# Necessary condition for the convergence of an improper integral.

My calculus professor mentioned the other day that whenever we separate an improper integral into smaller integrals, the improper integral is convergent iff the two parts of the integral are convergent. For example:

$$\int_0^{+\infty} \frac{\log t}{t} \mathrm{d}t = \underbrace{\int_0^1 \frac{\log t}{t} \mathrm{d}t}_{\text{Diverges to }-\infty} + \underbrace{\int_1^{+\infty} \frac{\log t}{t} \mathrm{d}t}_{\text{Diverges to }+\infty}$$

So the integral would not converge, because one of the parts of the integral is divergent (or both in this case).

However, I don't see why the part that diverges to $-\infty$ and the part that diverges to $+\infty$ cannot cancel out and make it converge, how ever counterintuitive it may seem. It's what happens with the Dirichlet Integral to some extent, although the areas are bounded.

It could be that the problem arises when things tend to $\pm\infty$ but if I recall correctly this is not a problem for the sum of an infinite series, e.g.:

$$\sum_{n=1}^{\infty}A_n+B_n= \underbrace{\sum_{n=1}^{\infty}A_n}_{\text{Diverges to }+\infty} + \underbrace{\sum_{n=1}^{\infty}B_n}_{\text{Diverges to }-\infty} \nRightarrow \nexists \sum_{n=1}^{\infty}A_n+B_n \vee \sum_{n=1}^{\infty}A_n+B_n= \pm \infty$$

Where does the problem arise?

Also, if my use of symbology is incorrect (which I suspect it is) please tell me so. I'm trying to writing more formally and efficiently.

This is a good question! It is all a matter of how to define an improper integral. The traditional way is to write

$$\int_a^b f(t)\, dt = \lim_{x \to a} \lim_{y \to b} \int_x^y f(t)\, dt$$

whenever one of the bounds is worrisome. In particular, we can write

$$\int_a^b f(t) \,dt = \lim_{x \to a} \lim_{y \to b} \left(\int_x^1 f(t) \, dt + \int_1^y f(t)\, dt \right)$$ or $$\lim_{x \to a} \lim_{y \to b} \left(\int_x^1 f(t) \,dt \right) + \lim_{x \to a} \lim_{y \to b} \left( \int_1^y f(t) \,dt \right)$$

so that

$$\int_a^b f(t)\, dt = \lim_{x \to a} \left(\int_x^1 f(t) \,dt \right) + \lim_{y \to b} \left(\int_1^y f(t)\, dt \right).$$

Now convergence means that the integral is finite and the only way for the sum of two numbers to be finite is if both are finite. Hence your professor's comment. All of the above assumes $a < 1 < b$ and that there is no trouble with $f$ near $t=1$.

But you are right that you can get cancelation if you take the limits differently. Instead of taking the limit to a and then the limit to b sequentially, if you took them at the same time--that is to say you combined them--you could possibly get cancellation.

Here's an example. If define instead the improper integral as

$$\int_{-\pi/2}^{\pi/2} \tan t\, dt = \lim_{x \to \pi/2} \int_{-x}^x \tan t\,dt$$

then

$$\int_{-\pi/2}^{\pi/2} \tan t\, dt = \lim_{x \to \pi/2} \left(- \ln \cos(x) + \ln \cos(-x) \right) = 0.$$

What you are thinking of is a useful concept and it often comes up with the term Principal Value in more advanced calculus courses.

• $\int_{-\infty}^{+\infty} \tan t \ \mathrm{d}t$ would also equal zero following your reasoning, would it not? Commented Feb 15, 2015 at 1:45
• Well, not exactly, or perhaps yes. Remember you need to take a limit every time you have a vertical asymptote and $tan(t)$ has infinitely many of them. But $tan(t)$ is periodic and each period things cancel. So you can't use the new definition I have in this case. But you could "show" that $\int_{-\infty}^{\infty} t^3 dt = 0$ with a similar idea. Commented Feb 15, 2015 at 1:58

It's a matter of convention, and the precise notation of which kinds of infinities are allowed to cancel out and which aren't depends on the precise notion of "integral" you're using. For Riemann integrals, the usual conventions are this

1. For proper integrals, i.e. integrals $\int_a^b f(x)\,dx$ with finite $a$ and $b$, no infinities are allowed to cancel. Thus, if either the area above the $x$-axis, or the area below it, are infinite, the integral does not exist.

