Inverse Fourier transform of $(1\pm\hat{f}(w))^{-1}$ Assume I have a real function $f(t)$ with Fourier transform of $\hat{f}(w)$. 
Can one say anything about the inverse Fourier transform of $\frac{1}{1\pm\hat{f}(w)}$?
 A: Heuristically, if $\hat{f}(w)$ was a reasonably well behaved and confined transformable function (imagine e.g. a Gaussian) such that $$\lim_{\omega=\pm\infty}\hat{f}(w)=0,\tag{1}$$ then:
$$\lim_{\omega=\pm\infty}\left[\hat{g}(w)=\dfrac{1}{1\pm \hat{f}(w)}\right]=1,$$
and we note that condition $(1)$ is a reasonable (if not necessary) assumption when $\hat{f}(w)$ is the spectrum of any physical system where  the energy is finite.
We might then write:
\begin{aligned}g(t)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left[\dfrac{1}{1\pm \hat{f}(w)}-1\right]e^{i\omega t}{d\omega}+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}1e^{i\omega t}{d\omega},\\&=-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left[\dfrac{\pm \hat{f}(w)}{1\pm \hat{f}(w)}\right]e^{i\omega t}{d\omega}+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}1e^{i\omega t}{d\omega},\\&=-h(t)+\sqrt{2\pi}\delta(t),\end{aligned}
and because:
$$\lim_{\omega=\pm\infty}\left[\hat{h}(w)=\dfrac{\pm \hat{f}(w)}{1\pm \hat{f}(w)}\right]=0,$$
we may expect the transformed function $h(t)$ to be reasonably well behaved (some caution required of course concerning the convergence of the integral). So, what may we say about the inverse Fourier transform of $\hat{g}(w)$? We may say that we expect it to contain a $\delta$ function contribution at the origin when $(1)$ holds and $h(t)$ exists.
Can we say anything about $h(t)$? At least in one important circumstance, yes. Take the plus sign and suppose $\hat{f}(w)$ is an approximately bandlimited function such that:
$$\hat{f}(w)\gg1,\,\,|\omega| \le \frac{B}{2},$$ $$\hat{f}(w)\approx 0,\,\,|\omega| \ge \frac{B}{2}, $$
then:
$$\hat{h}(w)\approx1,\,\,|\omega| \le \frac{B}{2},$$
$$h(t) \approx\frac{1}{\sqrt{2\pi}}\int_{-\frac{B}{2}}^{\frac{B}{2}}1e^{i\omega t}{d\omega}=\frac{B}{\sqrt{2\pi}}\text{sinc} \left( \dfrac{Bt}{2} \right), $$ 
which is a $\text{sinc}$ function who's peak value and oscillation frequency are proportional to the bandwidth of $\hat{f}(w)$. Of course we may not expect the edges of $\hat{h}(w)$ to be so sharp; allowing for a slower decay in $\hat{f}(w)$ (which may possibly not even be strictly decreasing), would allow $\hat{h}(w)$ to spread in the spectral domain which confines $h(t)$ more and encompasses the $\text{sinc}$ function in a damping envelope, but provided the approximate bandwidth of $\hat{f}(w)$ is preserved, so to is the peak value and oscillation frequency of $h(t)$, (see e.g. the Raised-cosine filter).
A: If you assume that this inverse transformation exists and $\left\|\hat{f}\right\|<1 $, you will get
$$\int\limits_{-\infty}^{+\infty}\frac{1}{1-\hat{f}(w)}e^{-iwx}dw=\int\limits_{-\infty}^{+\infty}\sum\limits_{j=0}^{+\infty}\left(\hat{f}(w)\right)^j e^{-iwx}dw=\sum\limits_{j=0}^{+\infty}\int\limits_{-\infty}^{+\infty}\left(\hat{f}(w)\right)^je^{-iwx}dw$$ 
There are no typical transforms of $\left(\hat{f}(w)\right)^j$ (WolframAlpha is silent, there is no such transform in "Tables of integral transforms (Bateman,Erdelyi)")
(upd)But we can say that $\left(\hat{f}\left(w\right)\right)^{j}=\hat{f}\left(w\right)\cdot\hat{f}\left(w\right)...\hat{f}\left(w\right)$, therefore,
$$ \int\limits_{-\infty}^{+\infty}\left(\hat{f}(w)\right)^je^{-iwx}dw=f^{j*}(x),$$ where $n*$ is n-th convolution of $f(x)$ with itself. So, the final answer is
$$\int\limits_{-\infty}^{+\infty}\frac{1}{1-\hat{f}(w)}e^{-iwx}dw=\sum\limits_{j=0}^{+\infty}f^{j*}(x)$$
The same you can do with 
$$\int\limits_{-\infty}^{+\infty}\frac{1}{1+\hat{f}(w)}e^{-iwx}dw=\sum\limits_{j=0}^{+\infty}\left(-1\right)^{j}f^{j*}(x)$$
