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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 \[\int_0^1 f_n(x)\,dx~?\]
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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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[\dfrac{(1)^n}{n^2(n+1)}\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, it reduces nicely into an infinite series. I'm curious to see how one might find the exact value of it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0^without prior knowledge of the polylogarithm. I've seen the answer on WA.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I thought the integral evaluates to that expression.. hmm let me go read it again :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int_0^1f_n(x)\,dx=\sum_{n=1}^\infty\left(\frac{(1)^n}{n^2}+\frac{(1)^{n+1}}{n(n+1)}\right)\](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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int_0^1f_n(x)dx=\sum_{n=1}^{\infty}\frac{(1)^n}{n^2(n+1)}\]\[=\sum_{n=1}^\infty(\frac{1}{(2n)^2}\frac{1}{2n}+\frac{1}{2n+1})\sum_{n=1}^\infty(\frac{1}{(2n1)^2}\frac{1}{2n1}+\frac{1}{2n})\]\[=\frac{\pi^2}{24}+\ln(2)1(\frac{\pi^2}{8}\ln(2))=\ln(4)\frac{\pi^2}{12}1\]Not sure if you can follow my thought process, but I THINK I got it haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well 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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