Dx before the f(x) in integrals

Demystifier
Gold Member
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.

weirdoguy
fresh_42
Mentor
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.

Demystifier
(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##

Demystifier
stevendaryl
Staff Emeritus
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##.

Demystifier
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##.

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
Gold Member
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?

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

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.

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
Gold Member
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?

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
Gold Member
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
Mentor
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?

Well, #59 comes pretty close, doesn't it?
Well, it says Riemann = Lebesgue.

Btw., does anybody know why the Riemann integral isn't called Aristoteles integral?
Why would it be called that?

Demystifier
Gold Member
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
Gold Member
Well, it says Riemann = Lebesgue.
See my post above.

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.

fresh_42
Mentor
Why would it be called that?
Wasn't he the first who came up with this idea?

Wasn't he the first who came up with this idea?
Aristoteles! Do you mean Archimedes?

fresh_42
Mentor
Aristoteles! Do you mean Archimedes?
Sure. How embarrassing ...

Demystifier