FourierConvolutionTheorem

Let's consider a function $h(x)$ defined as:

$$h(x) = \int_0^\infty f(x-y)g(y) \, dy$$

This is essentially a convolution of $f(x)$ and $g(y)$, often denoted as $f * g$. The convolution theorem states:

$$\mathcal{F}\{f * g\} = \mathcal{F}\{f\} \times \mathcal{F}\{g\}$$

where the Fourier transforms of $f(x)$ and $g(x)$ are given by:

$$F(k) = \int_{0}^{\infty} f(x) e^{-2\pi ixk} \, dx$$ $$G(k) = \int_{0}^{\infty} g(x) e^{-2\pi ixk} \, dx$$

Therefore, the Fourier transform of $h(x)$ is:

$$\mathcal{F}\{h(x)\} = F(k) \times G(k)$$

The original function $h(x)$ can be recovered through the inverse Fourier transform of this product:

$$h(x) = \mathcal{F}^{-1}\{ F(k) \times G(k) \}$$