Why Do Physicists Write Integrals as ##\int dx f(x)##?

[Demystifier]

This was already pointed out, but it doesn't explain the choice. You say they are the same, but in some cases you prefer ##dxf(x)##. I am simply curious why.

I think the best explanation why is given in my #18. Your objection in #30 does not make much sense when ##dx## is not viewed as a right bracket.

How about integrals not in the sense of Riemann? There the infinitesimals that commute is not very meaningful.

Could you be more specific about that? What kind of integral do you have in mind? If you talk about Grassmann/Berezin integral, there ##d\eta## cannot be interpreted as an infinitesimal at all.

 

[Demystifier]

Perhaps the whole problem stems from the fact that it is sacrilegious for mathematicians to think of ##dx## as an infinitesimal. For if it is not an infinitesimal, then what is it? The only remaining option seems to be that ##dx## is a kind of a right bracket, so it must be on the right. But sloppy physicists do not have problems with thinking of ##dx## as an infinitesimal, so they have a freedom to put it either on the right or on the left. Then they make a final choice by other criteria, such as those in post #18.

 

[Demystifier]

There is also a good cognitive reason to prefer ##\int dx\, f(x)## over ##\int f(x)dx##. In plain English, the first would be expressed as "integrate over x the function f", while the second would be "integrate the function f over x". But when one just says "integrate", the first question that comes to one's mind is "integrate over what?", especially if there are many variables involved. So it seems cognitively more natural to say "integrate over x the function f", rather than "integrate the function f over x".

EDIT: Linguistically, "integrate the function f over x" sounds better than "integrate over x the function f". But when I translate it to my native Croatian language, it is no longer the case. How about other languages?

 

[Demystifier]

Another thought. Sometimes the order of integration in multiple integrals matters. But if I write this as
$$\int\int f(x,y)dxdy \neq \int\int f(x,y)dydx$$
it looks confusing to me because my first thought is that it implies ##dxdy \neq dydx##, which of course is wrong. On the other hand, if I write this as
$$\int dx\int dy \, f(x,y) \neq \int dy\int dx \, f(x,y)$$
it does not make me confused. Indeed, if I think of ##\int dx## as an operator that may act on a function, then I can interpret the above as non-commutativity of operators
$$\int dx\int dy \neq \int dy\int dx $$
which is quite true.

 

[fresh_42]

But when I translate it to my native Croatian language, it is no longer the case. How about other languages?

German doesn't have the strict SPO rule, so you could say:

• I integrate f over x.
• Over x I integrate f.
• f w.r.t.x must be integrated.
• f must be integrated over x.
• I integrate over x the function f.

although some of them sound a bit constructed.

 

[weirdoguy]

EDIT: Linguistically, "integrate the function f over x" sounds better than "integrate over x the function f".

In polish "integrate the function f over x" sounds better and actually I don't see any other way to say it.

 

[Demystifier]

And also when did that start?

So far we said nothing about that. I have just checked out several old physics books and they all use ##\int f dx##. The oldest physics book with ##\int dx\, f## that I found is Schweber (1961). Can someone find an older example?

 

[martinbn]

Could you be more specific about that? What kind of integral do you have in mind?

Well, any integral really that comes from a measure. For example how do you think of Lebesgue's integral as an infinite sum of infinitesimals?

EDIT: Linguistically, "integrate the function f over x" sounds better than "integrate over x the function f". But when I translate it to my native Croatian language, it is no longer the case. How about other languages?

For me nether sounds right. You don't integrate over x, you integrate with respect to x and over a set. By the way how is it in Croatian?

it looks confusing to me because my first thought is that it implies ##dxdxdy \neq dydx##, which of course is wrong.

One would have to say what meaning one puts in this notation otherwise you cannot simply say that it is wrong. For example if these are suppose to be differential forms, then certainly ##dx\wedge dy \neq dy\wedge dx##.

So far we said nothing about that. I have just checked out several old physics books and they all use ∫fdx∫fdx\int f dx. The oldest physics book with ∫dxf∫dxf\int dx\, f that I found is Schweber (1961). Can someone find an older example?

Finally something towards my actual questions. How did Dirac write his integrals? Somewhere I saw a statement that Leibniz was using both notations! In any case there must have been a switch, I still want to know why. Also every one who writes it the "physicists" way must at some point in his life switch because it is more likely that he started in school or university studying integrals in a math course.

 

[Demystifier]

Finally something towards my actual questions. How did Dirac write his integrals?

He did it like a mathematician.

Also every one who writes it the "physicists" way must at some point in his life switch because it is more likely that he started in school or university studying integrals in a math course.

When I reread my undergraduate textbooks, it seems that I was first exposed to the physicist-like notation at the 3rd year, in Jackson's Classical Electrodynamics.

 

[Demystifier]

By the way how is it in Croatian?

Integral po x od funkcije f.

 

[fresh_42]

So far we said nothing about that. I have just checked out several old physics books and they all use ##\int f dx##. The oldest physics book with ##\int dx\, f## that I found is Schweber (1961). Can someone find an older example?

