# bringing a limit inside of an integral

How do we justify that $\displaystyle \lim_{a \to 0^{+}} \int_{0}^{\infty} f(x) e^{-ax} \ dx = \int_{0}^{\infty} f(x) \ dx$?

For example,$\displaystyle \int_{0}^{\infty} J_{0}(x) e^{-ax} \ dx = \frac{1}{\sqrt{1+a^{2}}}$ for $a>0$.

I've seen it stated in a couple of places without any justification that $\displaystyle \lim_{a \to 0^{+}} \int_{0}^{\infty} J_{0}(x) e^{-ax} \ dx = \int_{0}^{\infty} J_{0}(x) \ dx = 1$.

It was suggested to me that the justification comes from Abel's theorem, but I'm not sure.

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Usually this is achieved through the use of the dominated convergence theorem. –  Antonio Vargas Mar 16 '12 at 8:19
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.
I think your $g$ should be $f$. Also, you want your $a_n$:s to be positive for the inequality to hold. –  mrf Mar 19 '12 at 22:19
But $\int_{0}^{\infty} J_{0}(x) \ dx$ does not converge absolutely. –  Random Variable Mar 23 '12 at 20:41