Let $Λ$ be the von-Mangoldt function. then
What is the estimate for the sum $\sum_{1\leq x\leq n}\Lambda(x)^{4}$?
Is this $\sum_{1\leq x\leq n}\Lambda(x)^{4}\sim n\log^3n$
also what can we say about this when $x\neq y$?
$ \sum_{1\leq x, y\leq n}\Lambda(x)^{2}\cdot \Lambda(y)^{2}$?
Can any one help.
I'm trying this for a long time
Best Answer
A simple way to get an asymptotic formula. We can use the Prime Number Theorem in the form $$\theta\left(x\right):=\sum_{p\leq x}\log\left(p\right)=x+O\left(\frac{x}{\log\left(x\right)}\right)$$ the trivial estimate $$\sum_{n\leq x}\Lambda\left(n\right)^{4}=\sum_{p\leq x}\log^{4}\left(p\right)+O\left(\sqrt{x}\log^{3}\left(x\right)\right)$$ and the Abel summation formula $$\sum_{p\leq x}\log^{4}\left(p\right)=\theta\left(x\right)\log^{3}\left(x\right)-3\int_{2}^{x}\theta\left(t\right)\frac{\log^{2}\left(t\right)}{t}dt$$ to obtain $$\sum_{n\leq x}\Lambda\left(n\right)^{4}=x\log^{3}\left(x\right)+O\left(x\log^{2}\left(x\right)\right).$$