Conditions to apply the inverse Laplace transform Let $f: \mathbb{R}_+ \rightarrow \mathbb{R}$ be a continous function such that $\exists \alpha \in \mathbb{R}$: $t \mapsto e^{\alpha t}f(t) \in L^1(\mathbb{R}_+)$. Let $\hat{f}(\lambda) := \int_0^\infty{e^{-\lambda t}f(t) dt}$ be its Laplace transform defined for $Re(\lambda) \geq \alpha$.
My question is: what  natural conditions should we add on $f$ such that the inverse Laplace formula holds, that is:
$$ \forall t \geq 0, \forall \sigma > \alpha, f(t) = \lim_{T \rightarrow \infty} \frac{1}{2\pi i} \int_{\sigma - i T}^{\sigma + i T}{e^{zt} \hat{f}(z) dz}$$ 
What are the standard references on this question?
 A: If $\int_{0}^{\infty}e^{-\alpha t}|f(t)|dt < \infty$ for some real $\alpha$, then the Laplace transform $\mathscr{L}\{f\}(\lambda)=\int_{0}^{\infty}e^{-\lambda t}f(t)dt$ is defined and continuous for $\Re\lambda \ge \alpha$, and it is holomorphic in the open region as well. The proposed inversion integral is the limit of the following as $R\uparrow\infty$, assuming $\sigma \ge \alpha$:
\begin{align}
    & \frac{1}{2\pi i}\int_{\sigma-iR}^{\sigma+iR}e^{\lambda x}\int_{0}^{\infty}e^{-\lambda t}f(t)dt d\lambda \\
   & = \int_{0}^{\infty}\frac{1}{2\pi i}\int_{\sigma-iR}^{\sigma+iR}e^{-\lambda (t-x)}d\lambda f(t)dt \\
   & = \int_{0}^{\infty}\frac{1}{2\pi i}(e^{-(\sigma+iR)(t-x)}-e^{-(\sigma-iR)(t-x)})f(t)dt \\
   & = \int_{0}^{\infty}\frac{1}{\pi}e^{-\sigma(t-x)}\frac{\sin(R(t-x))}{t-x}f(t)dt \\
   & = e^{\sigma x}\frac{1}{\pi}\int_{0}^{\infty}\frac{\sin(R(t-x))}{t-x}e^{-\sigma t}f(t)dt.
\end{align}
This is a well-known integral from Fourier Analysis. Because $e^{-\alpha t}f(t)$ is absolutely integrable, so is $e^{-\sigma t}f(t)$ because $\sigma \ge \alpha$ by assumption. If $f$ is piecewise smooth, or satisfies other similar Fourier conditions, then the above converges as $R\rightarrow\infty$ to
$$
     e^{\sigma x} \left[e^{-\sigma x}\frac{f(x-0)+f(x+0)}{2}\right].
$$
And that's what you want. And the result does not depend on the choice of $\sigma$, provided $\sigma \ge \alpha$.
