How does one solve this integral equation $1+ax=\int_{-\infty}^xf(x-t)dt$ I've run into having to solve this equation for $f(x)$:
$$1+ax=\int_{-\infty}^xf(x-t)dt$$
Unfortunately, I am not familiar with solving integral equations.  Can anyone help? Is is even soluble?
Edit: Fixed a typo in the upper limit in the integral.

Note that rhs of this equation is constant. Indeed,
$$
\int\limits_{-\infty}^{x} f(x-t)dt=
\{\tau=x-t\}=
\int\limits_{0}^{+\infty}f(\tau)d\tau=
\int\limits_{0}^{+\infty}f(t)dt
$$
Therefore the lhs of this equation must be constant. But this is possible only if $a=0$. For the case when $a=0$, we have $1=\int\limits_{0}^{+\infty}f(t)dt$, otherwise there is no solution.
Finally if $a=0$ the solution of this equation is any integrable function $f$ such that $\int\limits_{0}^{+\infty}f(t)dt=1$. If $a\neq 0$, solution doesn't exist.
