I have a function $f$ defined by
$$f(x)=\begin{cases} xe^{-x} \textrm{ if } x>0,\\
0,\textrm{otherwise}. 
\end{cases}$$

I wish to know the Fourier transform of $f$, i.e,
$${\cal F}(f(x))=\frac1{\sqrt{2\pi}}\int\_{-\infty}^{\infty}f(x)e^{-ixy}dx.$$

I could do this with two different methods using the fact that$${\cal F}(f'(x))=iy{\cal F}(f(x)):$$

1. Note that $f'(x)=e^{-x}-xe^{-x}=e^{-x}-f(x).$ So,$$iy{\cal F}(f(x))={\cal F}(f'(x))={\cal F}(e^{-x})-{\cal F}(f(x))$$ that leads to
$$
{\cal F}(f(x))=\frac{{\cal F}(e^{-x})}{1+iy}=\frac1{\sqrt{2\pi}(1+iy)^2}.
$$
2. Put $g(x)=-e^{-x}-xe^{-x}$, then $g'(x)=f(x).$ So,$${\cal F}(f(x))={\cal F}(g'(x))=iy{\cal F}(g(x))=-iy{\cal F}(e^{-x})-iy{\cal F}(f(x))$$ that leads to
$$
{\cal F}(f(x))=-iy\frac{{\cal F}(e^{-x})}{1+iy}=-iy\frac1{\sqrt{2\pi}(1+iy)^2}.
$$

My question is: why are they different? Isn't the image of $f$ unique?