I'm wondering whether this expression has some significance:

$$\int_{-1}^1 \frac{dx}{\sqrt{|x|}}$$

And, in general, if expressions in the following form make sense:

$$\int_a^b f(x) dx$$

Where the set $(a, b)$ contains points out of the domain of $f(x)$.

According to the definitions I know, these are neither definite nor improper integrals, so the expression shouldn't make any sense.

But in case the integral exists, could you please tell me:

  1. Which kind is it?
  2. Is the following equation true? $\displaystyle \int_{-1}^1 \frac{dx}{\sqrt{|x|}} = \int_{-1}^0 \frac{dx}{\sqrt{|x|}} + \int_0^1 \frac{dx}{\sqrt{|x|}}$
  3. Is it an area?
  4. If not, what is the area of $\frac1{\sqrt{|x|}}$?

UPDATE 1: I'm saying that $\displaystyle \int_{-1}^1 \frac{dx}{\sqrt{|x|}}$ is not an improper integral is because of the following definition (taken from Wikipedia):

[...] an improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number or ∞ or −∞ [...]

-1 and 1 are the endpoints of my integral, but here the problem is 0. Perhaps, is that definition wrong or incomplete?

UPDATE 2: I've replaced $\frac1x$ with $\frac1{\sqrt{|x|}}$, so that my questions about the area make more sense.

UPDATE 3: I'm not particularly interested in calculating the value of that integral, what I really want to know is what is that stuff. An ideal answer would include a valid and coherent mathematical definition of integrals of that kind.

  • $\begingroup$ It does make sense: it is a divergent improper integral. I don't understand why you say it is not improper as it is the same as $$\int\limits_{-1}^0\frac{dx}{x}+\int\limits_0^1\frac{dx}{x}$$ $\endgroup$ – DonAntonio Feb 2 '13 at 11:06
  • $\begingroup$ Ut is undefined as long as you don't set its value of zero to something. If we set $f(x) = \frac{1}{x}$ for all $x \not = 0$ and then set $f(0) = A$ for some real number $A$, the integral $\int_{-1}^1 f(x) dx$ makes perfect sense. $\endgroup$ – malin Feb 2 '13 at 11:08
  • $\begingroup$ @DonAntonio: I've updated my answer explaining why I think it's not an improper integral. $\endgroup$ – hey hey Feb 2 '13 at 13:13
  • $\begingroup$ @malin: true, that's clear, but I'm interested in what happens when 0 is out of the domain. $\endgroup$ – hey hey Feb 2 '13 at 13:14
  • $\begingroup$ @heyhey, I see what you mean, yet your in order your integral is even defined I think it must be put, in fact, in the form of the sum of those two improper integrals as you did in(2)...before you changed the function, of course. $\endgroup$ – DonAntonio Feb 2 '13 at 16:06

The word "improper" refers to an extension of the Riemann integral, so I am assuming that is what you are talking about. The Riemann integral is only defined for functions on a finite interval (such as $[-1,1]$) which are bounded (which $1/\sqrt{|x|}$ is not). If your function or interval is unbounded, the only way to make sense of it using Riemann integrals is to split the interval into finitely many subintervals (such as $[-1,1]$ to $[-1,0)\cup (0,1]$) such that you get an improper Riemann integral on each interval, where here by an improper integral I mean $$\int_a^b f(x)dx:=\lim_{\alpha\to a+}\lim_{\beta\to b-}\int_\alpha^\beta f(x)dx.$$ This seems to be rarely stated or defined in textbooks, despite exercises about evaluating integrals such as $\int_{-1}^1\dfrac1{\sqrt{|x|}}dx$ commonly being given on exams. The common consensus is to call these integrals improper as well, but if you want to be able to answer with a definite yes or no, you have to choose a textbook and use the definitions it uses. Different textbooks often define things in a slightly different way. In particular, not all Wikipedia articles are written by the same person, so they can be inconsistent at times (although your Wikipedia article does state " Integrals are also improper if the integrand is undefined at an interior point of the domain of integration, or at multiple such points." in the introduction).

