Hi ↑ Marian:,
we can use an old trick here: Since
, we get, using Fubini's theorem that
. The inner integral can be evaluated using standard trigonometric substitution and some arctan identities to
. Then, the only applicable substitution
yields the result
Everything works for
, in particular, for
, we get
One could also use the Taylor series of ln, switch the sums and compute integral from a sine's power, but I don't feel like writing down the binomial expansions..
Construction of your own?