 Odd moment of a prime cosine sum

For odd $j$, expanding $({\sum }_{p\le X}{p}^{-\sigma }\cos (t\log p){)}^j$ gives products with different numbers of prime factors on the two sides. Unique prime factorization rules out every diagonal equality. The integral over $[T,2T]$ is therefore $O_j(X^j({\sum }_{n\le X^j}{n}^{-\sigma }{)}^2)$.