Asymptotic behaviour of a double sum 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.
 A: 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 $S_{n,\alpha,p}\simeq n^{(1-\alpha)p}I(\alpha,p)$.
A: 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 $\alpha p$.
