# Integral representations of $\zeta(s)$ using the floor/frac functions. How could this one be derived?

Browsing the web, I found quite a few integral representations for $\zeta(s)$ that use the Fractional part {x} or the Floor-function $\lfloor x\rfloor$ e.g.:

$$\zeta(s) = \dfrac{s}{s-1} - \frac12+s \int_1^\infty \frac{1/2-\{x\}}{x^{s+1}}\,\mathrm{d} x \qquad \Re(s)\gt 0\qquad(1)$$

or

$$\zeta(s) = \dfrac{s}{s-1} - s \int_1^\infty \frac{\{x\}}{x^{s+1}}\,\mathrm{d} x \qquad \Re(s)\gt 0\qquad(2)$$

Numerical evidence also suggests that the following expression holds:

$$\zeta(s) = \dfrac{1}{s-1} + \int_1^\infty \left(\frac{1}{\lfloor x \rfloor^{s}}-\frac{1}{x^s}\right)\,\mathrm{d} x \qquad \Re(s)\gt 0 \qquad(3)$$

That also leads to the very simple (trivial) integral:

$$\zeta(s) = \int_1^\infty \frac{1}{\lfloor x \rfloor^{s}}\,\mathrm{d} x \qquad \Re(s)\gt 1$$

Is there a way to derive (3) from (1) or (2)?

Thanks.

• See relation $(2.1.4)$ here. – Lucian Sep 12 '15 at 2:14