# A floor integral

Evaluate : $\int\limits_0^1\left\lfloor\frac{2}{x}\right\rfloor-2\left\lfloor\frac{1}{x}\right\rfloor\ \mathrm{d}x.$

Note by Haroun Meghaichi
4 years, 5 months ago

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With the substitution $$\dfrac{1}{x}=t$$, the integral is:

$$\displaystyle \int_1^{\infty} \frac{\lfloor 2t \rfloor-2\lfloor t \rfloor}{t^2}\,dt = \sum_{n=1}^{\infty} \left(\int_n^{n+1/2} \frac{(2n)-2(n)}{t^2}\,dt+\int_{n+1/2}^{n+1} \frac{(2n+1)-2(n)}{t^2}\,dt \right)$$

$$\displaystyle =\sum_{n=1}^{\infty} \int_{n+1/2}^{n+1} \frac{(2n+1)-2(n)}{t^2}\,dt =\sum_{n=1}^{\infty} \left(\frac{2}{2n+1}-\dfrac{2}{2n+2}\right)$$

$$\displaystyle =2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots \right)=2\left(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots \right)-\frac{1}{2}\right)=\boxed{2\ln 2-1}$$

- 4 years, 5 months ago

I'm a little confused. First I split the integral into two

$$\int_{0}^{1} \left \lfloor \frac{2}{x} \right \rfloor dx - 2\int_{0}^{1} \left \lfloor \frac{1}{x} \right \rfloor dx$$

In the first one I substituted x = 2t, to get

$$2\int_{0}^{1/2} \left \lfloor \frac{1}{t} \right \rfloor dt - 2\int_{0}^{1} \left \lfloor \frac{1}{x} \right \rfloor dx$$

which equals

$$-2 \int_{1/2}^{1} \left \lfloor \frac{1}{t} \right \rfloor dt$$

Now since t lies between 1/2 and 1, 1/t will lie between 1 and 2, therefore the integral will change to

$$-2 \int_{1/2}^{1} 1 dt$$

which will evaluate to -1

- 4 years, 5 months ago

Exactly my method to do it. Good job. ^ There is a small remark, if $$n$$ is an integer we may not always get : $\int_a^{\infty} f(t)\ \mathrm{d}t = \lim_{n\to \infty} \int_a^n f(t) \ \mathrm{d}t$ For example $$f(t)=\sin(2\pi t)$$, the first step should be checking whether it converges, if it does we can use this trick.

- 4 years, 5 months ago

I'm a little confused. First I split the integral into two

$$\int_{0}^{1} \left \lfloor \frac{2}{x} \right \rfloor dx - 2\int_{0}^{1} \left \lfloor \frac{1}{x} \right \rfloor dx$$

In the first one I substituted x = 2t, to get

$$2\int_{0}^{1/2} \left \lfloor \frac{1}{t} \right \rfloor dt - 2\int_{0}^{1} \left \lfloor \frac{1}{x} \right \rfloor dx$$

which equals

$$-2 \int_{1/2}^{1} \left \lfloor \frac{1}{t} \right \rfloor dt$$

Now since t lies between 1/2 and 1, 1/t will lie between 1 and 2, therefore the integral will change to

$$-2 \int_{1/2}^{1} 1 dt$$

which will evaluate to -1

- 4 years, 5 months ago

Looking at the rectangular areas under the line segments given by the two floor functions:

$$\int_0^1 \lfloor \frac{2}{x} \rfloor \mathrm{d}x \\ = 2\left(\frac{2}{2}-\frac{2}{3}\right)+3\left(\frac{2}{3}-\frac{2}{4}\right)+4\left(\frac{2}{4}-\frac{2}{5}\right)+5\left(\frac{2}{5}-\frac{2}{6}\right)+... \\ = \sum_{n=2}^{\infty} n\left(\frac{2}{n}-\frac{2}{n+1}\right) \\ = 2 \sum_{n=2}^{\infty} \frac{1}{n+1} \\ \int_0^1 \lfloor \frac{1}{x} \rfloor \mathrm{d}x \\ = 1\left(\frac{1}{1}-\frac{1}{2}\right)+2\left(\frac{1}{2}-\frac{1}{3}\right)+3\left(\frac{1}{3}-\frac{1}{4}\right)+4\left(\frac{1}{4}-\frac{1}{5}\right) ... \\ = \sum_{n=1}^{\infty} n \left( \frac{1}{n}-\frac{1}{n+1} \right) \\ = \sum_{n=1}^{\infty} \frac{1}{n+1} 2 \sum_{n=2}^{\infty} \frac{1}{n+1} - 2 \sum_{n=1}^{\infty} \frac{1}{n+1} = -2 \left( \frac{1}{2} \right) = -1$$

- 4 years, 5 months ago

The exact answer is $$\log(4)-1=0.386294...$$, which is from the infinite sum

$$\displaystyle\sum _{ n=1 }^{ \infty }{ \left(\dfrac { 2 }{ 2n+1 } -\dfrac { 2 }{ 2n+2 } \right) }$$

Edited your $$\LaTeX$$

- 4 years, 5 months ago

C

- 4 years, 4 months ago

2*log(2) -1

- 4 years, 5 months ago

My approach went like this:

First, note that

$$\displaystyle \int_0^1 \left\lfloor\dfrac{1}{x}\right\rfloor\;dx = \displaystyle\lim_{N\to\infty} \int_{1/N}^1 \left\lfloor\dfrac{1}{x}\right\rfloor\;dx$$

$$\left\lfloor\dfrac{1}{x}\right\rfloor = n$$ whenever $$n\leq \dfrac{1}{x} < n+1$$, or $$\dfrac{1}{n+1}<x\leq\dfrac{1}{n}$$. This means that

$$\displaystyle \int_{1/(n+1)}^{1/n} \left\lfloor\dfrac{1}{x}\right\rfloor\;dx = \displaystyle \int_{1/(n+1)}^{1/n} n\;dx = n\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$$ for every $$n\in\mathbb{N}$$

This now leads us to split the floor integral into a sum, with intervals $$(1/(n+1),1/n]$$:

$$\displaystyle\lim_{N\to\infty} \int_{1/N}^1 \left\lfloor\dfrac{1}{x}\right\rfloor\;dx =\displaystyle\lim_{N\to\infty} \sum_{n=1}^{N-1} \displaystyle \int_{1/(n+1)}^{1/n} n\;dx$$.

Working in a similar way with $$\left\lfloor\dfrac{2}{x}\right\rfloor$$, we find that our integral is supposed to be equal to

$$\displaystyle\lim_{N\to\infty}\left( \sum_{m=2}^{N-1} \displaystyle \int_{2/(m+1)}^{2/m} m\;dx-2\sum_{n=1}^{N-1} \displaystyle \int_{1/(n+1)}^{1/n} n\;dx \right)$$.

After making the lower limit of the sums equal, the variables similar and expanding the integrals, the sums should cancel out and what is left is one term

$$-2\displaystyle \int_{1/(n+1)}^{1/n} n\;dx$$ when $$n=1$$

as a result of making the lower limits equal. This is equal to $$\boxed{-1}$$

- 4 years, 5 months ago

The upper bound in the $$m$$ sum should be $$2N-1$$.

- 4 years, 5 months ago

thanks haha, ok let me rewrite. I was thinking, "wait this is answer is too nice"

- 4 years, 5 months ago