(Fourier Transform) I'm trying to solve an integral equation using Fourier Transforms, but the thing that I'm not used to/am not sure how to grapple with is that the integral equation itself is piecewise-defined; should I put it in the form of a linear combination of Heaviside Functions or something, or how do I deal with this? Original prompt below shortly.

Question

mendicant_bias · Answer

$$\text{Solve the \integral equation}$$
$$ \int\limits_{0}^{\infty}f(t)\sin(\alpha t)dt = \left\{
     \begin{array}{lr}
       -\alpha, & 0<\alpha<1\
       0, & 1<\alpha< \infty
     \end{array}
   \right. $$

mendicant_bias · Answer

@dan815

amistre64 · Answer

how would you work this if it wasnt pieced?

mendicant_bias · Answer

I'd take the appropriate fourier transforms, or maybe inverse fourier transforms, I guess. I just realized that the LHS is identical to a Fourier Sine transform, so I guess I could take the inverse of both sides. That would give me the LHS just being the original function, so that's good, but I have no idea what the RHS would be.

amistre64 · Answer

are you comfortable with switches?

amistre64 · Answer

$$\int f=\mu_{(0,1)}(-\alpha)$$
now when we get to 1 and onwards, switch 2 turns on so that we add to 0
$$\int f=\mu_{(0,1)}(-\alpha)+\mu_{(1,\infty)}(\alpha)$$

mendicant_bias · Answer

e.g. Heaviside functions, right? I've never seen that notation nor heard them called switches, but that's what it sounds like, which is what I mentioned in my original post.

dan815 · Answer

|dw:1428928468204:dw|