Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 of places without any justification that $$ \lim_{a \to 0^{+}} \int_{0}^{\infty} J_{0}(x) e^{-ax} \ dx = \int_{0}^{\infty} J_{0}(x) \ dx = \lim_{a \to 0^{+}} \frac{1}{\sqrt{1+a^{2}}} = 1 .$$

EDIT:

I decided to resurrect this thread since in the answer below it is assumed that $\displaystyle \int_{0}^{\infty} f(x) \ dx$ converges absolutely.

But in the example above, $ \displaystyle \int_{0}^{\infty} J_{0}(x) \ dx$ only converges conditionally.

And there are other examples like

$$ \int_{0}^{\infty} \frac{\sin x}{x}e^{-ax} \ dx = \arctan \left(\frac{1}{a} \right) \ .$$

So do we only need $ \displaystyle \int_{0}^{\infty} f(x) \ dx $ to converge conditionally?

share|improve this question
3  
Usually this is achieved through the use of the dominated convergence theorem. –  Antonio Vargas Mar 16 '12 at 8:19
add comment

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
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
1  
But $\int_{0}^{\infty} J_{0}(x) \ dx$ does not converge absolutely. –  Random Variable Mar 23 '12 at 20:41
add comment

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.

share|improve this answer
    
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 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 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 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 at 15:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.