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Consider $\mathbb{R}$.

$f(x)=(-1)^nn,\ x\in A_n,\text{ and }A_n=\left(\frac{1}{n+1},\frac{1}{n}\right].$ Compute $\displaystyle\int_{\bigcup_{n=1}^\infty A_n}f \, dx $

$\sum_{k=1}^\infty \int_{A_n}f=-1(1/2)+2(1/6)-3(1/2)+\cdots,\tag{1}$ This series converges, so I would say the the function is integrable. Is there a way to evaluate (1) without resorting to geometry? Can one be more thorough? Is there a closed form integral for $f$?

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$ \int_{\bigcup_{n=1}^\infty A_n}f=\sum_{n=1}^\infty\int_{A_n}f=\sum_{n=1}^\infty(-1)^nn\left(\frac{1}{n}-\frac{1}{n+1}\right) =\sum_{n=1}^\infty\frac{(-1)^n}{n+1}=g(1), $ where $ g:(-1,1] \to \mathbb{R},\ g(x)=\sum_{n=1}^\infty\frac{(-1)^n}{n+1}x^{n+1}=-x+\sum_{n=0}^\infty\frac{(-1)^n}{n+1}x^{n+1}. $ For every $x \in (-1,1]$ we have \begin{eqnarray} g(x)&=&-x+\sum_{n=0}^\infty\int_0^x(-1)^nt^n\, dt=-x+\int_0^x\frac{1}{1+t}\, dt=-x+\ln(1+x). \end{eqnarray} Hence $ \int_{\bigcup_{n=1}^\infty A_n}f=g(1)=\ln2-1. $