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I need to find the asymptotic behaviour of the following double sum: $$ S_{n,\alpha,p}:=\sum_{k_1=1}^n\sum_{k_2=1}^n \frac{(k_1k_2)^{p-2}}{(k_1+k_2)^{\alpha p}}, $$ depending on the parameters $\alpha>0$ and $1<p<\infty$, i.e., i need to find $f(n):=f_{\alpha ,p}(n)$ such that $$ C_1 f(n)\leq S_{n,\alpha ,p}\leq C_2 f(n), \qquad C_1,C_2>0. $$

In some cases I have succeeded, but in some others not. My problem is that the lower and higher estimates I am getting are too rough, and they don't coincide. For example, let us consider $p-2-\alpha p=-1$. Then, I can get this higher estimate: $$ S_{n,\alpha ,p}\approx\sum_{k_1=1}^n k_1^{p-2}\int_1^n\frac{x^{p-2}}{(k_1+x)^{\alpha p}}\, dx\leq \sum_{k_1=1}^n k_1^{p-2} k_1^{-\alpha p}\int_1^n x^{p-2}\, dx\approx n^{p-1}\sum_{k_1=1}^n k_1^{p-2-\alpha p}\approx n^{p-1}\log n. $$ However, for the lower estimate, $$ S_{n,\alpha, p} \approx\sum_{k_1=1}^n k_1^{p-2}\int_1^n\frac{x^{p-2}}{(k_1+x)^{\alpha p}}\, dx \geq \sum_{k_1=1}^n k_1^{p-2} (k_1+n)^{-\alpha p}\int_1^n x^{p-2}\, dx \approx n^{p-1-\alpha p}n^{p-1}=n^{p}n^{p-2-\alpha p}=n^{p-1}. $$ Joining everything together, we obtain $$ n^{p-1}\lesssim S_{n,\alpha , p}\lesssim n^{p-1}\log n, $$ which does not solve the problem. I am conscious that this loss is caused by the estimates of the denominators of my integrals, which are too rough, so I think i may be missing something. I would really appreciate if you could give a hint on how to estimate these integrals, instead of solving the whole problem. Thank you in advance.

PS: The case $p=2$ is already done.

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The main issue, in my humble opinion, is that you are destroying the symmetry. We have:

$$ \sum_{a,b=1}^{n}\frac{(ab)^{p-2}}{(a+b)^{\alpha p}} = \sum_{s=2}^{n+1}\frac{1}{s^{\alpha p}}\sum_{h=1}^{s-1}\left(h(s-h)\right)^{p-2} + \sum_{s=n+2}^{2n}\frac{1}{s^{\alpha p}}\sum_{h=s-n-1}^{n}\left(h(s-h)\right)^{p-2}$$ where the first sum can be estimated with: $$B(p-1,p-1)\sum_{s=2}^{n+1}\frac{s^{2p-3}}{s^{\alpha p}}$$ and the second sum can be estimated by replacing $\frac{1}{s^{\alpha p}}$ with $\frac{1}{n^{\alpha p}}$. The general asymptotics will depend on $p-2$ being positive or not, and $2p-4$ being greater or not than $\alpha p$.

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  • $\begingroup$ Thanks for your answer. I've worked on it and I have a question: shouldn't the first sum be estimated by $B(p-1,p-1)\sum_{s=2}^{n+1} s^{2p-4}/s^{\alpha p}$? The $2p-3$ exponent seems to appear when working on the second sum after rearranging it to $\sum_{h=2}^{n}h^{p-2}\sum_{s=n+2}^{n+h}(s-h)^{p-2}$ and performing the corresponding integral. $\endgroup$ – Alberto Debernardi May 19 '15 at 10:22
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    $\begingroup$ @AlbertoDebernardi: the extra "s" factor come from the change of variable $h = s\cdot t$, leading to $dh = s\cdot dt$. $\endgroup$ – Jack D'Aurizio May 19 '15 at 12:33
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Rewrite the sum as $$S_{n,\alpha,p}=n^{(1-\alpha)p}\sum_{0\leq k_1,k_2\leq n} \frac{\left(\frac{k_1}{n}\frac{k_2}{n}\right)^{p-2}}{\left(\frac{k_1}{n}+\frac{k_2}{n}\right)^{\alpha p}}\frac1{n^2}.$$ The sum converges to the Riemann integral $$I(\alpha,p)=\int_0^1\mathrm dx_1\int_0^1\frac{x_1^{p-2}x_2^{p-2}}{(x_1+x_2)^{\alpha p}}\mathrm dx_2.$$ According to Mathematica, if $(2-\alpha)p>2$ we have (sorry for the notations, I'm afraid you will have to check their exact definitions) $$I(\alpha,p)= \frac{-\Gamma(p-1)\left(\Gamma(\alpha p) \,_2\tilde F_1\left(\begin{array}{cc}p-1,&\alpha p\\p&\end{array}\Bigg\rvert-1\right)+\Gamma\left((\alpha-1)p+1\right)\right) } {\left((\alpha-2)p+2\right)\Gamma(\alpha p)} -(-1)^{p(1-\alpha)}\frac{B_{-1}\left((\alpha -1)p+1,1-\alpha p\right)}{(\alpha-2)p+2}$$

So the sum is asymptotically $S_{n,\alpha,p}\simeq n^{(1-\alpha)p}I(\alpha,p)$.

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