Integral of floor function: $\int \,\left\lfloor\frac{1}{x}\right\rfloor\, dx$ How would you go about solving integral of a floor? The particular problem I have is:
$$\int \,\left\lfloor\frac{1}{x}\right\rfloor\, dx$$
 A: The function:
$$\left\lfloor\frac{1}{x}\right\rfloor$$
is equal to $n$ on the interval $\left(\frac{1}{n+1},\frac{1}{n}\right)$,so if we try to determine the integral from $t>0$ to $1$, we can let $n=\left\lfloor\frac{1}{t}\right\rfloor$ and we have constant value $1$ on range $(\frac{1}{2},1)$, constant value $2$ on range $(\frac{1}{3},\frac{1}{2})$, etc.  So since $t<\frac{1}{n}$, we get terms for each interval $(\frac{1}{k+1},\frac{1}{k})$ when $k<n.$  The length of the $k$th interval is $\frac{1}{k(k+1)}$ and the value of the function is $k$ on this interval, so the integral on this interval is $\frac{1}{k+1}$.  So the integral from $\frac{1}{n}$ to $1$ is $1/2 + 1/3 + 1/4 + ... + 1/n$.  Then then integral from $t$ to $\frac{1}{n}$ is the length of the interval times $n$, which is $n(\frac{1}{n} - t) = 1-nt$.  So the total is:
$$\int_t^1 \,\left\lfloor\frac{1}{x}\right\rfloor\, dx = 1 - t{\left\lfloor\frac{1}{t}\right\rfloor} + 
\sum_{i=2}^{\left\lfloor\frac{1}{t}\right\rfloor}\frac{1}{i}$$
The indefinite integral, then, is the opposite of this:
$$x\left\lfloor\frac{1}{x}\right\rfloor - \sum_{i=2}^{\left\lfloor\frac{1}{x}\right\rfloor}\frac{1}{i} + C$$
