I have to solve the following integral equation \begin{align*} \int_{-\infty}^\infty e^{-y^2} \log \left( \int_{-\infty}^\infty e^{-(y-x-t)^2} f(t) dt\right) dy=-cx^2 \end{align*} where $c$ is some constant. We have to solve for $f(t)$. Also, assume, if needed, that $\int_{-\infty}^\infty f(t) dt < \infty$
The only thin I noticed that integral inside log is a convolution and so is the outside integral.
I don't know much about integral equations so any reference would be great too. Thanks
Clearly, this problem is meant to be solved using Fourier Transform methods.
I'm using the definition that the Fourier transform of $f(t)$ is $F(\omega) = \displaystyle\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt$.
Accordingly, the inverse Fourier transform of $F(\omega)$ is $f(t) = \dfrac{1}{2\pi}\displaystyle\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega$.
Define $g(z) = \log \left( \displaystyle\int_{-\infty}^\infty e^{-(-z-t)^2} f(t) dt\right)$. Then, we have $\displaystyle\int_{-\infty}^\infty e^{-y^2}g(x-y)\,dy=-cx^2$.
The left hand side is simply the convolution of $e^{-x^2}$ and $g(x)$.
The Fourier transform of $e^{-x^2}$ is $\sqrt{\pi}e^{-\omega^2/4}$ and the Fourier transform of $-cx^2$ is $2\pi c\delta''(\omega)$.
By the convolution property, we get $\sqrt{\pi}e^{-\omega^2/4}G(\omega) = 2\pi c\delta''(\omega)$, i.e. $G(\omega) = 2c\sqrt{\pi}e^{\omega^2/4}\delta''(\omega)$.
From Wikipedia, the Dirac Delta function satisfies $x\delta'(x) = -\delta(x)$. Differentiation yields $x\delta''(x)+\delta'(x) = -\delta'(x)$, i.e. $x\delta''(x) = -2\delta'(x)$. Therefore, $x^2\delta''(x) = -2x\delta'(x) = 2\delta(x)$. Then, since $x^n\delta(x) = 0$ for integers $n \ge 1$, we have $x^n\delta''(x) = 0$ for integers $n \ge 3$.
Using Taylor expansion on $e^{\omega^2/4}$, we get $G(\omega) = 2c\sqrt{\pi}e^{\omega^2/4}\delta''(\omega) = c\sqrt{\pi}(2\delta''(\omega)+\delta(\omega))$.
Taking the inverse Fourier transform yields $g(z) = -\dfrac{c}{\sqrt{\pi}}\left(z^2-\dfrac{1}{2}\right)$.
Now, we've reduced the problem to solving $\displaystyle\int_{-\infty}^\infty e^{-(-z-t)^2} f(t)\,dt = e^{-\tfrac{c}{\sqrt{\pi}}\left(z^2-\tfrac{1}{2}\right)}$.
Using the substitution $z \to -z$, we get $\displaystyle\int_{-\infty}^\infty e^{-(z-t)^2} f(t)\,dt = e^{\tfrac{c}{2\sqrt{\pi}}}e^{-\tfrac{c}{\sqrt{\pi}}z^2}$.
The left hand side is simply the convolution of $e^{-z^2}$ and $f(z)$.
The Fourier transform of $e^{-z^2}$ is $\sqrt{\pi}e^{-\tfrac{\omega^2}{4}}$ and the Fourier transform of $e^{\tfrac{c}{2\sqrt{\pi}}}e^{-\tfrac{c}{\sqrt{\pi}}z^2}$ is $e^{\tfrac{c}{2\sqrt{\pi}}}\sqrt{\tfrac{\pi\sqrt{\pi}}{c}}e^{-\tfrac{\sqrt{\pi}}{c}\tfrac{\omega^2}{4}}$
By the convolution property, we get $\sqrt{\pi}e^{-\tfrac{\omega^2}{4}}F(\omega) = e^{\tfrac{c}{2\sqrt{\pi}}}\sqrt{\tfrac{\pi\sqrt{\pi}}{c}}e^{-\tfrac{\sqrt{\pi}}{c}\tfrac{\omega^2}{4}}$.
Now, solve for $F(\omega)$, and take the inverse Fourier transform to get the answer.