If we use the Lebesgue integral instead, then the integral of unbounded functions on unbounded intervals can already be defined, so $\int_{-1}^1\dfrac1{\sqrt{|x|}}dx$ has a Lebesgue integral and does not need to be called improper. The Lebesgue integral even allows for your function to be undefined or infinite at a small set of points (which $1/\sqrt{|x|}$ is at $0$).

$\int_{-1}^1\dfrac1x dx$ however, diverges both as an improper Riemann integral and Lebesgue integral. As others have mentioned, there is another extension of the Riemann integral called the principal value at $0$, which gives it the value $0$. My understanding is that principal values are rarely used except sometimes in physics.

  • $\begingroup$ Thanks for your answer. So you are saying that my equation (2) is right, but why does $\int_{-1}^1 \frac{dx}x$ diverge? Shouldn't it be 0? Also, what about my questions about the areas? $\endgroup$ – hey hey Feb 4 '13 at 13:48
  • $\begingroup$ @heyhey: $\int_{-1}^1\frac{dx}x$ diverges because it is a sum of two divergent integrals. Answers to your questions: 1. Improper. 2. Yes. 3. Yes, the area under the curve $1/\sqrt{|x|}$ between $-1$ and $1$. 4. The value of the integral. $\endgroup$ – Samuel Feb 4 '13 at 20:15

Consider the Cauchy principal value

$$ \lim_{\epsilon \to 0}\left(\int_{-1}^{-\epsilon}\frac{1}{x}+ \int_{\epsilon}^{1}\frac{1}{x}\right)=0 $$

  • $\begingroup$ would we get same value from Lebesgue integral? $\endgroup$ – user58512 Feb 2 '13 at 11:21
  • 3
    $\begingroup$ @user58512: No. A function $f$ is Lebesgue integrable if and only if $|f|$ is Lebesgue integrable. $\endgroup$ – Samuel Feb 2 '13 at 11:42
  • $\begingroup$ @Samuel: Thanks for answering him. $\endgroup$ – Mhenni Benghorbal Feb 3 '13 at 0:22

Edit: what follows is partly incorrect. The only way to give a meaning to the integral is to use the concept of Cauchy principal value.

Your specific example. First, I don't see any reason for not calling it an improper integral; second, surely the decomposition you write in 2. is perfectly valid: $$ \int_{-1}^1 \frac1x dx= \lim_{\epsilon\rightarrow 0}\Bigg(\int_{-1}^{-\epsilon}\frac1x dx + \int_\epsilon^1 \frac1x dx\Bigg) $$ Changing variable from $x$ to $-x$ in the first integral you see that the parenthesis is $0$. Due to the symmetry of the integrand, we can say that the result is exactly $0$. (maybe I'm being not rigorous here?)

Accordingly, the area is $0$ as well, in fact it is a signed area.

On the contrary, $\displaystyle\int_{-1}^1\left|\frac1x\right|\; dx$ is a divergent integral.

In general, it still makes sense to define integrals of functions which are singular on the path of integration. Then it may well be that the integral is divergent and the best you can do is to define its principal value, taking a limit analogue to the one which I wrote above.

  • 2
    $\begingroup$ This is incorrect. The integral needs to be decomposed as $\lim_{\varepsilon\to0-}\int_{-1}^{\varepsilon}\dfrac1xdx + \lim_{\delta\to0+}\int_{\delta}^{1}\dfrac1xdx$, which does not exist. $\endgroup$ – Samuel Feb 2 '13 at 11:45
  • $\begingroup$ @Samuel: Thanks for your comment. In which sense, then, the intuition about the symmetry of the function (and the cancellation of the divergent contributes) holds? $\endgroup$ – Andrea Orta Feb 2 '13 at 11:52
  • $\begingroup$ There's something called the cauchy principal value of an integral, which is written $P.V.\int_{-1}^1\dfrac1xdx$, which is defined as your limit. $\endgroup$ – Samuel Feb 2 '13 at 12:32

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