I was looking for a polylog-free solution, but I guess this puts the nail on the coffin^ That was good practice!
I need to go out for dinner v soon, so will supply gory details later. In short, the answer can be expressed solely in terms of Apery's constant zeta(3).
1. IBP to essentially reduce the integral to log^2(1+x)/x. (The bulk of the difficulty of the originally posted integral lies in this integral as I see it.)
2. Expand this out to obtain an expression in terms of polylogs of order 2 and 3, but we basically only need to know the value of these polylogs at 1/2 and 1.
3. Special functions funtimes proving some reflection formulae etc to compute these guys.
Eh I wouldn't say that necessarily, I did the integral in a not particularly imaginative way...there could be a clever shortcut.I was looking for a polylog-free solution, but I guess this puts the nail on the coffin
because I like a nice balance of low tech and efficient solutions to integralsEh I wouldn't say that necessarily, I did the integral in a not particularly imaginative way...there could be a clever shortcut.
To me it does seem like computing this integral is roughly the same mathematical "depth" as computing some particular values of the di/trilogarithm though.
Anyway, is there a reason why you were seeking to avoid them?
I think it was there before. I remember answering it in my head but then forgetting to comment on it in my post. HahaOh didn't see you edited it, was the divergence thing always there and I just blind before?
In any case, we can still represent the quadratic form by a symmetric, hence diagonalisable matrix Q. If the quadratic form does not represent an ellipse, then at least one eigenvalue is non-positive, which implies that at least one of the integrals in the above product of single-variable integrals is infinite and we are done.