Alpha(t) = e^(2ipit) for 0 <or= t <or= 1
Prove ABS(Integral over Alpha of [sin(x)/x^2)]dx) <or= 2epi
The Attempt at a Solution
Looking at the book I've come across what I believe I need to use to derive the relation, namely: ABS[Integral over alpha( f(x)dx)) <or= Cl(alpha) where
C >or= ABS(f(x)) for all x in the Image of alpha.
What I want to show then is that ABS(sin(x)/x^2) < C for all complex numbers with magnitude one, since alpha is just the unit circle.
My basic question is then, how can I show sin(x)/x^2 is bounded above given complex arguments?
Any help would be appreciated