# This integral is defined ? $\displaystyle\int_0^0\frac 1x\:dx$

$\displaystyle\int_0^0\frac 1x\:dx$ This integral is defined ? If it define, what is the value ?

• Riemann, Lebesgue or what other type of integral do you mean? – GEdgar May 14 '16 at 20:32
• The function $f(x) = 1/x$ is defined everywhere except $x = 0$. Since the definition of a function at a finite number of points don't matter in integration (both Riemann and Lebesgue), it does not matter that $1/x$ is not defined at $x = 0$. What matters here is that the interval of integration is of length $0$ (also of measure $0$) and hence the integral is $0$ in both Riemann and Lebesgue sense. – Paramanand Singh May 15 '16 at 4:39
• @Paramanand Singh Doesn't the fact that f blows up to infinity matter to Riemann? For the integral I define in my answer,any lower Riemann sum will be zero, but upper Riemann sums will not have well-defined values, and if we insisted, we would take it to be infinite.. – Noe Blassel May 15 '16 at 10:44
• @NoeBlassel: By definition, Riemann integral over interval of length $0$ is $0$. This fact has nothing to do with upper lower Riemann sums. These sums come into picture when interval of integration has finite positive length. – Paramanand Singh May 15 '16 at 12:09
• @NoeBlassel: We can however try to introduce Riemann sums for intervals of length $0$ also. In this case there will be only one partition of interval $[a,a]$ and upper, lower sum will be $0$. – Paramanand Singh May 15 '16 at 12:14

It is defined, and has value $0$: write $I=\int_{0}^{0}\frac{dx}{x}=\int_{-1}^{1}\frac{\chi_{\{0\}}dx}{x}$, where $\chi_{A}$ is the indicator function on set A, $\chi_{A}(x)=1$ if $x$ belongs to $A$, and is $0$ otherwise. Now $\{0\}$ has Lebesgue measure $0$, and outside of $\{0\},\frac{\chi_{\{0\}}}{x}=0$. Hence $I=\int_{-1}^{1}0=0$. Thus $I$ is defined, at least in the sense of Lebesgue.

• I guess in the sense of Lebesgue theory, $\infty .0=0$ , i.e., value $\infty$ assumed in a set of measure $0$. – gary Jul 30 '17 at 20:10

I will take the case of Riemann Integral because the Lebesgue part is nicely done by Noe Blassel.

The definition of Riemann $\int_{a}^{b}f(x)\,dx$ has a pre-condition that $f$ is bounded on $[a, b]$. This itself requires that $f$ must be defined for all points in $[a, b]$. However we are still able to talk about Riemann integrals like $\int_{0}^{1}\sin(1/x)\,dx$. How??

Note that Riemann integrals have a nice property that the value of integral $\int_{a}^{b}f(x)\,dx$ does not depend on values of $f$ on a finite number points in $[a, b]$. Thus we simply define the function $f(x) = \sin (1/x)$ at $x = 0$ in any arbitrary manner say $f(0) = k$ and the function remains bounded in interval $[0, 1]$ and since its only discontinuity is at $x = 0$ the function is Riemann integrable and the integral makes sense.

Next we come to the function $f(x) = 1/x$ on interval $[0, 0]$. Here $f$ is not defined on the interval under consideration. Hence it does not make sense to talk about its integral unless we define it on that interval. So we can define it in any manner by setting $f(0) = k$ and then $f$ is bounded on $[0, 0]$ and the integral is equal to $0$ (if we use upper/lower sums both are $0$).

Note that the argument of the previous case applies to any function on any interval of length $0$ and we expect the integral to be $0$ (via upper / lower sums approach). Hence while studying Riemann integrals we add the following extra definitions:

1) If $f$ is any function then we define $\int_{a}^{a}f(x)\,dx = 0$.

2) If $f$ is Riemann integrable on $[a, b]$ (with $a < b$) then we define the symbol $\int_{b}^{a}f(x)\,dx$ as $-\int_{a}^{b}f(x)\,dx$

Together these two definitions help us to write the following theorem:

Theorem: If $a, b, c$ are any points lying in some closed interval $I$ on which $f$ is Riemann integrable then $$\int_{a}^{b}f(x)\,dx + \int_{b}^{c}f(x)\,dx = \int_{a}^{c}f(x)\,dx$$ irrespective of any order relations between $a, b, c$.

• How do we talk about integrals like $sin(1/x)$? This is not comparable, to $\int 0/0dx$ since above is bounded. – gary Jul 30 '17 at 20:11
• @gary: I have explained in my answer that such functions are easily handled by defining them at troublesome points in arbitrary manner. – Paramanand Singh Jul 30 '17 at 20:47
• I don't agree here, if I understood your argument. You can handle a product times a bounded expression, since the partition $x_{i+1}-x_i$ comprising a bounded value can be made indefinitely small, making that individual value in the Riemann sum zero. The problem is a different one when you have something like $a\infty$, since , in Riemann integration ( unlike with the Lebesgue case) $0 \times \infty$ is not defined. – gary Jul 30 '17 at 21:12
• @gary: Riemann integration is only defined for bounded functions. – Paramanand Singh Jul 31 '17 at 0:10
• That is precisely why you can't integrate $1/x$ at $x=0$; that was precisely my point. Only way I see is just redefining $1/x$ to have a different value at $x=0$. But then you are not really integrating $1/x$. – gary Jul 31 '17 at 0:16