$\int_a^af(x) \, dx$ always $0$?

Question

I was studying integrals and just out of curiosity,

Does there exist any 'continuous' functions such that $\int_a^af(x) \, dx$ ($a$ is any number) equals a value other than $0$?

Since continuous functions are Riemann integrable, so I think it should be $0$. Is this correct?

Also, with out the condition 'continuous', does there exist any function such that $\int_a^af(x) \, dx$ isn't $0$?

EDIT

I'm looking for any function that $\int_a ^a\ f(x) \neq 0$. Can anyone find me one?

@imranfat Okay, so the value of the integral should be 0 when the function is continuous. What about discontinuous ones? — zxcvber, Jul 09 '16 at 00:31
If you use lebesgue integral and the measure is the counting measure, the answer is yes. — Xianjin Yang, Jul 09 '16 at 00:31
@zxcvber It is always zero when the function $f$ is integrable. A more interesting question is something like $\int_0^0 \frac 1x \ dx$, which has been asked before (though I can't seem to find it now). — MathematicsStudent1122, Jul 09 '16 at 00:32
Here is a heuristic argument (not rigorous) for why $\int \limits_{a}^{a} f(x),dx = 0$ from a calculus point of view: Let's just say $f(a) > 0$. Then $\int \limits_{a}^{a} f(x) ,dx$ stands for the area under the curve $f(x)$, but only over the region/interval $[a,a]$. The "interval" $[a,a]$ is just the point $a$, so any rectangle you draw over this interval will have $0$ width. Now, if I gave you a rectangle and asked for its area, and told you the length is $5$ in. and the width is $0$ in., you would calculate the area as $5 \cdot 0 = 0$. So a rectangle with $0$ width has $0$ area. — layman, Jul 09 '16 at 00:58
(cont.) Since any rectangle drawn over the region $[a,a]$ has width $0$, then it has area $0$. Since the integral depends on these rectangles, the integral will also be $0$. Hope that makes sense. — layman, Jul 09 '16 at 00:59
@user46944 Thanks, and yes it makes sense. I understand the concept 'area' but I was just wondering if there were any weird functions that I don't know that satisfy the property I want. — zxcvber, Jul 09 '16 at 01:00
@zxcvber No, not with the Riemann integral you learn about in Calculus, and by extension not with the Lebesgue integral you learn about later in Real Analysis, since the interval $[a,a]$ has what's called measure $0$. However, if you learn something called measure theory, and define integration with respect to a different measure than the Lebesgue measure, say like counting measure, or any measure where the interval $[a,a]$ has nonzero measure, then the integral does not have to be $0$. — layman, Jul 09 '16 at 01:02
@user46944 So for continuous functions, the integral is always $0$. Can you show me an example of what you mentioned? When does it not have to be $0$? — zxcvber, Jul 09 '16 at 01:07
@zxcvber To your first point about continuous functions, if you don't already know, remember that the Riemann integral we learn in calculus is only finite if the integrand function is continuous almost everywhere (i.e., its set of discontinuities is so small it has "measure 0"). So, no the function doesn't have to be completely continuous, but it has to be mostly continuous to even talk about the Riemann integral, and for these mostly continuous functions, yes the integral above will always be $0$. — layman, Jul 09 '16 at 01:09
@zxcvber To your question: If you know any measure theory, then define $\mu$ to be counting measure, which means for any set $A \subseteq \Bbb R$, $\mu(A)$ (the measure of $A$) is the number of elements in $A$. So the interval $[a,a]$ has one element, so $\mu([a,a]) = 1$. Now, take $f(x) =1$ for every $x \in \Bbb R$. So $f$ is the constant function $1$. Then $\int \limits_{a}^{a} f(x) ,d\mu = 1$. Notice the $dx$ from earlier is replaced with $d\mu$ here because $dx$ usually means the integral is with respect to Lebesgue measure. $d\mu$ means it is with respect to the measure $\mu$. — layman, Jul 09 '16 at 01:11

score 7 · Answer 1 · edited Jul 09 '16 at 00:28

You are right, when talking about the Riemann Integral, then $$ \int_a^a f(x)\; dx $$ always equal to zero. If this integral is defined, then it is zero. To see this, you just have to write out the definition of the integral: $$ \int_a^a f(x)\; dx = \lim_{n\to \infty}\sum_{i=0}^{n} f(x_i^*) \, \Delta x $$ Here $\Delta x = \frac{a - a}{n} = 0$, so all the sums will be zero, and so the limit will be zero.

score 3 · Accepted Answer · edited Jul 09 '16 at 00:34

3

No, for $f(x)$ any continuous function, $\int_a^a f(x)dx= 0$. As for non-continuous functions, my first thought was a "Dirac delta function" for which $\int_C \delta(x)dx= 1$ for any set, C, containing 0. However, a "delta function" is not a true function- it is a "generalized function" or "distribution".

edited Jul 09 '16 at 00:34

Thomas

43,555

answered Jul 09 '16 at 00:33

user247327

18,710

So, $\int_0^0 \delta (x) dx$ equals $1$?, and why is a "generalized function" not a true function? – zxcvber Jul 09 '16 at 00:36
1

@zxcvber There is no function with the properties that $\delta$ has, so it's not a function. The notation $\int \delta(x)f(x)\mathrm{d}x$ is misleading to the uninitiated, since it's not defined as an integral at all, but the application of a linear functional called delta to the function $f(x)$. – anon Jul 09 '16 at 00:40
1

The delta function can sort of be a counterexample if you think of it as a measure instead of a distribution. Consider the Lebesgue integral $\int_0^01d\delta:=\int_{{0}}1d\delta=1$. – Funktorality Jul 09 '16 at 01:01

$\int_a^af(x) \, dx$ always $0$?

2 Answers2