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...
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?