I am looking for an intuitive explanation that explains the difference between absolutely and conditionally convergent (Riemann) improper integrals. Please note I understand how one goes about determining this difference and is not the question being asked here.

The case for series

To paraphrase from Wikipedia, absolute convergence is important in the study of infinite series since its definition is strong enough to have properties of finite sums that not all convergent series possess, yet is broad enough to occur commonly. One such property is the rearrangment of terms in an absolutely convergent series. Doing so does not change the value of its sum - something which is not true for conditionally convergent series. So in a way one can say absolutely convergent series behave "nicely" without undue "conditions" being placed upon them when summed.

The case for improper integrals

So what is the case for improper integrals? Does a rearrangement of terms in the case of infinite series perhaps correspond to a rearrangement of various intervals of integration if partitioned?


2 Answers 2


Just like infinite series, if the improper integral is conditionally convergent then it may not be uniquely defined unless the limiting process is specified in a very restricted way. A conditionally convergent improper integral is the beneficiary of some very fortunate cancellation (like the series) and this can be disrupted by "rearrangement."

For example, consider the conditionally convergent improper integral

$$\int_0^\infty \frac{\sin x }{x}\, dx = \frac{\pi}{2},$$

where convergence to that well-known value is obtained by defining the improper integral as

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

Intuitively, we might expect to obtain a unique limit for

$$\lim_{n \to \infty} \int_{A_n} \frac{ \sin x}{x} \, dx,$$

where $(A_n)$ is a sequence of nested sets converging to $[0,\infty)$, that are finite unions of intervals. In other words, $A_n \subset A_{n+1}$ and $\bigcup_{n=1}^\infty A_n = [0,\infty).$

Surprisingly, there is no unique limit independent of the choice for $(A_n)$ as consequence of conditional convergence. If $A_n = [0,n]$ then we obtain the limit $\pi/2$. However, if we take

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

then it is easy to show that $A_n \subset A_{n+1}$ for all $n$, and $\bigcup_{n=1}^\infty A_n = [0,\infty)$, but

$$\lim_{n \to \infty} \int_{A_n}\frac{\sin x }{x} \, dx > \frac{\pi}{2} + \frac{\log 2}{\pi}.$$

The integral over $A_n$ is

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

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

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


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


The typical example I've seen is something like $\int_{-\infty}^\infty \frac{x}{x^2+1} dx$. The integral converges to zero if we take it to mean $\lim_{N \rightarrow \infty} \int_{-N}^N \frac{x}{x^2+1} dx$, but we can make it converge to a non-zero value by e.g. doing $\lim_{N \rightarrow \infty} \int_{-N}^{2N} \frac{x}{x^2+1} dx$, and obviously it's possible to make it diverge as well.


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