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

Question

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.

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

Answer 1

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.