# Fourier cosine integral

1. Sep 16, 2012

1. The problem statement, all variables and given/known data
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

2. Relevant equations

f(x)=integral from 0 to infinity [A(w)cos(wx)] dw
3. 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 dont see how do i prove 2 sides which have 2 integrals in them equal each other?

2. Sep 16, 2012

>_> I know I didn't do alot 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

3. Sep 16, 2012

should I repost this in engineering section?

4. Sep 17, 2012

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

(20)(a2)

#### Attached Files:

File size:
42.4 KB
Views:
57
• ###### problem 2.jpg
File size:
37.5 KB
Views:
63
Last edited: Sep 17, 2012