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~?$$

Question

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~?$$

ganeshie8 · Answer

$$\dfrac{(-1)^n}{n^2(n+1)}$$

?

anonymous · Answer

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

anonymous · Answer

^without prior knowledge of the polylogarithm. I've seen the answer on W|A.

ganeshie8 · Answer

I thought the integral evaluates to that expression.. hmm let me go read it again :)

anonymous · Answer

$$\int_0^1f_n(x)\,dx=\sum_{n=1}^\infty\left(\frac{(-1)^n}{n^2}+\frac{(-1)^{n+1}}{n(n+1)}\right)$$(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 · Answer

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anonymous · Answer

$$\int_0^1f_n(x)dx=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2(n+1)}$$$$=\sum_{n=1}^\infty(\frac{1}{(2n)^2}-\frac{1}{2n}+\frac{1}{2n+1})-\sum_{n=1}^\infty(\frac{1}{(2n-1)^2}-\frac{1}{2n-1}+\frac{1}{2n})$$$$=\frac{\pi^2}{24}+\ln(2)-1-(\frac{\pi^2}{8}-\ln(2))=\ln(4)-\frac{\pi^2}{12}-1$$Not sure if you can follow my thought process, but I THINK I got it haha

anonymous · Answer

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$$\sum_{n=1}^\infty \frac{1}{n^2}=\dfrac{\pi^2}{6}$$