Sufficient conditions to conclude that $\lim_{a \to 0^{+}} \int_{0}^{\infty} f(x) e^{-ax} \, dx = \int_{0}^{\infty} f(x) \, dx$ What are sufficient conditions to conclude that $$ \lim_{a \to 0^{+}} \int_{0}^{\infty} f(x) e^{-ax} \, dx = \int_{0}^{\infty} f(x) \, dx \ ?$$
For example, for $a>0$, $$ \int_{0}^{\infty} J_{0}(x) e^{-ax} \, dx = \frac{1}{\sqrt{1+a^{2}}} \, ,$$
where $J_{0}(x)$ is the Bessel function of the first kind of order zero.
But I've seen it stated in a couple places without any justification that $$ \int_{0}^{\infty} J_{0}(x) \, dx = \lim_{a \to 0^{+}} \int_{0}^{\infty} J_{0}(x) e^{-ax} \, dx =  \lim_{a \to 0^{+}} \frac{1}{\sqrt{1+a^{2}}} =  1 .$$
EDIT:
In user12014's answer, it is assumed that $ \int_{0}^{\infty} f(x) \, dx$ converges absolutely.
But in the example above, $ \int_{0}^{\infty} J_{0}(x) \, dx$ does not converge absolutely.
And there are other examples like
$$ \int_{0}^{\infty} \frac{\sin x}{x} \, dx  = \lim_{a \to 0^{+}} \int_{0}^{\infty} \frac{\sin x}{x}e^{-ax} \, dx =  \lim_{a \to 0^{+}} \arctan \left(\frac{1}{a} \right) = \frac{\pi}{2} $$
and 
$$ \int_{0}^{\infty} \text{Ci}(x) \, dx = \lim_{a \to 0^{+}} \int_{0}^{\infty} \text{Ci}(x) e^{-ax} \, dx =  - \lim_{a \to 0^{+}} \frac{\log(1+a^{2})}{2a} =0 \, ,$$ where $\text{Ci}(x)$ is the cosine integral.

SECOND EDIT:
Combining Daniel Fischer's answer below with his answer to my follow-up question shows that if  $\int_{0}^{\infty} f(x) \, dx$ exists as an improper Riemann integral, then $$\lim_{a \to 0^{+}} \int_{0}^{\infty} f(x) e^{-ax} \, dx = \int_{0}^{\infty} f(x) \, dx.$$
 A: 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.
A: 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.
