Matematické Fórum

Skrytý text:

we can use an old trick here: Since , we get, using Fubini's theorem that , where . 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?