Sum with the von-Mangoldt function: $\sum_{1\leq x\leq n}\Lambda(x)^4$

Question

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

Answer 1

$$\sum_{n\le x}\Lambda(n)^4 = \sum_{n\le x} \Lambda(n)\log^3 n +O(x^{1/2}\log^4 x)$$ From there you can use the PNT and a partial summation, obtaining $$\sum_{n\le x}\Lambda(n)^4 \sim \sum_{n\le x}\Lambda(n)\log^3 n\sim \sum_{n\le x}\log^3 n\sim x \log^3 x$$ The exact asymptotic is found from $$\sum_n\Lambda(n) n^{-s}\log^3 n=\frac{d^3}{ds^3}\frac{\zeta'(s)}{\zeta(s)}$$ Following the same method as in the PNT we obtain $$\sum_{n\le x} \Lambda(n) \log^3 n = Res((\frac{d^3}{ds^3}\frac{\zeta'(s)}{\zeta(s)} )\frac{x^s}{s},1)+O(x\log^{-k} x) = \sum_{m=1}^4 c_m x \frac{\log^{m-1} x}{(m-1)!} +O(x\log^{-k} x)$$ Where $c_m$ is the coefficient of $(s-1)^{-m}$ in the Laurent expansion of $(\frac{d^3}{ds^3}\frac{\zeta'(s)}{\zeta(s)} )\frac1s$.

Under the RH the error term can be improved to $O(x^{1/2+\epsilon})$.

Answer 2

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 $$\sum_{n\leq x}\Lambda\left(n\right)^{4}=x\log^{3}\left(x\right)+O\left(x\log^{2}\left(x\right)\right).$$