Could you please help me to solve the following integral equation?
$$y(x)=e^{-x^2/2}+\lambda\int^{+\infty}_{-\infty}e^{-i(x-z)}y(z)dz$$ I tried to turn the exponentiential term into its trigonometric form and tried to solve the equation obtain in term of new constants $c_1=\int^{+\infty}_{-\infty}cos(z)y(z)dz$ and $c_2=\int^{+\infty}_{-\infty}sinz(z)y(z)dz$. But I obtained for $c_1$ $$c_1=\int^{+\infty}_{-\infty}cos(z)e^{-z^2/2}dz+\lambda (c_1+ic_2)\int^{+\infty}_{-\infty}cos^2(z)dz+\lambda(c_2-c_1i)\times\int^{+\infty}_{-\infty}cos(z)sinzdz.$$ I can't go further.
I also applied the Fourier transform on both sides and obtained that this equation has a unique solution iff $$1-\lambda\hat{K}(\alpha)\neq 0$$ where $$\hat{K}(\alpha)$$ is the Fourier transform of $e^{-ix}$. But I think (I am not sure) the Fourier transform of such a function is the dirac function $\delta_{-\alpha}$, and I don't know how to proceed.
Please need help. Thanks in advance.
As far as I can see the equation has a solution only if $\lambda=0$.
Let $y(x)$ be a solution. Multiply both sides with $e^{ix}$. Then we have \begin{equation} w(x)=e^{ix-x^2/2}+\lambda\int_{-\infty}^{\infty}w(z)\,\mathrm{d}z,\tag {1} \end{equation} where $w(x)=e^{ix}y(x)$. Put \begin{equation*} C =\int_{-\infty}^{\infty}w(z)\,\mathrm{d}z. \end{equation*} Then $C$ is a constant and $w(x)= e^{ix-x^2/2}+\lambda C $. However, the integral in $(1)$ is convergent. Thus $\lambda C = 0$.
If $C=0$ then $w(x)= e^{ix-x^2/2}$ and \begin{equation*} C=\int_{-\infty}^{\infty}w(z)\,\mathrm{d}z= \int_{-\infty}^{\infty}e^{iz-z^2/2}\,\mathrm{d}z =\sqrt{\dfrac{2\pi}{e}} \neq 0. \end{equation*} What remains is the alternative $\lambda= 0$.