The question is pretty straightforward. In class, I have seen $\int_{\Bbb R}$ and $\int_{-\infty}^\infty$ being used interchangeably. However $\int_{\Bbb R}$ contains no implicit sense of direction, so technically isn't $\int_\infty^{-\infty}$ also the same as $\int_{\Bbb R}$?

Or $\int_{\infty}^{0}+\int_{-\infty}^{0}\equiv\int_{\Bbb R}$, provided all integrals exist?

There could be several such possibilities. Is that why we have attached a conventional direction to $\int_{\Bbb R}$, i.e. from $-\infty$ to $\infty$?

  • 2
    $\begingroup$ In Riemann integration integral over $\mathbb R$ is taken as integral from $-\infty$ to $\infty$. In Lebesgue integration direction has no relevance. $\endgroup$ Aug 31, 2019 at 6:33
  • $\begingroup$ More generally, if $a,\,b$ are extended reals with $a\le b$, $\int_{[a,\,b]}$ is synonymous with $\int_a^b$. $\endgroup$
    – J.G.
    Aug 31, 2019 at 7:45

4 Answers 4


The notation $\int_a^b$ (or, in an extended sense, $\int_{-\infty}^{\infty}$) is meant to be suggestive of Riemann integration, which, as you said, entails an orientation in your Riemann sums and hence, your integrals.

The notation $\int_{\mathbb{R}}$ is a more general notation from the theory of Lesbegue integration, where integrals are non oriented and over a measure space $X$. You are, indeed, correct that $\int_{-\infty}^{\infty}$ and $\int_{\mathbb{R}}$ are used interchangably, but this is due to the fact that the Lesbegue integrals and the Riemann integrals agree in pretty much all relevant cases. The notation $\int_{\infty}^{-\infty}$ refers, conventionally, to "summing in the wrong direction" and it's not true that this could be substituted with $\int_{\mathbb{R}},$ since the corresponding Riemann and Lesbegue integrals will have a sign difference.


As it has been pointed out, $\int_{\Bbb R}$ usually refers to Lebesgue integral over $\Bbb R$, whereas $\int_{-\infty}^{\infty}$ usually refers to the improper Riemann integral over $\Bbb R$. A nuance of which, in my opinion, all writers should be mindful is that the condition of improper-Riemann integrability and Lebesgue integrability for functions that change sign are slightly different, and they sometimes play a role.

In short, given $a\in\Bbb R$, $\int_{-\infty}^{\infty}f(x)\,dx=a$ means that $f$ is Riemann integrable on all subintervals and that either of the following two equivalent conditions hold:

  1. $\lim_{x\to -\infty}\lim_{y\to\infty} \int_x^y f(t)\,dt=a$

  2. for all $\varepsilon>0$ there is some $M>0$ such that $\left\lvert a-\int_x^y f(t)\,dt\right\rvert<\varepsilon$ for all $x<-M$ and for all $y>M$.

Whereas $\int_{\Bbb R} f(x)\,dx=a$ means that, as Lebesgue integrals, $\int_{\Bbb R} f^+(x)\,dx<\infty$, $\int_{\Bbb R} f^-(x)\,dx<\infty$ and $a=\int_{\Bbb R}f^+(x)\,dx-\int_{\Bbb R}f^{-}(x)\,dx$.

There are some obvious modifications to (1) for the cases $a=\infty$ and $a=-\infty$, whereas for Lebesgue integration the cases $a=\infty$ and $a=-\infty$ are adapted to when $\lvert f\rvert$ has infinite integral and exactly one of $f^+$ or $f^-$ has infinite integral.

In the aforementione instances, it is always the case that $\int_{\Bbb R}f(x)\,dx=a$ plus $f$ Riemann integrable on compact intervals implies $\int_{-\infty}^\infty f(x)\,dx=a$.

On the other hand, it is possible that $\int_{\Bbb R}f^+(x)\,dx=\int_{\Bbb R}f^-(x)\,dx=\infty$, so that $\int_{\Bbb R}f(x)\,dx$ doesn't make sense, while $\int_{-\infty}^\infty f(t)\,dt=a$ for some real number $a$. A famous instance is \begin{align}&\int_\Bbb R\max\left\{-\frac{\sin x}{x}, 0\right\}\,dx=\infty \\&\int_\Bbb R\max\left\{\frac{\sin x}{x}, 0\right\}\,dx=\infty\\ &\int_{-\infty}^\infty\frac{\sin x}{x}\,dx=\pi\end{align}


As a Riemann integral, the notation $\int_\mathbb{R}$ necessarily implies a positive orientation, so the only interpretation is $\int_{-\infty}^\infty.$

This type of notation is more common in Riemann integrals when we talk about the multiple integrals $\iint_D f(x,y)\,dA$ and $\iiint_R f(x,y,z)\,dV$ where $D,R$ are domains of integration for the integrals. In these cases, as in the one-dimensional case, the integral always is interpreted with a positive orientation unless otherwise specified.


No, the two things are not equivalent.

The integral


is a Lebesgue integral, while


is a Riemann integral.

