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This integral was posted on AopS but so far it didn't receive an answer, is there a closed form for this integral? $$\int_{0}^{1}\int_{0}^{1}\sin\bigg\{ \frac{x}{y} \bigg\}\sin\bigg\{ \frac{y}{x} \bigg\}dxdy$$ I dont have an approach that actually lead to something relevant, but I would love to see a closed form, and I suspect it has a nice one since another similarly looking one can be evaluated to be: $$\int_{0}^{1}\int_{0}^{1}\bigg\{ \frac{x}{y} \bigg\}\bigg\{ \frac{y}{x} \bigg\}dxdy=1-\frac{\zeta(2)}{2}$$ Also the proposer there mentioned that the integral is interesting I suspected there is one.

Of course there $\{a\}$ denotes the fractional part of $a$.

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  • $\begingroup$ could you use the fact that $\frac{x}{y}+\frac{y}{x}=\frac{x^2+y^2}{xy}$ by using a compound angle formula then see what you have left over. Polar coordinates may also be an option? $\endgroup$ – Henry Lee May 29 '18 at 8:42
  • $\begingroup$ Is anything known about the general case $\int_{0}^{1}\int_{0}^{1}\{ \frac{x}{y} \}^n\{ \frac{y}{x} \}^mdxdy$ ? $\endgroup$ – Benedict W. J. Irwin May 29 '18 at 10:46
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Here's how I would get started: $$\begin{align} \int_0^1\int_0^1 \sin\bigg\{\frac{x}{y}\bigg\}\sin\bigg\{\frac{y}{x}\bigg\}dxdy &=2\int_0^1\int_0^y \sin\bigg\{\frac{x}{y}\bigg\}\sin\bigg\{\frac{y}{x}\bigg\}dxdy\\ &=2\int_0^1\int_0^1 \sin\bigg\{\frac{xy}{y}\bigg\}\sin\bigg\{\frac{y}{xy}\bigg\}ydxdy\\ &=2\int_0^1\int_0^1 y\sin\{x\}\sin\bigg\{\frac{1}{x}\bigg\}dxdy\\ &=2\bigg(\int_0^1 ydy\bigg)\int_0^1 \sin\{x\}\sin\bigg\{\frac{1}{x}\bigg\}dx\\ &=\int_0^1 \sin\{x\}\sin\bigg\{\frac{1}{x}\bigg\}dx\\ &=\int_1^\infty \frac{\sin\{x\}\sin(1/x)}{x^2}dx\\ \end{align}$$ For this integral, I would try converting it to a series... but a closed-form doesn't look too promising.

EXTRA: Concerning the comment from @BenedictWilliamJohnIrwin: by employing the above strategy to the integral he proposed, one would obtain $$\begin{align} \int_1^\infty \frac{\{x\}^m}{x^{n+2}}dx &=\sum_{k=1}^\infty \int_k^{k+1} \frac{(x-k)^m}{x^{n+2}}dx \end{align}$$ a general evaluation of this seems unlikely as well, but for certain cases we can evaluate it with ease. For example, consider $m=1$: $$\begin{align} \int_1^\infty \frac{\{x\}}{x^{n+2}}dx &=\sum_{k=1}^\infty \int_k^{k+1} \frac{x-k}{x^{n+2}}dx\\ &=\sum_{k=1}^\infty \frac{1}{nk^n}-\frac{1}{n(k+1)^n}-\frac{1}{(n+1)k^{n}}+\frac{k}{(n+1)(k+1)^{n+1}}\\ &=\frac{1}{n}-\frac{1}{n+1}\sum_{k=1}^\infty \frac{1}{k^{n}}-\frac{k}{(k+1)^{n+1}}\\ &=\frac{1}{n}-\frac{1}{n+1}\sum_{k=1}^\infty \frac{1}{k^{n}}-\frac{1}{(k+1)^{n}}+\frac{1}{(k+1)^{n+1}}\\ &=\frac{1}{n}-\frac{\zeta(n+1)}{n+1}\\ \end{align}$$ The case for $m=2$ is probably similar, but more messy - perhaps I shall come back and do it later.

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