All the integrals I'm familiar with have the form:

$\int f(x)\mathrm{d}x$.

And I understand these as the sum of infinite tiny rectangles with an area of: $f(x_i)\cdot\mathrm{d}x$.

Is it valid to have integrals that do not have a differential, such as $\mathrm{d}x$, or that have the differential elsewhere than as a factor ? Let me give couple of examples on what I'm thinking of:

$\int 1$

If this is valid notation, I'd expect it to sum infinite ones together, thus to go inifinity.

$\int e^{\mathrm{d}x}$

Again, I'd expect this to go to infinity as $e^0 = 1$, assuming the notation is valid.

$\int (e^{\mathrm{d}x} - 1)$

This I could potentially imagine to have a finite value.

Are any such integrals valid? If so, are there any interesting / enlightening examples of such integrals?


First of all Non-standard Analysis makes nonsense like $\frac{\mathrm{d}y}{\mathrm{d}x}$ into a meaningful fraction rather than just a "symbol" (whatever that means).

The first and simplest example of differentials being used beyond the $\int f(x)\mathrm{d}x$ format is would be multidimensional integration:

$$ \int_V \mathrm{d}x \mathrm{d}y \mathrm{d}z = \int \left(\int \left(\int \mathrm{d}x \right) \mathrm{d}y \right) \mathrm{d}z $$

which gives the volume of the solid $V$.

The next example is Greens Theorem from vector calculus,

$$ \int A(x,y) \mathrm{d}x - B(x,y) \mathrm{d}y = \int (\partial_1 B - \partial_2 A) \mathrm{d}x \mathrm{d}y $$

It is also possible to forget integration completely and just use equations with differentials in them to solve calculus problems.

So you can see the standard format is not even close to the whole picture, If you want to integrate terms like $\int (e^{\mathrm{d}x} - 1)$ please do! There is absolutely nothing to stop you figuring out from scratch how to solve this sort of integral and making the theory rigorous.

  • $\begingroup$ Thanks for the answer. I think it's closest to answering my original question. Also, especially thanks for the link to the intro to non-standard analysis. I've been interested in learning about hyper real numbers for a while now, but every article I've looked at so far has been kind of terse. $\endgroup$ – Sami Aug 13 '10 at 3:00
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    $\begingroup$ You write «If you want to integrate terms like $\int(e^{dx}-1)$ please do!» but you don't mean integrate: you mean find a meaning for the notation. You say «how to solve this sort of integral» but that integral does not make sense according to standard definitions of the integral, so it simply does not make sense to 'solve' it. You can of course come up with a theory that does give sense to such notations, but coming up with such a thing is not computing integrals... $\endgroup$ – Mariano Suárez-Álvarez Oct 21 '10 at 14:20
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    $\begingroup$ @Mariano Suárez-Alvarez, that is splitting hairs, everything you said is formally correct but there's not much to be learned from it. $\endgroup$ – anon Oct 21 '10 at 16:34
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    $\begingroup$ @muad: one cannot compute things which do not make any sense under the definitions one uses. That is not splitting hairs! $\endgroup$ – Mariano Suárez-Álvarez Oct 21 '10 at 16:58
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    $\begingroup$ @muad: I have seen way too many students 'compute' things which do not make any sense with the definitions they have... I would say that recognizing when something is defined or not is one of the most important things one has to learn in order to do any kind of meaningful math. $\endgroup$ – Mariano Suárez-Álvarez Oct 22 '10 at 16:26

I think your question here shows that, while you have been using these symbols, you haven't really been given a proper motivation for where they came from.
Let's go back and consider how we came up with the idea of an integral. In a typical class, you will see a lot of pictures like this:
alt text
We find the area under the curve by summing up the area of all these little rectangles. If we wanted to write an expression for the area, it would look like:
alt text
(source: mathurl.com)
The Σ means that we are computing a sum. We are adding the areas of the rectangles, which we have numbered 1 through n, to get the complete area under the curve. The area of each rectangle is given by multiplying the height by the width. The height is given by f(xi) because the base of the rectangle is at 0, and the top of the rectangle is where it meets the function f. The Δx represents the width of each rectangle.
When we find the integral, we are taking the limit of this sum as the number of rectangles goes to infinity, and each individual rectangle becomes infinitesimally tiny. You can think of the dx as the equivalent of Δx: it represents the infinitesimally small width of each rectangle that we added up to get the area.

Once you realize this, we can see why integrals only make sense when written ∫f(x)dx. Because we are adding up the areas of rectangles that have height f(x) and width dx. If you try to interpret the expressions you wrote in this way, you will see that they do not really make sense as integrals: you are not summing up rectangles, so you are not finding an area under a curve.

You could, of course, define your own notation in which those expressions behave the way you expect them to, but all mathematical notation is driven based on what people find useful, and what people can agree on and easily understand. Your reuse of the integral sign and dx that people are used to seeing in a particular context will probably result in few people adopting your definition.

