RE-EDITED (Because the conjecture of $(1)$ was wrong when I first posted)
Note that $(1)$ is only valid for $n\le4$
Compute the closed form of $(1)$
$$\int_{-\infty}^{\infty}\prod_{k=1}^{n}(k^a-\cos{k^ax})\cdot{\mathrm dx\over x^2}\tag1$$
ORIGINAL POST
Consider the integral $(1)$
$$\int_{-\infty}^{\infty}\prod_{k=1}^{n}(k^a-\cos{k^ax})\cdot{\mathrm dx\over x^2}=(n!)^a\pi\tag1$$ $a\ge1$; Integers
How does one prove $(1)$?
An attempt:
Expand the products in $(1)$, then we have
$$\int_{-\infty}^{\infty}{(1-\cos x)(2^a-\cos 2^ax)(3^a-\cos 3^ax)\cdots(n^a-\cos n^ax)\over x^2}\mathrm dx\tag2$$
Suggested idea: I guess we could multiply out the brackets and using Contour integration
Or does this product $\prod_{k=1}^{n}(k^a-\cos{k^ax})$ has a closed form?
Recall this recurrence relation
$$\cos (nx)=2\cos[(n-1)x]\cos x-\cos[(n-2)x]\tag3$$
Problem $(1)$ seem tough, I can't see anyway to even changed a bit.
How else do we go about tackling $(1)$?