R.Courant / D.Hilbert, 1924, kept it consequently at the end. The first appearance I have found was in:
E. Madelung, 1950, Die Mathematischen Hilfsmittel des Physikers (The Mathematical Tools of the Physicist)
but only for those integrals with long integrands like ##F(t,v) = \int dv \int dt \varphi(x,t)e^{\{\,\ldots\,\}}##

 

[DrDu]

By analogy with ##\sum_n f_n## I would propose to use a third notation
$$\int_x f(x)$$

This form is used in the theory of differential forms!

 

[Demystifier]

This form is used in the theory of differential forms!

I think in differential forms it is more like
$$\int_V f$$
where ##V## is the domain of integration, not the variable of integration.

 

[Demystifier]

For example how do you think of Lebesgue's integral as an infinite sum of infinitesimals?

If Riemann's integral is a sum of infinitesimal vertical strips, then Lebesgue's integral is a sum of infinitesimal horizontal strips. See e.g. the picture in https://en.wikipedia.org/wiki/Lebesgue_integration .

 

[fresh_42]

If Riemann's integral is a sum of infinitesimal vertical strips, then Lebesgue's integral is a sum of infinitesimal horizontal strips. See e.g. the picture in https://en.wikipedia.org/wiki/Lebesgue_integration .

... or https://www.physicsforums.com/insights/omissions-mathematics-education-gauge-integration/

 

[Demystifier]

but only for those integrals with long integrands like ##F(t,v) = \int dv \int dt \varphi(x,t)e^{\{\,\ldots\,\}}##

Yes, somehow that also makes sense, even though I am not able to explain why. Anyway, since physicists more often than mathematicians use long integrands, perhaps it could explain why such notation was adopted by physicists and not by mathematicians.

 

[fresh_42]

Yes, somehow that also makes sense, even though I am not able to explain why. Anyway, since physicists more often than mathematicians use long integrands, perhaps it could explain why such notation was adopted by physicists and not by mathematicians.

It could also has been because of

If you have a double integral, then the notation
$$\int_{a}^{b}\int_{c}^{d}f(x,y)dxdy$$
is ambiguous, because it is not clear whether ##x\in[a,b]## or ##x\in[c,d]##. With notation
$$\int_{a}^{b}dx\int_{c}^{d}dy \,f(x,y)$$
there is no risk for such a confusion.

since I found this exception from his other notations in a double integral, which occur more often in physics, too, as mathematicians would probably write it usually as a single integral over an appropriate area. Maybe it took the way from double integrals to single integrals. An occurrence in 1950 appears reasonable, too.

 

[steenis]

(1) While contemplating about the use and position of ##dx## in ##\int f(x) dx## one must not forget history. For the inventor of this notation, Leibniz (1646-1716), the notation ##\int f(x) dx## literally meant the sum of the "infinitesimals" ##f(x) dx## (hence the S-notation ##\int##).

(2) It seems to me that in a physical context, ##\int f(x)## has another physical dimension than ##\int f(x) dx##

 

[stevendaryl]

Yes, this is the same. Mathematicians wouldn't put basis vectors first and the numbers second. It would be ##\sum a_i\vec{e}_i##, not ##\sum\vec{e}_i a_i##.

Yes, but the advantage of writing ##\sum_i |i\rangle \langle i |a\rangle## is that it can be interpreted as the identity operator ##I = \sum_i |i\rangle \langle i|## operating on the vector ##|a\rangle##.

 

[martinbn]

Yes, but the advantage of writing ##\sum_i |i\rangle \langle i |a\rangle## is that it can be interpreted as the identity operator ##I = \sum_i |i\rangle \langle i|## operating on the vector ##|a\rangle##.

Yes, I agree, but this is an advantage of the bra-ket notation. It is absent in the ##\sum \vec{e}_i a_i##.

 

[martinbn]

If Riemann's integral is a sum of infinitesimal vertical strips, then Lebesgue's integral is a sum of infinitesimal horizontal strips. See e.g. the picture in https://en.wikipedia.org/wiki/Lebesgue_integration .

Yes, and for Riemann the base of the strip is the infinitesimal ##dx## the height is the value of the function ##f(x)##, so the area can be written as ##f(x)dx## or equally (perhaps better) as ##dxf(x)##. For Lebesgue the base is the measure of some set, the height might be interpreted somehow as an infinitesimal. But I don't see how you can make the area, even just heuristically and non-rigorously as ##f(x)dx##.

 

[Demystifier]

Yes, and for Riemann the base of the strip is the infinitesimal ##dx## the height is the value of the function ##f(x)##, so the area can be written as ##f(x)dx## or equally (perhaps better) as ##dxf(x)##. For Lebesgue the base is the measure of some set, the height might be interpreted somehow as an infinitesimal. But I don't see how you can make the area, even just heuristically and non-rigorously as ##f(x)dx##.

