Is $\exp$ the only function satisfying $f(x)=\displaystyle \int_{-x}^{+\infty} f(-t) dt$?

Question

Today in class we first dealt with improper integrals, and as an example we found $ \displaystyle \int_0 ^{+\infty} e^{-x}dx=1$. Soon, I noticed that in fact $$e^x=\int_{-x}^{+\infty}e^{-t}dt. $$ Since I already know $\exp$ can be defined as the unique function such that $f'(x)=f(x)$ and $f(0)=1$, I wondered whether this might be yet another way of defining it. In search of counterexamples, I rewrote the equation $$f(x) = \int_{-x}^{+\infty}f(-t)dt $$ as $$\lim_{x\to-\infty}f(x)=f(x)-f'(x),$$ but still couldn't think of other solutions (besides $c \cdot e^x$ ). If there aren't, how to prove it?

(this question arised from a terribly silly oversight)

Answer 1

Note that the functional equation $$f(x) = \int\limits_{-x}^\infty f(-t)dt$$ implies that any (integrable) solution of this equation is differentiable (why?). Then, by differentiating, you get $$f^\prime(x) = f(x)$$

I leave it to you to continue the reasoning and fill any gaps you may suspect (note that you have still the freedom to assign an initial value).

Answer 2

If $f$ is contnuous, then by the fundamental theorem of calculus $$ \frac{\mathrm d}{\mathrm dx}\int_{-x}^{+\infty}f(-t)\,\mathrm dt = \frac{\mathrm d}{\mathrm dx}\int_{-\infty}^{x}f(t)\,\mathrm dt =f(x)$$ Hence $$ f(x)=\int_{-x}^{+\infty}f(-t)\,\mathrm dt$$ is equivalent to $f'=f$ and $\lim_{x\to-\infty}f(x)=0$. This would suit $f(x)=ce^x$ for any $c$.