A floor integral

Evaluate : ∫01⌊2x⌋−2⌊1x⌋ dx. \int\limits_0^1\left\lfloor\frac{2}{x}\right\rfloor-2\left\lfloor\frac{1}{x}\right\rfloor\ \mathrm{d}x.

#Calculus #Integration #Integrals

Note by Haroun Meghaichi
6 years, 11 months ago

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Comments

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

∫1∞⌊2t⌋−2⌊t⌋t2 dt=∑n=1∞(∫nn+1/2(2n)−2(n)t2 dt+∫n+1/2n+1(2n+1)−2(n)t2 dt)\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)

=∑n=1∞∫n+1/2n+1(2n+1)−2(n)t2 dt=∑n=1∞(22n+1−22n+2)\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)

=2(13−14+15−16+⋯ )=2((1−12+13−14+⋯ )−12)=2ln⁡2−1\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}

Pranav Arora - 6 years, 11 months ago

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Exactly my method to do it. Good job. ^ There is a small remark, if nn is an integer we may not always get : ∫a∞f(t) dt=lim⁡n→∞∫anf(t) dt\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⁡(2πt)f(t)=\sin(2\pi t), the first step should be checking whether it converges, if it does we can use this trick.

Haroun Meghaichi - 6 years, 11 months ago

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

∫01⌊2x⌋dx−2∫01⌊1x⌋dx \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∫01/2⌊1t⌋dt−2∫01⌊1x⌋dx 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∫1/21⌊1t⌋dt -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∫1/211dt -2 \int_{1/2}^{1} 1 dt

which will evaluate to -1

Vaibhav Sawhney - 6 years, 11 months ago

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

∫01⌊2x⌋dx−2∫01⌊1x⌋dx \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∫01/2⌊1t⌋dt−2∫01⌊1x⌋dx 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∫1/21⌊1t⌋dt -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∫1/211dt -2 \int_{1/2}^{1} 1 dt

which will evaluate to -1

Vaibhav Sawhney - 6 years, 11 months ago

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

∑n=1∞(22n+1−22n+2)\displaystyle\sum _{ n=1 }^{ \infty }{ \left(\dfrac { 2 }{ 2n+1 } -\dfrac { 2 }{ 2n+2 } \right) }

Edited your LaTeX \LaTeX

Michael Mendrin - 6 years, 11 months ago

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

∫01⌊2x⌋dx=2(22−23)+3(23−24)+4(24−25)+5(25−26)+...=∑n=2∞n(2n−2n+1)=2∑n=2∞1n+1∫01⌊1x⌋dx=1(11−12)+2(12−13)+3(13−14)+4(14−15)...=∑n=1∞n(1n−1n+1)=∑n=1∞1n+12∑n=2∞1n+1−2∑n=1∞1n+1=−2(12)=−1 \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

Bill Bell - 6 years, 11 months ago

My approach went like this:

First, note that

∫01⌊1x⌋  dx=lim⁡N→∞∫1/N1⌊1x⌋  dx\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

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

∫1/(n+1)1/n⌊1x⌋  dx=∫1/(n+1)1/nn  dx=n(1n−1n+1)\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∈Nn\in\mathbb{N}

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

lim⁡N→∞∫1/N1⌊1x⌋  dx=lim⁡N→∞∑n=1N−1∫1/(n+1)1/nn  dx\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 ⌊2x⌋\left\lfloor\dfrac{2}{x}\right\rfloor, we find that our integral is supposed to be equal to

lim⁡N→∞(∑m=2N−1∫2/(m+1)2/mm  dx−2∑n=1N−1∫1/(n+1)1/nn  dx)\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∫1/(n+1)1/nn  dx-2\displaystyle \int_{1/(n+1)}^{1/n} n\;dx when n=1n=1

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

Kevin Catbagan - 6 years, 11 months ago

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The upper bound in the mm sum should be 2N−12N-1.

Haroun Meghaichi - 6 years, 11 months ago

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

Kevin Catbagan - 6 years, 11 months ago

2*log(2) -1

Surajit Rajagopal - 6 years, 11 months ago

C

Akram Elbarsha - 6 years, 10 months ago
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