What is the Fourier Cosine Integral Identity for Deriving B* and A(w)?

[madah12]

Homework Statement

show that
xf(x)=integral from 0 to infinity of [B*(w)sin(wx)]dw , // B* is a function not B * w

where B* = -dA/dw
A(w) = 2/pi integral from 0 to infinity [f(v) cos(wv)] dv

Homework Equations

f(x)=integral from 0 to infinity [A(w)cos(wx)] dw

The Attempt at a Solution

working on right hand side
B*= 2/pi * integral from 0 to infinity [vf(v)cos(wv)dv]
=integral from 0 to infinity of [2/pi * integral from 0 to infinity [vf(v)cos(wv)dv]sin(wx)]dw
left side = integral from 0 to infinity [ A(w)xcos(wx)]dw

Even when i tried writing A as integral i don't see how do i prove 2 sides which have 2 integrals in them equal each other?

 

[madah12]

>_> I know I didn't do a lot of work but there is not much to work on from the book and I don't really know how to algebriacly manipulate integrals of the form g(x) = integral from 0 to infinity f(x,y)dy

 

[madah12]

should I repost this in engineering section?

 

[madah12]

those are pictures of the problem if it's not clear ><

(20)(a2)