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    $\begingroup$ The summation formula is a beautiful analogy. $\endgroup$ – Justin L. Jul 29 '10 at 5:18
  • $\begingroup$ Actually, the motivation for my question was me trying to understand, if there was a lim representation for integration like the one for differentiation. I imagined it to look something like: int A to B { f(x) dx } = lim b -> infinity { SUM a=1 to b { f(A + (B - A) * a / b) * (1 / b) } }. And within that definition, integrals such as int A to B { f(x) } = lim b -> infinity { SUM a=1 to b { f(A + (B - A) * a / b) } } would seem to make sense, too. $\endgroup$ – Sami Jul 29 '10 at 12:11
  • $\begingroup$ @Sami: There is a limit definition for the integral. However, even for the standard Riemann integral, the rigorous definition can appear quite arcane. A less general definition may look something like this. Like I said in my answer, you can adjust the expression as you have in your comment, but the result is not useful in the same way the standard integral is, it only shares a similarity in form. $\endgroup$ – Larry Wang Aug 3 '10 at 5:22
  • $\begingroup$ Justin L, indeed - in the Non-standard analysis it is a beautiful definition! $\endgroup$ – anon Aug 12 '10 at 20:21

When you write $\int f(x) dx$, the whole of $\int ... dx$ is an indivisible symbol, just as the $d/dx$ is an indivisible symbol when you write $df/dx$.

Of course, there are reasons why the notation is as it is, but trying to manipulate it like you suggest in $\int e^{dx}$, for example, is simply meaningless.

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    $\begingroup$ I could imagine e^{dx} being used as notation for a Riemann-Stieltjes integral, but it's sort of weird notation. $\endgroup$ – Qiaochu Yuan Jul 29 '10 at 0:06
  • $\begingroup$ We should just write $\int f$ or $\int \lambda x. f(x)$ or something, if the $\mathrm{d}x$ is meaningless and indivisible. $\endgroup$ – anon Aug 12 '10 at 19:50
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    $\begingroup$ @muad, why should we? Should we also drop the $df/dx$ notation? There is absolutely no reason why notations which are 'split' in this way should not be used. $\endgroup$ – Mariano Suárez-Álvarez Oct 21 '10 at 14:17
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    $\begingroup$ @Mariano Suárez-Alvarez, yes that notation is awful, we should drop it - obviously I don't write the thousands of calculus text books out there. $\endgroup$ – anon Oct 21 '10 at 16:33
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    $\begingroup$ Well, the immense majority of people disagree with you :) $\endgroup$ – Mariano Suárez-Álvarez Oct 21 '10 at 16:59

No, it's not valid. The dx in the integral is a representation of the fact that the integral is obtained as an area, so multiplying the "average" of the function value at each point by an infinitesimal interval.

As the manner in which we don't calculate the area does not change, the notation does not change.

There are different notations that are used when the integral is over a curve, or over more than variable (thus leading for example to volumes).

The d(variable) notation is also used as a reminder that the integral is against a specific variable and not another, e.g. that int x/y dx differs from int x/y dy.


I know I am late to the party, but while I see perfectly good answers to the question, I think it would be interesting to try to evaluate the integral which led to the question.

It must be noted that the integral in question is quite non-standard, because all integrals considered in mathematics take form of product with infinitesimal as one of its factors. There may be different ordering, such as $\int \mathrm{d}x f(x)$, and there may be various nesting and shorthand notations, but ultimately all integrals result from product with infinitesimal.

But, let's pretend we use non-standard analysis and integral given in question is valid. It was already correctly noted that the value would be finite, but actually it can be shown that:

$$\int (e^{\mathrm{d}x} - 1) = x$$

We will solve it by substituting $e^{\mathrm{d}x} - 1$ for something more useful. But where to find good substitution?

Let's recall using Leibniz's notation (d something is a very little bit of something, as Silvanus Thompson put) to take derivative of exponential function (that happens to be its own derivative):

$$e^x + \mathrm{d}e^x = e^{x+\mathrm{d}x}$$ $$e^x + \mathrm{d}e^x = e^xe^{\mathrm{d}x}$$ $$e^x + \mathrm{d}e^x = e^x\left(1+\mathrm{d}x\right)$$ $$\mathrm{d}e^x = e^x\mathrm{d}x$$ $$\frac{\mathrm{d}e^x}{\mathrm{d}x} = e^x$$

In the third step we invented a substitution $e^{\mathrm{d}x} = 1 + \mathrm{d}x$ to allow us to continue. While it led to finding the exponential function that is its own derivative and it can be used to find the value of $e$, it also allows us to evaluate the integral in question.

$$e^{\mathrm{d}x} - 1 = \mathrm{d}x$$ $$\int (e^{\mathrm{d}x} - 1) = \int {\mathrm{d}x} = x$$

But be careful with this.

First, such use of Leibniz's notation is not considered correct because of the notion of freely roaming infinitely small yet nonzero real numbers - that's the reason we use limits and epsilon-delta approach nowadays in calculus. If we still want to do the above trickery, it must be noted we are in the realm of non-standard analysis.

Second, while non-standard analysis may consider such integral valid (because it interprets integral as continuous sum) and evaluate it (as done above), it is still big question what that integral, as formulated, actually means and how it came to be.

Using known limits and identities we can also invent these non-standard integrals:

$$\int \sin{\mathrm{d}x} = x, \int \tan{\mathrm{d}x} = x$$

Obvious proof is left to the reader as an exercise. I also considered adding $\int (\cos{\mathrm{d}x} - 1) = 0$ to the list, but I am not sure about it when using infinitesimals that freely.

Disclaimer and opinion: I am not convinced about any sensible meaning and use of these integrals and I discourage their use. Using them may lead to confusion without any gain. The only result to me is some deep thinking about notation we use in math.


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