Integral equal to Riemann Zeta Function

Question

As part of a homework problem in Rudin, I need calculate

$$ \int_{1}^{N} \frac{[x]}{x^{s+1}} \,dx$$ where $[x]$ is the floor function. Clearly $[x]$ has derivative $0$ everywhere but the integers.

Is the following correct?:

$$\begin{align} \int_{1}^{N} \frac{[x]}{x^{s+1}} \,dx &= \Big( -\frac{[x]}{s}x^{-s} \Big)_{1}^{N} + \int_{1}^{N} \frac{d}{dx}([x])\frac{1}{x^{s+1}} \,dx \\ &= \Big( -\frac{[x]}{s}x^{-s} \Big)_{1}^{N} + \int_{1}^{2} \frac{d}{dx}([x])\frac{1}{x^{s+1}} \,dx + \cdots + \int_{N-1}^{N} \frac{d}{dx}([x])\frac{1}{x^{s+1}} \,dx \\ &= \Big( -\frac{[x]}{s}x^{-s} \Big)_{1}^{N} + 0 \\ &=\frac{-1}{s} \Big(\frac{N}{x^{-s}} -1 \Big) \end{align}$$

Is this a correct calculation? If not, please do not explicitly give the correct answer, rather a hint as to where I've gone wrong would be nice.

Answer 1

You can not apply integration by parts here because one of the function is not continue, as Hint the right way is to write your intgral as: $$\int_{1}^{N} \frac{[x]}{x^{s+1}} \,dx=\sum_{i=1}^{N-1}\int_{i}^{i+1} \frac{[x]}{x^{s+1}} $$

now what is the value of $[x]$ in the interval $[i,i+1]$ and continue your calculations.

Answer 2

We have \begin{align} \int_1^N \dfrac{\lfloor x \rfloor}{x^{s+1}}dx & = \sum_{k=1}^{N-1} \int_k^{k+1} \dfrac{\lfloor x \rfloor}{x^{s+1}}dx = \sum_{k=1}^{N-1}k \int_k^{k+1} \dfrac{dx}{x^{s+1}} = \sum_{k=1}^{N-1}\dfrac{k}s \left(\dfrac1{k^s} - \dfrac1{(k+1)^s}\right)\\ & = \dfrac1s \sum_{k=1}^{N-1}\left(\dfrac1{k^{s-1}} - \dfrac{k}{(k+1)^s}\right) = \dfrac1s \sum_{k=1}^{N-1}\left(\dfrac1{k^{s-1}} - \dfrac{k+1}{(k+1)^s} + \dfrac1{(k+1)^s}\right)\\ & = \dfrac1s \sum_{k=1}^{N-1}\left(\dfrac1{k^{s-1}} - \dfrac1{(k+1)^{s-1}} + \dfrac1{(k+1)^s}\right) = \dfrac1s\left(1-\dfrac1{N^{s-1}} + \sum_{k=1}^{N-1} \dfrac1{(k+1)^s}\right)\\ & = \dfrac1s\left(\sum_{k=1}^N \dfrac1{k^s} - \dfrac1{N^{s-1}}\right) \end{align}

Answer 3

Your approach is incorrect. Under the framework of Riemann integral, one can not integrate by part an integrand which has jump discontinuities.

If you really want to use integrate by part, you can interpret the integral as a Riemann–Stieltjes integral. One has

$$\int_1^N \frac{\lfloor x \rfloor}{x^{s+1}} dx = \int_{1^-}^{N^+} \frac{\lfloor x \rfloor}{x^{s+1}} dx = -\frac1s\int_{1^-}^{N^+} \lfloor x \rfloor d\frac{1}{x^s}\\ = -\frac1s\left( \left[\frac{\lfloor x \rfloor}{x^s}\right]_{1^-}^{N^+} - \int_{1^-}^{N^+} \frac{1}{x^s} d\lfloor x \rfloor \right)\tag{*1} $$ Notice $\lfloor N^+ \rfloor = N$ and $\lfloor 1^- \rfloor = 0$, we have $$\left[\frac{\lfloor x \rfloor}{x^s}\right]_{1^-}^{N^+} = \frac{1}{N^{s-1}}$$

For the integral in last expression of $(*1)$, notice $\lfloor x \rfloor$ is constant over intervals of the form $(k,k+1), k \in \mathbb{Z}$ and has a jump discontinuity $1$ at $k \in \mathbb{Z}$, i.e

$$\lfloor k^+ \rfloor - \lfloor k^- \rfloor = \lim_{\epsilon\to 0^+} \left(\lfloor k+\epsilon \rfloor - \lfloor k-\epsilon \rfloor \right) = k - (k-1) = 1$$ The integral in last expression of $(*1)$ reduces to a sum: $$\int_{1^-}^{N+} \frac{1}{x^s} d\lfloor x \rfloor = \sum_{k=1}^N \frac{1}{k^s}$$ Combine these, we have

$$\int_1^N \frac{\lfloor x \rfloor}{x^{s+1}} dx = \frac1s \left( \sum_{k=1}^N\frac{1}{k^s} - \frac{1}{N^{s-1}}\right)$$

The same answer you will get if you continue to treat the integral as Riemann integral but split it as a sum of integrals over intervals of the form $(k,k+1)$, $1 \le k < N$.