Q:

Fourier transform of $(-\Delta+\lambda)^{-1}$

I was supposed to solve an exercise and I actually did it, but I had to use a step that I did not comprehend at all. 
I know that the fourier transform of $F((-\Delta)f)(k)=|k|^2 F(f)(k).$
But now I had to use that $F((-\Delta+\lambda)^{-1}f)(k)=((|k|^2+\lambda)^{-1})F(f)(k),$ where $\lambda>0$ is a constant. 
So I now that the Fourier transform changes differentiation into multiplication and vice versa, but I would have expected that since the Laplacian does not enter this formula directly, but via this $(.+\lambda)^{-1}$ construct, that this should somehow affect the Fourier transform.  Could anybody explain to me whether this is actually true and why this holds? 

A:

$$F[f](\xi) = F[(-\Delta+\lambda)((-\Delta+\lambda)^{-1}f)](\xi) = (|\xi|^2+\lambda)F[(-\Delta+\lambda)^{-1}f](\xi),$$
hence
$$F[(-\Delta+\lambda)^{-1}f](\xi)=\frac{1}{|\xi|^2+\lambda}F[f](\xi)$$

