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Suppose a function $f$ is Riemann integrable over any interval $[0,b]$. By definition the improper integral is convergent if there is a real number $I$ such that

$$\lim_{b \to \infty}\int_0^b f(x) dx= I := \int_0^\infty f(x)dx.$$

I have shown that if $f$ is nonnegative then this is equivalent for $n \in \mathbb{N}$ to

$$\lim_{n \to \infty}\int_0^n f(x) dx = I.$$

Now I would like to know for general $f$ (not just nonnegative) if the following proposition holds.

Suppose the improper integral of $f$ over $[0,\infty)$ is convergent. Let $A_n \subset [0, \infty)$ be a nested sequence of compact sets such that $A_1 \subset A_2 \subset A_3 \subset ...$ and $\cup_n A_n =[0,\infty)$, and where the Riemann integral is defined on each $A_n$. Then

$$\lim_{n \to \infty} \int_{A_n}f(x) dx = \int_0^\infty f(x) dx.$$

This question is not about Lebesgue integration. Thank you for any help you can give me.

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  • $\begingroup$ +1 for thinking of alternative definition of improper integral. $\endgroup$ – Paramanand Singh Feb 11 '17 at 3:50
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If the improper integral converges conditionally and not absolutely, then the limit of $\int_{A_n} f$ need not exist. This is somewhat surprising since $A_n \subset A_{n+1}$ and $A_n \uparrow [0,\infty).$

For example, it is well known that the improper integral of $f(x) = \sin x / x$ converges conditionally with

$$\lim_{c \to \infty}\int_0^c \frac{\sin x}{x} \, dx = \frac{\pi}{2}.$$

A counterexample to your conjecture is provided by the following sequence $A_n$ where each set is a finite union of intervals with gaps, of the form

$$A_n = [0, 2n\pi - \pi] \cup \bigcup_{k=n}^{2n}[2k\pi,2k\pi + \pi].$$

It is easy to show that $A_n \subset A_{n+1}$ for all $n$. Furthermore for any $c > 0$ there exists $n$ such that $2n\pi - \pi > c$ and $[0,c] \subset A_n$. This implies $\cup_n A_n = [0,\infty)$.

The integral over $A_n$ is

$$\int_{A_n}\frac{\sin x }{x} \, dx = \int_0^{2n\pi - \pi}\frac{\sin x }{x} \, dx + \sum_{k=n}^{2n} \int_{2k \pi}^{2k \pi + \pi} \frac{\sin x} {x} \, dx,$$

which can be shown to converge to a value greater than $\pi/2 + \log 2 /\pi$.

The first integral on the right-hand side converges to $\pi/2$ and, since $\sin x \geqslant 0$ for $x \in [2k \pi,2k \pi + \pi]$, it follows that

$$\int_{2k \pi}^{2k \pi + \pi} \frac{\sin x} {x} \, dx > \frac{1}{2k\pi + \pi}\int_{2k \pi}^{2k \pi + \pi} \sin x \, dx = \frac{2}{2k\pi + \pi} > \frac{1}{\pi}\frac{1}{k+1}.$$

Thus,

$$\lim_{n \to \infty}\int_{A_n} \frac{\sin x }{x} \, dx > \frac{\pi}{2} + \lim_{n \to \infty} \frac{1}{\pi} \sum_{k = n}^{2n}\frac{1}{k+1}. $$

There can be no unique value of the limit of the integral for every choice of sequence $(A_n).$

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  • $\begingroup$ Very good answer to a very good question. +1 $\endgroup$ – Paramanand Singh Feb 11 '17 at 3:49
  • $\begingroup$ Is this still true if you replace Riemann integrals with Lebesgue integrals? $\endgroup$ – Keshav Srinivasan Oct 16 '18 at 4:37
  • $\begingroup$ @KeshavSrinivasan: Notice that the counterexample involves $\frac{\sin x}{x}$ which is not Lebesgue integrable on $[0,\infty)$. This is very much a consequence conditional convergence of an improper Riemann integral. Lebesgue integrable means absolutely integrable and this issue should not come up. $\endgroup$ – RRL Oct 16 '18 at 18:13
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Not true. You can choose the $A_n$ in a biased way. Consider $f(x)=\sin(x)$ and $A_n=[0,n\pi/2]$.

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  • $\begingroup$ Sorry. This is not helpful. The improper integral $\int_0^\infty \sin x \, dx$ does not exist. $\endgroup$ – SAS Feb 10 '17 at 21:15
  • $\begingroup$ Under this new hypothesis, it is still not true. See RRL's answer, it has a good explanation and example. $\endgroup$ – Fimpellizieri Feb 10 '17 at 21:33

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