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I'm trying to prove that the following integral $\int_{0}^{\infty}(-1)^{[ x^2]}dx$ is conditional converges, ( The brackets stand for the floor function). I can't use the integral test since this is clearly not a positive function, but I'd like somehow to do an analog to Leibniz series and use it, I'm not sure how.

Any suggestions?

Thank you!

2 Answers 2

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Note that $ \int\limits_{[0,+\infty)}(-1)^{[x^2]}dx= \sum\limits_{k=0}^\infty\int\limits_{[\sqrt{k},\sqrt{k+1})}(-1)^{[x^2]}dx= \sum\limits_{k=0}^\infty\int\limits_{[\sqrt{k},\sqrt{k+1})}(-1)^{k}dx= $ $ \sum\limits_{k=0}^\infty(-1)^{k}(\sqrt{k+1}-\sqrt{k})= \sum\limits_{k=0}^\infty\frac{(-1)^{k}}{\sqrt{k+1}+\sqrt{k}} $ Using Leibniz test we see that this sum converges, hence converges the integral

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Let $f(x)=(-1)^{[x^2]}$. Then $f(x)$ is constant on the intervals $[\sqrt{n-1},\sqrt{n})$, when $n$ is a positive integer. The value of $\int_{\sqrt{n-1}}^{\sqrt{n}} (-1)^{[x^2]}dx$ is $(-1)^{n-1}(\sqrt{n}-\sqrt{n-1})$

So, for any $n$, $\int_0^{\sqrt{n}} = \int_{\sqrt 0}^{\sqrt{1}} + \int_{\sqrt{1}}^{\sqrt 2}... + \int_{\sqrt{n-1}}^{\sqrt{n}}$

So $\int_{0}^{\sqrt n} (-1)^{[x^2]} dx = \sum_{k=0}^{n-1} (-1)^k (\sqrt{k+1}-\sqrt{k})$

But $a_k = \sqrt{k+1}-\sqrt{k}$ is strictly decreasing and converges to 0, and therefore $\sum (-1)^k a_k$ converges.

The only question then is what to do with the $\int_0^{z}$ when $z$ is not a perfect square root.

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    Thank you very much for the helpful explanation.2012-02-07