Sufficient conditions to conclude that $\lim_{a \to 0^{+}} \int_{0}^{\infty} f(x) e^{-ax} \, dx = \int_{0}^{\infty} f(x) \, dx$

The easy case is of course when $\lvert f\rvert$ is integrable, then the dominated convergence theorem asserts

$$\lim_{a \downarrow 0} \int_0^\infty f(x) e^{-ax}\,dx = \int_0^\infty f(x)\,dx.$$

If $f$ isn't absolutely integrable, the most useful condition that I'm aware of is that for every $\varepsilon > 0$ there exists a $K(\varepsilon) \in (0,\infty)$ such that

$$\left\lvert \int_{K(\varepsilon)}^\infty f(x) e^{-ax}\,dx\right\rvert < \varepsilon$$

for all $a \in [0,\delta]$ for some $\delta > 0$. That condition is also known as "uniform convergence" of the integrals $\int_0^\infty f(x)e^{-ax}\,dx$. (I'm not sure how standard that terminology is, I didn't encounter it until recently.)

In that case, splitting the integral yields

$$\begin{align} \left\lvert \int_0^\infty f(x)\bigl(1-e^{-ax}\bigr)\,dx\right\rvert & \leqslant \left\lvert \int_0^{K(\varepsilon)} f(x) \bigl(1 - e^{-ax}\bigr)\,dx\right\rvert + \left\lvert \int_{K(\varepsilon)}^\infty f(x)\,dx\right\rvert + \left\lvert \int_{K(\varepsilon)}^\infty f(x) e^{-ax}\,dx\right\rvert\\ &\leqslant \left\lvert \int_0^{K(\varepsilon)} f(x) \bigl(1 - e^{-ax}\bigr)\,dx\right\rvert + 2\varepsilon \end{align}$$

for all $a \leqslant \delta$, and since $e^{-ax}$ converges to $1$ uniformly on $[0,K(\varepsilon)]$, there is an $A(\varepsilon) > 0$ such that the remaining integral has absolute value less than $\varepsilon$ for all $a < A(\varepsilon)$. Thus the interchangeability of limit and integral is established in that case.

The condition is fulfilled for the Bessel function $J_0$ as well as for $\frac{\sin x}{x}$. These functions oscillate with decreasing amplitude and periodic resp. nearly periodic zeros, and the same holds for the product of these with $e^{-ax}$. Therefore (for the Bessel function, that is not as easy to show as for $\frac{\sin x}{x}$) the absolute integrals between two consecutive zeros

$$I_k = \int_{z_k}^{z_{k+1}}\lvert f(x)\rvert\,dx$$

form a monotonically decreasing (at least from some point on) sequence converging to $0$, and we have

$$\left\lvert \int_{z_k}^\infty f(x) e^{-ax}\,dx\right\rvert \leqslant I_k$$

since the signs alternate.

As suggested in the comments, the easiest way to see this is with the dominated convergence theorem. Suppose $f \in L^1(0,\infty)$, i.e. $$\int_0^\infty \! |f| \, dx < \infty$$ Let $a_n \in \mathbb{R}$ be some sequence such that $a_n \geq 0$ and $a_n \to 0$. Define $f_n(x) = f(x)e^{-a_nx}$. Then we have that $$|f_n(x)| \le |f(x)|$$ for all $x \in [0,\infty)$ and it is clearly true that $$\lim_{n \to \infty} f_n(x) = f(x)$$ for all $x \in [0,\infty)$. Thus by the dominated convergence theorem we have $$\lim_{n \to \infty} \int_0^\infty \! f_n \, dx = \int_0^\infty \! f \, dx$$ But this says that for every non-negative sequence $a_n$ with $a_n \to 0$ we have

$$\lim_{n \to \infty} \int_0^\infty \! fe^{-a_nx} \, dx = \int_0^\infty \! f \, dx$$ which, by the general properties of metric spaces implies that, $$\lim_{a \to 0^+} \int_0^\infty \! fe^{-ax} \, dx = \int_0^\infty \! f \, dx$$ is also true.