I want to know why we use dummy variables in integral? thanks so much.

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    $\begingroup$ Note that in measure theory, we can ''not'' use them. We use a measure $\mu$ (in your case the measure would be ''$dx$'', i.e. the Lebesgue measure, even though you probably know Riemann integration... doesn't matter) and we write $\int f \, d\mu$. $\endgroup$ – Patrick Da Silva Sep 21 '14 at 20:06
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    $\begingroup$ @PatrickDaSilva It's not uncommon to find $\int f(x)\,d\mu(x)$, particularly in connection with convolutions: $f*g(x)=\int f(y)g(x-y)\,d\mu(y)$, in order to single out the variable of integration. $\endgroup$ – egreg Sep 21 '14 at 23:16
  • $\begingroup$ @egreg : I know, but I said we ''can'' not use them (now I need to switch my '' of position to put emphasis on something else...) I actually expected your comment somehow. $\endgroup$ – Patrick Da Silva Sep 22 '14 at 1:13
  • $\begingroup$ @PatrickDaSilva It's more a linguistic issue, then; maybe “we are allowed not to use them” would be better. $\endgroup$ – egreg Sep 22 '14 at 9:04
  • $\begingroup$ "We are allowed not to use them" sounds different. The imitation of a voice emphasis on the not with " " was the closest to what I meant. $\endgroup$ – Patrick Da Silva Sep 22 '14 at 17:11

Bear with me.

Let's say we want to model a function that represents the number of boxes a factory outputs per hour. The input, of course would be number of hours, and the output would be number of boxes. Now, we need to name our function, because we don't want to keep referring to the function by its definition -- that would take up a lot of hour time. Let's call it Efficiency.

So, $\text{Efficiency}(\text{number of hours)} = \text{number of boxes}. \tag1$

We could do stuff with this function. We can differentiate it for optimization, or integrate it to find the average values, or anything in between.

Of course though, this is a time killer, since the function in $(1)$ is tedious to write out. We can introduce other variables, like $f$ for the name of our function, $t$ for our time (in hours), and $y$ for the output (in boxes).

Now, $(1)$ is equivalent to $$f(t) = y.$$

But look, we didn't really change anything. Our function still models the original problem in the same exact way. The variables are just abbreviations. We abbreviate them for convenience.

Now suppose the factory makes different types of boxes. We can call these new types of boxes $y$ also, but that would be awfully confusing. The same reasoning is behind the use of dummy variables in definite integrals.

When we introduce dummy variables into integrals, we're just doing it for convenience. The meaning behind the algebra does not change the original problem; it just makes it nicer to analyze.


I'll interpret the question as

Why are we not using the orignal one instead of a dummy variable?
Why use $\int_0^xf(x')dx'$ and not $\int_0^xf(x)dx$?

(If this is the case, please clarify you question with an marked(!) edit, others interpreted in differently.)

Because it has a different meaning and there is a problem with how to define it.
To show the difference define

$$g(x):= \int_0^xf(x,x')dx'$$ for some $f:\mathbb{R}^2\to \mathbb{R},(x,y)\mapsto f(x,y)$ continuously differentiable with respect to both arguments (being only continuous instead of differentiable would be enough).

Now compute

$$\frac{dg(x)}{dx}= \frac{d}{dx}\int_0^xf(x,x')dx' =f(x,x) +\int_0^x\frac{df(x,x')}{dx}dx' =f(x,x) + \int_0^x f_1(x,x')dx',$$ where $f_1$ means the first partial derivative of $f$.
The right part happens, because the inner part of the integral was dependend of the outher variable $x$.
So $\int_0^xf(x,x')dx'$ means something different from $\int_0^xf(x',x')dx'$

This is the reason why $$\int_0^xf(x,x)dx$$ is somewhat not well defined.

  • $\begingroup$ Can you clarify the steps/rules between ∫x0f(x,x′)dx′=f(x,x)+∫x0df(x,x′)dxdx′ and then =f(x,x)+∫x0f1(x,x′)dx′ ? I'm not sure I follow it. $\endgroup$ – Chris2048 Feb 1 '17 at 11:58
  • $\begingroup$ @Chris2048 This is just notation, I added the information in the answer.:) $\endgroup$ – Antitheos Feb 1 '17 at 18:55
  • $\begingroup$ I understand the notation, I just don't know the rules being applied, after having read the body of the text. $\endgroup$ – Chris2048 Feb 3 '17 at 18:21
  • $\begingroup$ @Chris2048 Try reading this: en.wikipedia.org/w/… The $t$ is my $x$, $x$ is my $x'$ and $b(t)=t$ is my special case. $\endgroup$ – Antitheos Feb 3 '17 at 19:34

Because the definite integral depends only on the function and the limits of integration. So writing $\int_{a}^{b} f(x)dt$, $\int_{a}^{b} f(t)dt$, $\int_{a}^{b} f(s)ds$ makes no difference as long as $x$, $t$ and $s$ are "integrated out" in your final answer.

For more information look here and here (section 7.2).


If you're looking at say something like a Riemann integral, you're "adding up pieces". You need some way of keeping track of which piece you're talking about (i.e. of referring to where you are), but it doesn't actually have anything to do with the value of the integral -- it's just a placeholder so you can add up correctly.

It's essentially the same idea as working with a sum like $a_1+a_2+...+a_n$. If we write it as $\sum_{i=1}^n a_i$, we introduce a variable, $i$ to be able to refer to a particular item ($a_i$ in this case) as we run through the sum. But using $i$ is arbitrary.

The variable could be anything (as long as its not a variable you're already using), it's effectively an accounting convenience - a dummy variable.


In principle we don't need the dummy variables in an integral of a function $f$ like $\int f(x) dx$; it's just a historical convention. By definition, a function a rule of computation with fixed domain and codomain, so it would not be ambiguous if one wrote $\int f$ instead of $\int f(x) dx$ or $\int_a^b f$ instead of $\int_a^b f(x) dx$. The notation with $dx$ is quite convenient if wants to write out the expression under the integral instead of the name of a function, but that could be done as well by applying $\int f$ notation: e.g. if $f(x)=x^2$ or in other words, $f=(x \mapsto x^2)$, then $\int x^2dx = \int f=\int (x \mapsto x^2)$. However, I have never seen anyone write $\int (x \mapsto x^2)$; the main reason is probably that readers are not used to it (and $\int x^2 dx$ is also a little shorter).

Actually, in addition to convention and brevity of writing expressions, I can think of two more reasons.

The first reason is that sometimes people abuse notation and denote different functions with the same name (letter). Then the name of the dummy variable may help the reader resolve which function is meant. E.g. if functions $f$ and $g$ are defined so that $f(b)$ is the price of a box after a box factory has produced $b$ boxes and $g(t)$ is the price at time $t$, then $f$ and $g$ are in general different functions, i.e. rules of computation: one converts the number of boxes to price and the other converts time to price. However, both can be used to compute the same physical quantity (price) and it is not uncommon to see people denote both by the same letter by sometimes writing $p(b)$ for $f(b)$ and sometimes $p(t)$ for $g(t)$. In that case, $\int p$ is ambiguous, as it could mean either $\int f$ or $\int g$, but an expression like $\int p(b)db$ or $\int p(t)dt$ would hint the reader which of the two is meant. (In particular, $\int_0^B p(b)db = \int_0^B f$ is the total revenue for the first $B$ boxes.)

The second reason is easiness of memoization of the rule for change of variables: $\int f(x)dx = \int f(y) \frac{dx}{dy}dy$. (Here I have made use of the same kind of abuse of notation that I mentioned in the first reason.)


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