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Under what conditions does $$ \lim_{a \to 0^{+}} \int_{0}^{\infty} f(x) e^{-ax} \, dx = \int_{0}^{\infty} f(x) \, dx \ ?$$

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 $$ \int_{0}^{\infty} J_{0}(x) \, dx = \lim_{a \to 0^{+}} \int_{0}^{\infty} J_{0}(x) e^{-ax} \, dx = \lim_{a \to 0^{+}} \frac{1}{\sqrt{1+a^{2}}} = 1 .$$


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

$$ \int_{0}^{\infty} \frac{\sin x}{x} \, dx = \lim_{a \to 0^{+}} \int_{0}^{\infty} \frac{\sin x}{x}e^{-ax} \, dx = \lim_{a \to 0^{+}} \arctan \left(\frac{1}{a} \right) = \frac{\pi}{2} $$


$$ \int_{0}^{\infty} \text{Ci}(x) \, dx = \lim_{a \to 0^{+}} \int_{0}^{\infty} \text{Ci}(x) e^{-ax} \, dx = - \lim_{a \to 0^{+}} \frac{\log(1+a^{2})}{2a} =0 \, ,$$ where $\text{Ci}(x)$ is the cosine integral.

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Usually this is achieved through the use of the dominated convergence theorem. – Antonio Vargas Mar 16 '12 at 8:19
up vote 7 down vote accepted

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.

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Thanks. This was the very first question I had ever asked on here. But it remained only partially answered for 2+ years. Can the M-test for checking the uniform convergence of a series be extended to checking the uniform convergence of an integral? I sort of remember reading about that. – Random Variable Jun 8 '14 at 21:17
As far as I can see, that would only work for absolutely integrable functions. But I might be overlooking something. – Daniel Fischer Jun 8 '14 at 21:25
If $f(t)$ is continuous for $t>0$ and $\int_{0}^{\infty} f(t) \ dt$ converges conditionally, will this argument always justify bringing the limit inside of the integral? I can't think of a counterexample. – Random Variable Jun 9 '14 at 15:35
I don't know. It looks a little "too good to be true", but I don't see a way for it to fail yet. Maybe the specific behaviour of $e^{-ax}$ makes it true, gotta think about it. – Daniel Fischer Jun 9 '14 at 15:44

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.

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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
Ah, of course. Will edit. – user12014 Mar 19 '12 at 22:35
But $\int_{0}^{\infty} J_{0}(x) \ dx$ does not converge absolutely. – Random Variable Mar 23 '12 at 20:41

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