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.

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 \(\boxed{-1}\)

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## Comments

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

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

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

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

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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.

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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

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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

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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 \( \LaTeX \)Log in to reply

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 \)

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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 \(\boxed{-1}\)

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

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

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2*log(2) -1

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C

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