Suppose that $f\colon [0,\infty) \to \mathbb{R}$ and, for some $\sigma > 0$, $f$ satisfies $$\left|\int_0^\infty e^{(\sigma + i\omega)t}f(t)\,dt\right|<\infty$$ for all $\omega \in \mathbb{R}$, where $i$ is the imaginary unit, and $|\cdot|$ denotes the modulus. Is it true that $f$ satisfies $$\int_0^\infty |f(t)|\,dt < \infty?$$
I have been trying to prove that $|f(t)| < Ce^{-\sigma t}$ for some $C > 0$ but have been unsuccessful so far.
\begin{align*} \left|\int_{0}^{\infty}e^{(\sigma+i\omega)t}f(t)dt\right|^{2}=\left|\int_{0}^{\infty}e^{\sigma t}f(t)dt\right|^{2}+\left|\int_{0}^{\infty}e^{\omega t}f(t)dt\right|^{2}<\infty, \end{align*} in particular, \begin{align*} \left|\int_{0}^{\infty}e^{\sigma t}f(t)dt\right|<\infty. \end{align*} If we view the integrals taken here are all the Lebesgue one, then the above integral entails that \begin{align*} \int_{0}^{\infty}e^{\sigma t}|f(t)|dt<\infty. \end{align*} Now we see that $e^{\sigma t}|f(t)|\geq|f(t)|$ for $t\geq 0$, note that $\sigma>0$ is in the assumption, so the result follows.