Is Solving an Integral Equation as Easy as a Fourier Transform?

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In summary, the conversation discusses solving an integral equation by converting it to a Fourier transform. The equation is g(x)=Int(-1,1)exp(-xy)f(y)dy and can be written as g(x)=Int(-infinite,infinite)exp(-xy)f(y)W(y)dy, where W(y) is defined as 1 if -1<y<1 and 0 elsewhere. The person in the conversation expresses surprise at how easy it seems to solve the integral equation. However, there is confusion about whether the Fourier transform requires an additional term of exp(-ixy).
  • #1
eljose79
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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...
 
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  • #2
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
think he forgot the i in the exp
but i don't know if you can add the W(y)
 

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