@Pranav Arora
–
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.
–
Haroun Meghaichi
·
3 years, 2 months ago

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

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 \(\boxed{-1}\)
–
Kevin Catbagan
·
3 years, 2 months ago

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TopNewestWith the substitution \(\dfrac{1}{x}=t\), the integral is:

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

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

\(\displaystyle =2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots \right)=2\left(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots \right)-\frac{1}{2}\right)=\boxed{2\ln 2-1} \) – Pranav Arora · 3 years, 2 months ago

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

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 – Vaibhav Sawhney · 3 years, 2 months ago

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– Haroun Meghaichi · 3 years, 2 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.Log in to reply

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 – Vaibhav Sawhney · 3 years, 2 months ago

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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 \) – Bill Bell · 3 years, 2 months ago

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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 \( \LaTeX \)– Michael Mendrin · 3 years, 2 months agoLog in to reply

C – Akram Elbarsha · 3 years, 1 month ago

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2*log(2) -1 – Surajit Rajagopal · 3 years, 2 months ago

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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 \(\boxed{-1}\) – Kevin Catbagan · 3 years, 2 months ago

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– Haroun Meghaichi · 3 years, 2 months ago

The upper bound in the \(m\) sum should be \(2N-1\).Log in to reply

– Kevin Catbagan · 3 years, 2 months ago

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