that leads to which is easily killed by makin' the substitution Finally, your "nice" integral equals which holds for and we're done.
Exactly the same method as Krizalid's just a slightly different start...it is how I solved this integral the first time I saw it... sometimes if its not apparent to you just remember that if you ever see an integral with a second variable expression in it try letting it equal to its own function.
So
So the same method as Krizalid's just a different way of seeing it.