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How would you go about solving integral of a floor? The particular problem I have is:

$$\int \,\left\lfloor\frac{1}{x}\right\rfloor\, dx$$

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A simple picture should help with this one: –  JavaMan Apr 17 '11 at 21:37
If $\left\lfloor x\right\rfloor$ is constant on $[n,n+1)$, then $\left\lfloor\frac{1}{x}\right\rfloor$ is constanct on ... –  Rasmus Apr 17 '11 at 21:56

1 Answer 1

up vote 10 down vote accepted

The function:


is equal to $n$ on the interval $(\frac{1}{n},\frac{1}{n+1})$,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 $(\frac{1}{2},\frac{1}{3})$, etc. So since $t<\frac{1}{n}$, we get terms for each interval $(\frac{1}{k},\frac{1}{k+1})$ 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:

$$x\left\lfloor\frac{1}{x}\right\rfloor - \sum_{i=2}^{\left\lfloor\frac{1}{x}\right\rfloor}\frac{1}{i} + C$$

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Ah I get it, so we basically interpret the floor function as a constant. Thanks for the detailed explanation. :) –  Paul Manta Apr 19 '11 at 14:54

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