Question:
The function $p(x)$ is defined as $p(x) = e^{-x}$ for $x>0$ and $p(x) = 0$ for $x<0$. Find the Fourier Transform of $p(x)$ and use the convolution theorem to find $f(x)$ such that: $$\int_0^{\infty} p(y)f(x-y) dy = xf(x)$$
My attempt:
I have found $\hat{p}(k) = \frac{1}{1 + ik}$. I took the convolution of both sides and obtained the ODE $$\hat{f}(k) = i(1 + ik)\hat{f}'(k)$$ which I solved to yield $$\hat{f}(k) = \frac{1}{A(1 + ik)}$$ where $A$ is an arbitrary constant. After inverting this using the expression for $\hat{p}(k)$ earlier, and substituting into the original question, I get the left hand side integral diverging to infinity.
I cannot find my mistake and am unsure how to progress. Could someone please point me in the right direction?Thank you.
Taking the Fourier transform* of the functional identity $(p*f)(x)=xf(x)$, we obtain $\widehat{p*f} =\widehat{\text{id}\, f}$, where $\text{id}\, f$ is the mapping $x \mapsto xf(x)$. According to the convolution theorem, we have $$ \widehat{p*f}(\nu) = \hat{p}(\nu)\hat{f}(\nu) = \frac{\hat{f}(\nu)}{1+2\text i\pi \nu} . $$ According to a property of the Fourier transform, $\widehat{\text{id}\, f}(\nu) = \frac{\text i}{2\pi} \hat{f}' (\nu)$, so that we end up with the differential equation $$ \hat{f}'(\nu) + \frac{2 \text i\pi}{1+2\text i\pi \nu} \hat{f}(\nu) = 0 . $$ The solutions are $\hat{f}(\nu) = C/(1+2\text i\pi \nu)$, where $C$ is constant. The inverse Fourier transform is the function $f$ defined by $$ f(x) = C e^{-x} u(x) = C p(x) , $$ where $u$ is the Heaviside unit step function.
*The convention used is $\hat{f}(\nu) = \int_{\Bbb R} f(x) e^{-2\text i \pi \nu x}\,\text d x$.