# This can not be so easy

1. Apr 4, 2003

### eljose79

suppose we want to solve the integral equation

g(x)=Int(-1,1)exp(-xy)f(y)dy (2) then this is equal to

g(x)=Int(-infinite,infinite)exp(-xy)f(y)W(y)dy where the W(y) function is defined like that:

W(y) is 1 iif -1<y<1 and 0 elsewhere.

then we would have that the equation (2) is a fourier transform and inverting we can get f(x)W(x).

Is all that?..i can not believe that solving an integral equation could be so easy...

2. Apr 4, 2003

### HallsofIvy

Staff Emeritus
I'm afraid I don't quite understand your question.

You refer to g(x)=Int(-1,1)exp(-xy)f(y)dy as an "integral equation" so I presume you mean that g is given and you want to find f.

How does converting to the integral from -inf to inf make this a Fourier transform? Doesn't the Fourier transform have an exp(-ixy) in it?

3. Apr 5, 2003

### ottjes

think he forgot the i in the exp
but i don't know if you can add the W(y)