Fourier-like integral equation (This question inspired by question A specific 1st order PDE which looks almost like a linear PDE.)
Solve integral equation
$$
g(x)=\int\limits_{-\infty}^\infty \rho(\omega)\left[e^{i\omega x} - \frac{1}{1 + k i\omega e^{-i\omega}}\right]d\omega,
$$
where $g(x)\colon\mathbb R\to\mathbb R$ is given and $k$ is a real constant; $\rho$ is unknown function.
I tried expand $1/(1 + k\ldots)$ into series or differentiate w.r.t. $k$, but it's useless. Since first part (with $e^{i\omega x}$) is a inverse Fourier transform, I tried to apply Fourier transform, but what we should do with integral
$$
\int\limits_{-\infty}^\infty e^{-i\eta x}d\eta\int\limits_{-\infty}^\infty \frac{\rho(\omega)d\omega}{1 + ki\omega e^{-i\omega}}?
$$
Part with $k$ is just a number, but we can't denote $g(x)=\mathcal F^{-1}[\rho] + \mathrm{const}$. I have tried, and it is not a true constant anyway (depends on $\rho$).
 A: This is an attempt to answer to the tricky question of the determining of the constant :

This draw me to think that :


*

*If $\int_{-\infty}^{\infty}\frac{\varphi (\omega)}{1+k i\omega e^{-i\omega}}d\omega =0 $, then the problem has a solution, which is the inverse Fourier transform of $g(x)$.

*If not, there is no function $\rho(\omega)$ satisfying the given equation.
Since I am not quite sure of the validity of the argument, I think that the question remains open.
A: This is not a solution. 
It would have been desirable to have some understanding of the nature of the operation of multiplying by $\frac{1}{1 + k i\omega e^{-i\omega}}$ in frequency domain, which I don't. I tried to inverse transform it, it is probably going to be a distribution/generalized function and I am unable to find what it would be.
Numerically, it looks something like this

An example where I do (Using non unitary angular frequency transform and using $g(w)\rightarrow G(x),\rho (w) \rightarrow R(x)$):
$$G(x)=\int\limits_{-\infty}^\infty \rho(w)\left[e^{i w x}+ \frac{1}{k +  i w}\right]dw=2\pi R(x)+\int ^0 _{-\infty}R(x)e^{kx}dx$$
For the second term, I used the convolution theorem and the fact that the integral in one domain is equivalent to its transform evaluated at 0.
Lets pick $k=1$ and $G(x)=e^{-x^2/2}$
$$G(x)=2\pi R(x)+\int ^0 _{-\infty}R(x)e^{x}dx$$
G and R will differ by a simple constant. Say $$R(x)=\frac{e^{-x^2/2}-A}{2 \pi}$$
Now it is a trivial matter to find A, after that you would inverse transform $R$ to get $\rho$
$$e^{-x^2/2}=\int_{-\infty }^{\infty } \left(e^{i w x}+\frac{1}{1+i w}\right) \left(\frac{e^{-\frac{w^2}{2}}}{\sqrt{2 \pi }}-\sqrt{\frac{e \pi }{2}} \text{erfc}\left(\frac{1}{\sqrt{2}}\right) \delta (w)\right) \, dw$$
A: Okay, I mentioned that I was unable to inverse transform by hand (and I doubt that it is possible at all) however this process can be done numerically.
Using the same convention:
using non-unitary angular frequency transform and using $g(w)\rightarrow G(x),\rho (w) \rightarrow R(x)$
$$h(w)=-\frac{1}{1 + ik w e^{-iw}}$$
$H(x)$ is the (numerical) inverse frequency transform of this function. This is a known function. For $k=1/2$ it looks like:

Using convolution theorem and the fact that the integral in one domain is equivalent to its transform evaluated at 0.
$$G(x)=2\pi R(x)+\int^\infty _{-\infty}R(\xi)H(x-\xi)d\xi \bigg|_{x=0}$$
$$R(x)=\frac{G(x)-A}{2\pi}$$
$$A=\frac{1}{2\pi}\int^\infty _{-\infty}G(\xi)H(-\xi)d\xi -\frac{A}{2\pi}\int^\infty _{-\infty}H(-\xi)d\xi$$
All the integrals are known numbers, that can easily be found numerically. From those, you can solve for A. And then, you can easily find $\rho$
You can continue solving using these if you'd like
n1=18;
start1=2^7;
x=linspace(-start1,start1,2^n1+1);
x=x(1:2^n1);
G=sech(x);

w=AngularFFTIndex(x);
k=1/2;
h=-1./(1+ 1i* k *w.*exp(- 1i *w));
[~,H ]=EasyIFFT(w,h);

-
function [ x,f ] = EasyIFFT(  k, F )

x=AngularFFTIndex(k);
f=ifftshift(ifft(F));

Enf=sum(abs(f).^2)*(x(2)-x(1));
EnF=sum(abs(F).^2)/2/pi*(k(2)-k(1));

f=f*sqrt(EnF/Enf).*exp(-1i*k*x(1));


end

-
function [ k,F ] = EasyFFT(  x, f )%just make sure number of points is a power of 2

k=AngularFFTIndex(x);
F=fftshift(fft(f));

Enf=sum(abs(f).^2)*(x(2)-x(1));
EnF=sum(abs(F).^2)/2/pi*(k(2)-k(1));

F=F*sqrt(Enf/EnF).*exp(1i*k*x(1));


end

