I am given the following integral equation to solve for $f(x)$:
$$\int_{-\infty}^{\infty}f(x) \ f(x-u) \ du = \frac{1}{1+x^2}$$
Firstly, note that the LHS is not the convolution of $f(x)$ with itself, as may appear at a first glance. I tried to approach this using Fourier transform methods. Using the following definition:
$$\mathfrak{F}[f(x)] \equiv \frac{1}{(2 \pi)^\frac{1}{2}} \int_{-\infty}^{\infty}f(x)e^{iwx} \ dx$$
I get that the RHS transforms to:
$$\mathfrak{F}[\frac{1}{1+x^2}] = -(2 \pi)^{\frac{1}{2}} \ sinh(\omega)$$
However, I don't know how to tackle the LHS. It's the product of two functions of $x$, one of which is an integral. As stated above, the LHS is not a convolution and so it isn't simply $\mathfrak{F}[f(x)]^2$ (even if it was, the final solution would require the computation of the inverse fourier transform of $(-(2 \pi)^{\frac{1}{2}} \ sinh(\omega))^{\frac{1}{2}}$ which looks wholly intractable).
What is $f(x)$?
Since $f(x)$ can be factored out of the integral, making a change of variables leads to $$ \frac{1}{1+x^2}=f(x)\int_{-\infty}^{\infty}f(x-u)\;du=f(x)\int_{-\infty}^{\infty}f(t)\;dt=cf(x) $$ where $$ c=\int_{-\infty}^{\infty}f(t)\;dt$$ Now integrating both sides of $cf(x)=\frac{1}{1+x^2}$ yields $c^2=\pi$, so we can take $c=\sqrt{\pi}$. Therefore $$ f(x)=\frac{1}{\sqrt{\pi}}\frac{1}{1+x^2}$$ is a solution to the equation.