempt this? I can't seem to get it completely right.
Perron's formula gives $$ \lfloor x\rfloor = \frac1{2\pi i} \int_{2-i\infty}^{2+i\infty} \zeta(s) \frac{x^s}s\, ds = \frac1\pi \int_0^\infty \Re\bigg( \zeta(2+it) \frac{x^{2+it}}{2+it} \bigg)\,dt. $$ Note this only works for nonintegers $x$: when $x$ is a positive integer, the left-hand side is $x$, while the right-hand side is $x-\frac12$.