2. For improper integrals, infinities are allowed to cancel to some extent. The integral $\int_a^\infty f(x)\,dx$ is defined as $$\int_a^\infty f(x)\,dx := \lim_{b \to \infty} \int_a^b f(x)\,dx.$$ It may thus be that the total areas of $f$ above and below the $x$-axis are infinite, but within every finite range they must be finite. But both areas may grow larger and targer as $b$ increases - since the limit is taken after the positive and negative areas are subtracted, the limit can exist even though both parts grow arbitrarily large. This is what happens for the Dirichlet integral.

For proper integrals, there's the notion of the principal value of an (otherwise undefined) integral, written $\text{PV}\int\,\ldots$ which does allow infinite areas to cancel out. For example, $$\text{PV}\int_{-1}^1 \frac{1}{x} \,dx = \lim_{\delta\to 0} \int_{[-1,1]\setminus [-\delta,\delta]} \frac{1}{x}\,dx = \lim_{\delta\to 0} \left(\int_{-\infty}^\delta \frac{1}{x}\,dx + \int_\delta^\infty \frac{1}{x}\,dx\right) = 0,$$ even though $\int_{-1}^1\frac{1}{x}\,dx$ does not exists are a proper integral. This is the kind of cancelling out that you have in mind, I think. I consider the fact that improper integrals are usually considered to still be somewhat "normal" integrals, while calling integrals of the above $\frac{1}{x}$-type "principal values" more of a historical accident than anything else.

For lebesgue integrals, i.e. the integral definition used in measure theory, infinite areas are never allowed to cancel out. Lebesgue integrals can directly be defined even for infinite bounds, i.e. the definition of $\int_a^\infty \ldots$ as $\lim_{b\to\infty} \int_a^b \ldots$ is no longer necessary. But if both the positive and the negative areas under the integrand are infinite, the integral is undefined. The direchlet integral, for example, is undefined as a lebesgue integral. Such integrals are, btw, the only cases in which an integral can be defined as a (improper!) riemann integral, but be undefined as a lebesgue integral.

If you study the theory of lebesgue integrals, i.e. measure theory, it will become clear that never allowing infinities to cancel has huge benefits (although this is not the only different between the two integral definitions. Lebesgue integrals have lots of other advantages as well).

• Is the PV well-defined? By tweaking the "excised" interval cleverly instead of making it $[-\delta,\delta]$, couldn't I make it converge to any value, instead of 0? Commented Feb 15, 2015 at 2:09
• @user7530 You're correct that using a different "excised" interval may change the "principal value". It's still well-defined though - the usual definition of the (cauchy) principal value mandates that you pick a symmetric interval. Other definition may yield other values of course.
– fgp
Commented Feb 15, 2015 at 3:55

$$\int_{-\infty}^{+\infty} x\,dx = \text{what?}$$ $$\int_{-\infty}^{+\infty} x\,dx = \underbrace{\int_{-\infty}^0 x\,dx}_{\text{Diverges to }-\infty} + \underbrace{\int_0^\infty x\,dx}_{\text{Diverges to }+\infty}$$ But $$\lim_{a\to\infty} \int_{-a}^a x\,dx = 0.$$

Here's a more involved example: $$\lim_{a\to\infty}\int_{-a}^a \frac{x\,dx}{1+x^2} = 0 \tag 1$$ but $$\lim_{a\to\infty}\int_{-a}^{2a} \frac{x\,dx}{1+x^2} = \lim_{a\to\infty} \frac 1 2 \log_e \frac{1+4a^2}{1+a^2} = \log_e 2.$$

Re-arranging a sum or an integral can result in a different value only if positive and negative parts both diverge to infinity.

The result in $(1)$ is the "principal value" of this improper integral.

• Is the principal value only valid when the function is odd? Commented Feb 15, 2015 at 10:39
• @Gonate : Certainly not; there is no need for that assumption. Commented Feb 15, 2015 at 18:34