Let $\{\}$ denote the fractional part, does the following integral have a closed form ? $$\int_{0}^{1}\int_{0}^{1}\bigg\{\frac{1}{x\,y}\bigg\}^2dx\,dy$$
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Probably duplicate: https://math.stackexchange.com/questions/875076/evaluating-int-01-cdots-int-01-bigl-frac1x-1-cdots-x-n-big?rq=1 – Mariusz Iwaniuk Jul 19 '18 at 15:17
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@MariuszIwaniuk thank you for the update. I do not know that existed before already. – Kays Tomy Jul 19 '18 at 16:30
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This is an incomplete answer that only addresses the 1-dimensional case.
We split the integral into continuous pieces: $$\begin{align} \int_0^1 \left\{ \frac{1}{x} \right\}^2 \mathrm{d}x &= \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} \left\{ \frac{1}{x} \right\}^2 \mathrm{d}x \\\\ &= \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} \left( \frac{1}{x} - n\right)^2 \mathrm{d}x \\\\ &= \sum_{n=1}^{\infty} \left( n^2x - 2n \ln |x| - \frac{1}{x} \right)\biggr\rvert_{\frac{1}{n+1}}^{\frac{1}{n}} \\\\ &= \sum_{n=1}^{\infty} \left( n - 2n \ln \frac{1}{n} - n - \frac{n^2}{n+1} + 2n \ln \frac{1}{n+1} + n + 1 \right) \\\\ &= \sum_{n=1}^{\infty} \left( 2n \ln \frac{n}{n+1} + \frac{2n + 1}{n+1} \right) \end{align}$$
With the aid of computer algebra, we obtain that this series converges to $$\ln (2\pi) - \gamma - 1$$ where $\gamma$ is the Euler-Mascheroni constant.
Fytch
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$\int_0^1 \left{ \frac{1}{x} \right}^2 \mathrm{d}x=-1-\gamma +\ln (2)+\ln (\pi )$
:)– Mariusz Iwaniuk Jul 19 '18 at 14:39 -
@MariuszIwaniuk How did you arrive at that? Did you evaluate the emergent series using complex analysis? – Fytch Jul 19 '18 at 15:16
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@Fytch. I've used your answer exactly only added a square in integral. – Mariusz Iwaniuk Jul 19 '18 at 15:20
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@MariuszIwaniuk Yes, I did that too and got $\sum_{n=1}^{\infty} \left( 2n \ln \frac{n}{n+1} + \frac{2n + 1}{n+1} \right)$ but I don't know how to evaluate this sum to get your answer. Can you give me some pointers? – Fytch Jul 19 '18 at 15:21
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@Dears, to get the result of squared single variable integral you should use the Stirling formula – Kays Tomy Jul 19 '18 at 15:48