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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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4  
A simple picture should help with this one: en.wikipedia.org/wiki/File:Floor_function.svg – JavaMan Apr 17 '11 at 21:37
5  
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
up vote 12 down vote accepted

The function:

$$\left\lfloor\frac{1}{x}\right\rfloor$$

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

The integral is true for treating it as a constant. However, the series I obtained is $\frac 1n$ from $0$ to $[\frac 1x]$. It's important to realize that the lower boundary has to do with the definite integral. Also, I get an added series unlike the other answer here.

The final answer is:

$$ x[\frac 1x] + \sum^{[\frac 1x]}_{n=0} \frac 1n $$

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This answer does not make sense. Please clarify or delete the answer. – Rory Daulton Nov 27 '15 at 0:01
    
I'm not sure what you are saying isn't clear. – TheGreatDuck Nov 27 '15 at 23:19
    
Wait no. I got the jump series completely wrong! – TheGreatDuck Nov 29 '15 at 1:07
    
I see now that my answer is indeterminate as there is not a valid jump series for x[1/x] at this current point in time. However, my two-step algorithm for solving floor function based integrals is still valid as it is a simple question of how to remove the jumps from a broken graph. It just happens to be that the divergence surrounding x = 0 produces... Bad results. – TheGreatDuck Jan 14 at 18:48

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