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anonymous

  • one year ago

Consider the function defined sequentially by \[f_n(x)= \begin{cases} \dfrac{(-1)^n}{n} & \text{for }x\in\left(\dfrac{1}{n+1},\dfrac{1}{n}\right]\\[1ex] 0 & \text{else} \end{cases}\]where \(n\in\mathbb{N}\). What's the value of \[\int_0^1 f_n(x)\,dx~?\]

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  1. ganeshie8
    • one year ago
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    \[\dfrac{(-1)^n}{n^2(n+1)}\] ?

  2. anonymous
    • one year ago
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    Yep, it reduces nicely into an infinite series. I'm curious to see how one might find the exact value of it.

  3. anonymous
    • one year ago
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    ^without prior knowledge of the polylogarithm. I've seen the answer on W|A.

  4. ganeshie8
    • one year ago
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    I thought the integral evaluates to that expression.. hmm let me go read it again :)

  5. anonymous
    • one year ago
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    \[\int_0^1f_n(x)\,dx=\sum_{n=1}^\infty\left(\frac{(-1)^n}{n^2}+\frac{(-1)^{n+1}}{n(n+1)}\right)\](you can see this equivalence easily if you plot \(f_n\); rectangles of width \(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1}{n(n+1)}\) and height \(\dfrac{(-1)^n}{n}\)) The second term gives a telescoping sum that's easy to deal with if you use the power series for \(-\ln(1+x)\). The first term is trickier.

  6. anonymous
    • one year ago
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  7. anonymous
    • one year ago
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    \[\int_0^1f_n(x)dx=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2(n+1)}\]\[=\sum_{n=1}^\infty(\frac{1}{(2n)^2}-\frac{1}{2n}+\frac{1}{2n+1})-\sum_{n=1}^\infty(\frac{1}{(2n-1)^2}-\frac{1}{2n-1}+\frac{1}{2n})\]\[=\frac{\pi^2}{24}+\ln(2)-1-(\frac{\pi^2}{8}-\ln(2))=\ln(4)-\frac{\pi^2}{12}-1\]Not sure if you can follow my thought process, but I THINK I got it haha

  8. anonymous
    • one year ago
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    Well done! The thought to split into even and odd terms occurred to me a few moments ago. The rest is easy so long as we're familiar with the sum\[\sum_{n=1}^\infty \frac{1}{n^2}=\dfrac{\pi^2}{6}\]

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