## 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 $\int_0^1 f_n(x)\,dx~?$

1. ganeshie8

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

2. anonymous

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

3. anonymous

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

4. ganeshie8

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

5. anonymous

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

7. anonymous

$\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

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