11
$\begingroup$

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

  • 4
    Usually this is achieved through the use of the [dominated convergence theorem](http://en.wikipedia.org/wiki/Dominated_convergence_theorem).2012-03-16

2 Answers 2

8

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.

  • 0
    @RandomVariable The dominated convergence theorem gives you the conclusion on $(0,1]$, since on that interval $\bigl\lvert \frac{f(x)}{x}\bigr\rvert$ is an integrable dominating function for $f(x)x^{a-1}$. On $(1,+\infty)$ we have \lvert f(x)x^{a-1}\rvert >\lvert f(x)x^{-1}\rvert (except where $f(x)\neq 0$). But if you have (Lebesgue) integrability of $f(x) x^{b-1}$ for some b > 0, then you can of course use that as dominating function on $(1,+\infty)$. If you don't have Lebesgue integrability and all $\int_1^{+\infty} f(x)x^{a-1}\,dx$ only exist as improper Riemann integrals, it's harder.2016-05-29
4

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.

  • 2
    But $\int_{0}^{\infty} J_{0}(x) \ dx$ does not converge absolutely.2012-03-23