Determine Lebesgue integral of a function containing floor function

Question

I'm practicing for the real-analysis exam and I've got stuck at this integral... Could you help me, please?
Determine: $$ \int_{[0,\infty)}\dfrac{1}{\lfloor{x+1}\rfloor\cdot\lfloor{x+2}\rfloor}d\lambda(x).$$
It is ok to split in two integrals and to obtain two natural logarithms?

Answer 1

For $x\in[k, k+1)$, where $k\in\mathbb{N}$, we have $\lfloor{x}\rfloor=k$. Also, note that for any $l\in\mathbb{N}$ we have $\lfloor{x+l}\rfloor = \lfloor{x}\rfloor + l$. \begin{align} \int_0^n \frac{1}{\lfloor{x+1}\rfloor\lfloor{x+2}\rfloor}\ dx &= \sum_{k=0}^{n-1} \int_{k}^{k+1} \frac{1}{\lfloor{x+1}\rfloor\lfloor{x+2}\rfloor}\ dx\\ &= \sum_{k=0}^{n-1} \frac{1}{(k+1)(k+2)}\\ &= \sum_{k=0}^{n-1} \left( \frac{1}{k+1} - \frac{1}{k+2} \right)\\ &= \sum_{k=0}^{n-1} \frac{1}{k+1} - \sum_{k=1}^{n} \frac{1}{k+1}\\ &= 1 - \frac{1}{n+1} \text{.} \end{align}

Answer 2

Hint: the value of the integrand changes only at integer values of $x$. For $x\in[0,1)$ it's $\frac{1}{1\cdot 2}$. For $x\in[1,2)$, it's $\frac{1}{2\cdot 3}$. Can you figure out the value for $x\in [n, n+1)$? This means that you can express the integral as a sum (i.e., a series). I don't see a way to find the exact value of the series though.