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How to estimate $\sum\limits_{2\le n,m\le X}\frac{\Lambda(n)\Lambda(m)}{(nm)^\sigma\log(n)\log(m)}$ in terms of $X$, where $\Lambda$ is the Von-Mangoldt function, and $\sigma\ge\frac12$

Is it possible to say $\log\log(X)$ ?

If I neglect the prime powers and consider the conribution of pure primes then from Cauchy Schwarz I get

$\sum\limits_{2\le p_1,p_2\le X}\frac{1}{p_1^{\sigma}p_2^{\sigma}}\le\sum\limits_{p\le X}\left(\frac1{p^{2\sigma}}\right)\le\sum\limits_{p\le X}\left(\frac1{p}\right)\approx \log\log X$

Now I have to justify why the other terms can be discarded

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This is just the same as \[\left(\sum_{n \leq X} \frac{\Lambda(n)}{n^{\sigma} \log n}\right)^2.\] The sum is \[\sum_{k \leq \frac{\log X}{\log 2}} \sum_{p \leq X^{1/k}} \frac{1}{k p^{k\sigma}}.\] The main contribution comes from $k = 1$, and this can be large: partial summation and the prime number theorem imply that for $1/2 \leq \sigma < 1$, \[\sum_{p \leq X} \frac{1}{p^{\sigma}} = \int_{2}^{X} \frac{1}{t^{\sigma} \log t} \, dt \sim \frac{X^{1 - \sigma}}{(1 - \sigma) \log X}.\]

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