The Lebesgue integral is, in fact, despite it often being called such to the contrary in textbooks, not a straightforward generalization or extension of the Riemann integral, but rather to an extent a conceptually distinct entity which under some circumstances happens to agree with it. In particular, where the two integrals above both yield a finite result when applied to a given function defined on $\mathbb{R}$, those results will be equal, but that is not necessarily the case. For example,

$$\int_\mathbb{R} \mathrm{sinc}(x)\ dx$$

does not exist (effectively, it "equals" $\infty - \infty$, which is undefined), while

$$\int_{-\infty}^{\infty} \mathrm{sinc}(x)\ dx$$

does exist, and equals $\pi$.

To understand why this is, one needs to know that there are at least two different conceptual metaphors for integration, and these in their most mature forms actually give two entirely different theories thereof. The Riemann approach, as demonstrated in calculus, is based on the conceptual metaphor of the "accumulation of change" - effectively, we have a directed process which starts at some point and ends at another - those are the integration bounds - with an infinite number of steps in between that, at each point, adds an infinitely small bit $f(x)\ dx$ to the total sum. This is, in fact, what makes the Fundamental Theorem of Calculus work. Indeed, it is an entirely transparent result, in that the process of adding small increments is literally exactly how you would reverse differentiation: a derivative,


most reasonably represents the idea of the sensitivity to a small change in the function input: it tells you how much that the output value of the function will respond, as a proportion, to suitably-small perturbations on the input value. For example, if we consider $f_1(x) := x^2$, then $f_1'(x) = 2x$ and $f_1'(5) = 10$ means that around the point $x_1 := 5$, if I change that by a little teeny bit, say to $5.00000001$, i.e. $5 + 10^{-8}$, then I can assume that the change in $f_1(x_1)$ is 10 times larger, or on the order of $10^{-7}$. And indeed, a calculator with suitable precision will show that $f_1(5) = 25$ but $f_1(5.00000001) = 25 + (1.000000001 \times 10^{-7}) \approx 25 + 10^{-7}$, just as said.

Then the Riemann integral, or antidifferentiation, strictly speaking, mean that we are in effect, given a derivative of some function, attempting to reconstruct the original function by perturbing $x$ a very small amount, i.e. $dx$, adding this times the value of the derivative at $x$ to some arbitrary initial value to get the value of the reconstructed function at $x + dx$, then likewise do the same starting from this value just had, perturbing one step further by the same amount and in the same direction, to now get the value at $x + 2\ dx$, and so forth.

The Fundamental Theorem, then, simply states that the integral gives the original function, up to a constant shift - which, then, is exactly what we intended it to do. And conversely, we can start with the integral, and differentiate it to get the function integrated (that's the other half).

Now the Lebesgue approach is quite different: it belongs to the other conceptual metaphor of integration - that of area. The area integral

$$\int_{[a, b]} f(x)\ dx$$

note the change in notation, is now to be understood as an area underneath a curve - something which, at first sight, has nothing to do with the process described for Riemann integration and the Fundamental Theorem, which deals with change, not measurement. And, in fact, there are many ways we could try to find this area, and the Lebesgue one is the most mature of all of them, being applicable to any function we can actually construct explicitly, however badly behaved it might be. Of course, the Riemann integral

$$\int_{a}^{b} f(x)\ dx$$

can be proven to equal this area (which really needs a definition on its own), but only when $a < b$ - if it's the other way around, we get a strange sign flip. The reason for that is that Riemann's integral is inherently, as I just described, a directed process: it (in a non-rigorous sense) starts from $a$, goes to $a + dx$, then $a + 2\ dx$, $a + 3\ dx$, and so on, up to $b$. Whereas if I reverse the bounds, I have to make the stepping increment $dx$ negative.

The area integral, on the other hand, has no stepping increment - technically, we should not even be writing "$dx$" there (for Lebesgue, we often write $d\mu$ for arbitrary measure, but really, we should note it in a wholly different way altogether) - because area isn't something that comes about a little at a time, but is just "there", "all at once", holistically.

For what it's worth, the Riemann approach does "honestly" generalize, and when it does so, it becomes the theory of differential forms on a curved, non-Euclidean manifold with integration of a form along curves on such manifold, while the fullest form of the area approach is measure theory. We can, of course, "pull back" through the transition maps the area one as well, but this doesn't change the essential difference between the two approaches that calculus texts, like so many math texts in general, fail to do adequate justice to.

And yes, this is important. At the very least, you can switch the bounds, and you have to mind the sign, and this matters in manipulations - e.g. think of the identity

$$\int_{a}^{b} + \int_{b}^{c} = \int_{a}^{c}$$

familiar from first-year Calculus. The analogue for Lebesgue (area) integrals over intervals only works with the operands in the right order, something that can be easily seen by setting $c = a$. Even more, the "double" and "triple integrals" you may have heard for volumes and hypervolumes in the second year, really have this approach as their honest home - they are not simply generalized Riemann integrals, as their integration sets are two- and three-dimensional regions, which cannot be reasonably directed or ordered.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .