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

[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.

 

[fresh_42]

Why would it be called that?

Wasn't he the first who came up with this idea?

 

[martinbn]

Wasn't he the first who came up with this idea?

Aristoteles! Do you mean Archimedes?

 

[fresh_42]

Aristoteles! Do you mean Archimedes?

Sure. How embarrassing ...

 

[Demystifier]

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.

Not if you are a physicist with an intuitive feeling for limitations of any heuristic rules.

 

[martinbn]

Sure.

That is a good question on its own.

 

[Demystifier]

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

Due to the Stigler's law: https://en.wikipedia.org/wiki/Stigler's_law_of_eponymy

 

[stevendaryl]

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

But it does seem to write the Lebesgue integral in terms of infinitesimals.

 

[martinbn]

But it does seem to write the Lebesgue integral in terms of infinitesimals.

But those infinitesimals are for another function. The point was to be able to make sense,at least intuitively, of $\int dxf(x)$ as an infinite sum of infinitesimals.

 

[Demystifier]

Would it be correct to say that both Riemann and Lebesque integrals are just special cases of computing the area
$$\int\int dxdy$$
of a region on a 2-dimensional plane?

 

[martinbn]

And what do you mean by area? The Lebesgue measure of the set? The integral you wrote down? What exactly is that integral?

 

[stevendaryl]

And what do you mean by area? The Lebesgue measure of the set? The integral you wrote down? What exactly is that integral?

Let $f$ be a function from reals to reals (for simplicity, let it be nonnegative). Let $S(A,B)$ be the set of points $(x,y)$ such that $A \leq x \leq B$ and $0 \leq y \leq f(x)$. Then if the area of set $S(A,B)$ is defined, it will be equal to $\int_A^B f(x) dx$. Right? By "area" I mean a measure on $R^2$ such that every set of the form $\{ (x,y) | x_0 \leq x \leq x_1 \wedge y_0 \leq y \leq y_1\}$ has measure $|(x_1 - x_0) \cdot (y_1 - y_0)|$

 

[martinbn]

Let $f$ be a function from reals to reals (for simplicity, let it be nonnegative). Let $S(A,B)$ be the set of points $(x,y)$ such that $A \leq x \leq B$ and $0 \leq y \leq f(x)$. Then if the area of set $S(A,B)$ is defined, it will be equal to $\int_A^B f(x) dx$. Right? By "area" I mean a measure on $R^2$ such that every set of the form $\{ (x,y) | x_0 \leq x \leq x_1 \wedge y_0 \leq y \leq y_1\}$ has measure $|(x_1 - x_0) \cdot (y_1 - y_0)|$

May be, it depends on the unstated assumptions. Say you take area to mean the Lebesgue measure in the plane. It satisfies your requirement for area. Take a function $f(x)$ so that $S(A,B)$ is Lebesgue measurable and $f(x)$ is not Riemann integrable. Then if the integral you wrote is the Riemann integral, then the area is not it, the integral doesn't exists. I assume that is not what you mean, but you didn't say anything about the relationship between the measure in the plane and the integral on the line in you statement. But let's say that it is all fine. Then what? Where are you going with this?

 

[stevendaryl]

But let's say that it is all fine. Then what? Where are you going with this?

I was just trying to answer the question "What do you mean by area"?

 

[nuuskur]

$\int dx f(x) = (x+C)f(x)$

Would it be correct to say that both Riemann and Lebesque integrals are just special cases of computing the area
$$\int\int dxdy$$
of a region on a 2-dimensional plane?

I'm having difficulty in understanding how either of them would be special cases of finding an 'area'. An application of either theory would be finding the area/volume/measure of an $n$-dimensional, pre-determined to be measurable, set.

The expression $\int\int dx\,dy$ is just some function $f:\mathbb R ^2\to\mathbb R$ (hopefully).