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Consider a function $f(x)$ which is undefined / indeterminate at $x=a$. Is it still true that,

$$\int\limits_a^a f(a)\,\mathrm dx=0~?$$

asdsaf
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  • What do you mean by "indeterminate"? Under the usual definition, a function $f$ is defined at $a$ or not: $f(a)$ is either defined or undefined. – Rory Daulton Jun 04 '15 at 20:33
  • @RoryDaulton, for example, if $f(x)=x^x$, then $f(0)$ is said to be "indeterminate" by some. – asdsaf Jun 04 '15 at 20:58
  • A related question i came across this afternoon: http://math.stackexchange.com/questions/1311840/mathcali-int-limits-00-x-lfloor-x-rfloor-mathrm-dx-0-textrm-or-u?noredirect=1#comment2664667_1311840 – b00n heT Jun 04 '15 at 21:41
  • @b00nheT, that was me actually. :) I asked it again (general form this time) since the previous questions doesn't have any answers that satisfies me. – asdsaf Jun 04 '15 at 21:57

2 Answers2

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First note that the only possible partition of $[a,a]$ is $\{a\}$, which is again $a$. Therefore the Riemann sum is $$\sum_{n=1}^1f(a)(a-a)$$

This expression is not defined, so we really can't do anything with it, other than give it some prescribed value. Instead of going this far back though, we can directly assign a value to the original integral (also to not wind up with some inconsistency for discrete sums). The inspiration is that if $g$ is integrable over $[a,b]$ for some $b > a$, and $g$ is not defined at $a$, we have $$\int_{[a,b]} g(x)\,dx = \int_{(a,b]}g(x)\,dx$$

since we don't have to use $g(a)$ in taking the limit of Riemann sums (or in considering supremums/infimums). Therefore it's only sensible to define $\int_{[a,a]}g(x)\,dx$ as $0$, so that we have $$\int_{[a,b]} g(x)\,dx = \int_{[a,a]}g(x)\,dx + \int_{(a,b]}g(x)\,dx$$ and now the additivity of the integral holds in more generality. If such shameless and too-convenient-seeming definitions cause you discomfort (as they do me), note that you could always stop at every place in your theoretic deductions where you might have to integrate over degenerate intervals, and split into cases (where the case that the interval is degenerate most likely simplifies for some other trivial reason). No matter how much you want to avoid conventions, nobody in their right mind would take this approach to Riemann integration.

This mess is easily avoided with Lebesgue integration since we only care about equivalence classes of functions where a small amount of point-values don't matter.

GPerez
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I assume you intended to write $f(x)$ inside the integral; for instance $$\int_0^0 \frac{1}{x}\ dx\ \stackrel{?}{=}\ 0.$$

This is a matter of convention and the specific definition of integration that you are using. I believe that most rigorous definitions of the Riemann integral using tagged partitions etc. will require $f$ to be defined on $[0,0]$, but it wouldn't surprise me if some theorem somewhere requires $\int_0^0 f(x)\,dx = 0$ regardless of $f$ (in analogy to the empty sum).

user7530
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  • I'm working with elementary stuff, so I guess this is simple Riemann integral that I'm considering. I have been told that the integral will be $0$ regardless of the integrand if we're considering Lebesgue integral, but my concern is with Riemann integral. Would it be zero in that case?

    I need the answer to complete a generalization of a problem that I've doing.

     – asdsaf Jun 04 '15 at 21:03
  • @GPerez, So, there isn't any definitive answer? – asdsaf Jun 04 '15 at 21:13
  • @asdsaf I think I made a mistake, I'll think about it some more an add an answer. – GPerez Jun 04 '15 at 21:18
  • @asdsaf What definition of the Riemann integral are you using? Look at it to get your definitive answer (though NB that definitive here will mean with respect to your definition only.) – user7530 Jun 04 '15 at 21:20
  • @user7530, wait, there are different definitions of the Riemann Integral? – asdsaf Jun 04 '15 at 21:27
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    @asdsaf There is no stone tablet somewhere on which are engraved the definitions of mathematical objects: different mathematicians write different textbooks and papers and use different conventions. Of course all reasonable definitions of the Riemann integral will agree on the value of e.g. $\int_0^1 \sin x,dx$ but your question is about a very technical corner case that could well be handled differently in different texts. – user7530 Jun 04 '15 at 21:36