Well, physicists don't often use Lebesgue integrals in practical computations, so they don't invent their own notation for that. But if they did, I guess they would write it something like
$$\int df\,x(f)$$
where ##df## would be an infinitesimal. Here ##x(f)## is the inverse of ##f(x)## and the nontrivial thing to take into account is that ##x(f)## can be a multivalued "function", some branches of which have actually a negative contribution to the integral, which somehow should be incorporated into the notation too. Perhaps something like
$$\int d_s f\,x(f)$$
with
$$d_s f=-df \, {\rm sign}\left( x\frac{df(x)}{dx} \right)$$

EDIT: I have never seen such a representation of Lebesgue integrals before, I've just invented it. Is it really new, or has somebody somewhere already done something similar?

 

[stevendaryl]

Well, physicists don't often use Lebesgue integrals in practical computations, so they don't invent their own notation for that. But if they did, I guess they would write it something like
$$\int df\,x(f)$$
where ##df## would be an infinitesimal. Here ##x(f)## is the inverse of ##f(x)## and the nontrivial thing to take into account is that ##x(f)## can be a multivalued "function", some branches of which have actually a negative contribution to the integral, which somehow should be incorporated into the notation too. My proposal is
$$\int d_s f\,x(f)$$
where
$$d_s f=-df \, {\rm sign}\frac{df(x)}{dx}$$

EDIT: I have never seen such a representation of Lebesgue integrals before, I've just invented it. Is it really new, or has somebody somewhere already done something similar?

A Lebesgue integral can be converted into a Riemann type integral by switching the integration variable.

You have (for simplicity, assume it's nonnegative) a function ##f(x)##. You define a new function ##f^*(t) = \mu(\{ x | f(x) > t \})## (##\mu(A)## means the Lebesgue measure of set ##A##). Then the lebesgue integral of ##f## is the Riemann integral of ##\int_0^\infty f^*(t) dt##. I guess the reason it's better defined is that ##f^*(t)## will often be better-behaved than ##f##.

 

[Demystifier]

Speaking of physicists's way of doing Lebesque integrals, how about the following heuristics?
$${\rm Riemann}=\int dx\, f \cong \int dx \int df \cong \int df \int dx \cong \int df\, x ={\rm Lebesque} $$

 

[martinbn]

You don't need to write ##df## it is enough to write ##f##, and ##x(f)## is not just the set of all those values, you need the measure, so it is better to write ##\mu## or ##\mu(x)## so you integral becomes ##\int f\mu##, which of course is standard.

 

[martinbn]

A Lebesgue integral can be converted into a Riemann type integral by switching the integration variable.

You have (for simplicity, assume it's nonnegative) a function f(x)f(x)f(x). You define a new function f∗(t)=μ({x|f(x)>t})f∗(t)=μ({x|f(x)>t})f^*(t) = \mu(\{ x | f(x) > t \}) (μ(A)μ(A)\mu(A) means the Lebesgue measure of set AAA). Then the lebesgue integral of fff is the Riemann integral of ∫∞0f∗(t)dt∫0∞f∗(t)dt\int_0^\infty f^*(t) dt. I guess the reason it's better defined is that f∗(t)f∗(t)f^*(t) will often be better-behaved than fff.

Which doesn't say anything about the notation ##\int fdx##.

 

[Demystifier]

You don't need to write ##df## it is enough to write ##f##, and ##x(f)## is not just the set of all those values, you need the measure, so it is better to write ##\mu## or ##\mu(x)## so you integral becomes ##\int f\mu##, which of course is standard.

But you challenged me to write it somehow in terms of infinitesimals, didn't you?

 

[martinbn]

But you challenged me to write it somehow in terms of infinitesimals, didn't you?

Yes, but in a way that can be interpreted as ##dxf(x)##. That was the whole point, right?

 

[Demystifier]

Yes, but in a way that can be interpreted as ##dxf(x)##. That was the whole point, right?

Well, #59 comes pretty close, doesn't it?

 

[fresh_42]

Well, physicists don't often use Lebesgue integrals in practical computations

What do they use in their favorite Hilbert space ##L_2(M)## if not Lebesgue?

Speaking of physicists's way of doing Lebesque integrals, how about the following heuristics?
$${\rm Riemann}=\int dx\, f \cong \int dx \int df \cong \int df \int dx \cong \int df\, x ={\rm Lebesque} $$

I would be cautious here. Lebesgue ##\neq## Riemann. The above "equation" brings in automatically the physicists' chronic disregard of mathematical subtleties right from the start, rather than its usual occurrence later on.

Btw., does anybody know why the Riemann integral isn't called Aristoteles integral?

 

[martinbn]

Well, #59 comes pretty close, doesn't it?

Well, it says Riemann = Lebesgue.

 

[martinbn]

Btw., does anybody know why the Riemann integral isn't called Aristoteles integral?

Why would it be called that?

 

[Demystifier]

Lebesgue ##\neq## Riemann.

That's why I used ##\cong## instead of ##=##.
Here ##\cong## means something like "almost equal but not quite", or "equal in most cases of practical interest".

 

[Demystifier]

Well, it says Riemann = Lebesgue.

See my post above.

 

[martinbn]

What did you mean by ##\cong##?

But if you are going to use ##\int df=f## and ##\int dx = x## in that line, then you may as well say that ##\int fdx=\int dxf=xf=fx## with the appropriate ##\cong##'s.