Another even worse double fractional part integral

Question

These integrals are something terrible for me at the moment and the book where I pick them isn't very useful since the examples are quite repetitive in the chapter of Fractional Part Integrals… However, the boi is $$\int_0^1 \int_0^1 (xy)^n \left\{\frac xy\right\}\left\{\frac yx\right\}\,\mathrm{d}x\mathrm{d}y$$ with $n\in\mathbb{N}$ and all my attemps failed… The answer should be $$\frac1{(n+1)^2}-\frac{\zeta(n+2)}{(n+1)(n+2)}$$ Have you any idea?

Answer 1

Disclaimer: There may be a significantly more straightforward approach using some neat fractional-part tricks I'm not aware of. This is somewhat brute-force, but hopefully satisfactory.

Begin by noting that both the integration domain and the integrand are symmetric in $x$ and $y$, so we can split the unit square along its diagonal $y = x$ and consider $$I = 2 \int_0^1 \int_0^y (xy)^n \left\lbrace \frac{x}{y} \right\rbrace \left\lbrace \frac{y}{x} \right\rbrace dx dy$$ Let $f(x, y)$ be the product of the fractional parts contained in the integrand, and expand these fractional parts using $\lbrace x \rbrace = x - \lfloor x \rfloor$ to give $$f(x,y) = \left( \frac{x}{y} - \left \lfloor \frac{x}{y} \right \rfloor \right) \left( \frac{y}{x} - \left \lfloor \frac{y}{x} \right \rfloor \right) = 1 - \frac{y}{x} \left\lfloor \frac{x}{y} \right \rfloor - \frac{x}{y} \left\lfloor \frac{y}{x} \right \rfloor + \left\lfloor \frac{x}{y} \right \rfloor \left\lfloor \frac{y}{x} \right \rfloor$$ Now, in the (innermost) $x$ integral, we have $x < y$ (due to the integration limits) in the interior of the integration domain, so $$\left \lfloor \frac{x}{y} \right \rfloor = 0$$ Hence in the region of interest, $$f(x, y) = 1 - \frac{x}{y} \left\lfloor \frac{y}{x} \right\rfloor$$ which gives $$I = 2 \int_0^1 y^n \left[ \int_0^y x^n \left( 1 - \frac{x}{y} \left \lfloor \frac{y}{x} \right\rfloor \right) dx \right] dy$$ Let $I_1(y)$ be the contents of the square brackets in the line above, and substitute $x = \lambda y \implies dx = y \, d\lambda$, $$\implies I_1(y) = \int_0^1 (\lambda y)^n \left(1 - \lambda \left\lfloor \frac{1}{\lambda} \right\rfloor \right) y \, d\lambda = y^{n+1} \int_0^1 \lambda^n \left( 1 - \lambda \lfloor \lambda^{-1} \rfloor\right) \, d\lambda$$ The $\lambda^n$ integral is straightforward, but we need to evaluate $$I_2 = \int_0^1 \lambda^{n+1} \lfloor \lambda^{-1} \rfloor \, d\lambda$$ To make progress, let's transform $\lambda = \mu^{-1}$ so we get $$I_2 = \int_1^\infty \frac{\lfloor \mu \rfloor}{\mu^{n+3}} \; d\mu = \sum_{k=1}^\infty \int_k^{k+1} \frac{k}{\mu^{n+3}} \; d\mu = \sum_{k=1}^\infty \frac{k}{n+2} \left( k^{-(n+2)} - (k+1)^{-(n+2)} \right)$$ where we've split the integral up into segments where $\lfloor \mu \rfloor$ is constant (giving the $k$ in the numerator). Hence, $$I_2 = \frac{1}{n+2}\left[ \sum_{k=1}^\infty k^{-(n+1)} - \sum_{k=1}^\infty \frac{k}{(k+1)^{n+2}}\right]$$ The first sum is $\zeta(n+1)$. The latter sum (which I'll denote $S$) is somewhat problematic. Note however that we can extend to lower limit to $0$, and then reindex $k \to k + 1$ to get $$S = \sum_{k=0}^\infty \frac{k}{(k+1)^{n+2}} = \sum_{k=1}^\infty \frac{k-1}{k^{n+2}} = \sum_{k=1}^\infty \frac{1}{k^{n+1}} - \sum_{k=1}^\infty \frac{1}{k^{n+2}} = \zeta(n+1) - \zeta(n+2)$$ Returning to $I_2$, we see that the $\zeta(n+1)$ terms cancel, leaving $$I_2 = \frac{1}{n+2} \zeta(n+2)$$ Now we can return to $I_1(y)$ - performing the remaining $\lambda^n$ integration gives $$I_1(y) = y^{n+1} \left(\frac{1}{n+1} - \frac{\zeta(n+2)}{n+2} \right)$$ Finally, we can return to $I$ - substituting the result of $I_1$ gives $$I = 2 \left(\frac{1}{n+1} - \frac{\zeta(n+2)}{n+2} \right) \int_0^1 y^{2n+1} \; dy = \left(\frac{1}{n+1} - \frac{\zeta(n+2)}{n+2} \right) \cdot \frac{1}{n+1}$$ i.e. $$I = \frac{1}{(n+1)^2} - \frac{\zeta(n+2)}{(n+1)(n+2)}$$ as required.