U-Substitution Notation as division

[lightlightsup]

Homework Statement

I've been warned by a professor to not do u-substitution like this unless du is a constant (ex: a number, like du = 4 dx).
But, if g'(x) is a variable like x^2, I shouldn't use division like this due to division by zero issues?
I know that du can directly replace g'(x)dx in many cases.
But, is division really wrong here?
Any book references on this?

Relevant Equations

See image.

 

[fresh_42]

It is disturbingly correct, i.e. usually works. But if $g'(x)=0$ then you have divided by $0$ so you have to be careful. The infinitesimals $dx,du$ are not numbers, so they cannot be treated as such. They abbreviate limits. However, it works.

It is a reminder to check afterwards: Solve the integral and differentiate the result in order to check whether you end up where you started from.

 

[lightlightsup]

It is disturbingly correct, i.e. usually works. But if $g'(x)=0$ then you have divided by $0$ so you have to be careful. The infinitesimals $dx,du$ are not numbers, so they cannot be treated as such. They abbreviate limits. However, it works.

It is a reminder to check afterwards: Solve the integral and differentiate the result in order to check whether you end up where you started from.

It makes intuitive sense to me.
But, if g'(x) = 0, then the original g(x) within f(x) would be a constant, which would make f(x) a constant. This, in turn, would change our entire approach to the integral and perhaps even obviate u-substitution.

 

[fresh_42]

It makes intuitive sense to me.
But, if g'(x) = 0, then the original g(x) within f(x) would be a constant that would make f(x) a constant. So, that would change our approach to the problem?

I haven't checked all possibilities, so I was cautious in my answer. What we are doing is a kind of coordinate transformation: instead of walking the integration path along $x$, we walk along $g(x)$. This shouldn't change anything if $g(x)$ is one-to-one, but needs special care if not, e.g. splitting the integral.

The integration limits change, too, so that is also important to consider.

 

[lightlightsup]

Understood.
As a conclusion, I think this is the best way for a student to proceed/process u-substitution:

Manage constants by "division" / "adjustment" BUT directly replace all other variables.

 

[Infrared]

It seems fine to me. If $F(u)$ is an antiderivative for $f(u)$, then $F(g(x))$ is an antidervative for $f(g(x))g'(x)$ by the chain rule. I think this is all the statement is saying.

 

[Stephen Tashi]

I've been warned by a professor to not do u-substitution like this unless du is a constant (ex: a number, like du = 4 dx).

How does the professor want you to write the steps in finding antiderivatives like $\int (cos(x^2+1))(2x) dx$ ?

Presumably not:

$u = x^2$
$du = 2x dx$
$dx = du/2x$
$\int (cos(x^2+1))2x dx = \int( (cos(u+1))(2x)(du/2x) = \int cos(u+1) du = sin(u+1)$

 

[vanhees71]

Well, that's physicist' sloppy notation, which works well as mnemonics but it not well-defined in a rigorous mathematical sense.

You can simply write it as
$$\int \mathrm{d} x u'(x) f[u(x)] = \int \mathrm{d} u f(u).$$
Some mathematicians also don't like the way I write the "differentials" in the front directly after the integration symbol, but that makes life so much easier that I don't care ;-)).

 

[lightlightsup]

The professor wants what Paul's Online Notes shows.
Directly substitute all variables and differential symbols ("dx" and "du").
Constants, you can divide.
I have come to agree with his method.
The division thing, except in the case of a constant, seems to be a mathematical coincident/fluke?

 

[Stephen Tashi]

The division thing, except in the case of a constant, seems to be a mathematical coincident/fluke?

The topic of this thread can be discussed on two levels. On level is the practical question of what technique of symbolic manipulation is most likely to produce a correct answer. The next level is the question of what the symbolic manipulations actually mean - interpreted in words as complete sentences. To discuss the topic on that level is difficult. For example, does $\int f(x) dx$ denote a function, or does it denote a set of functions?

On the most mundane level, what method of u-substitution will produce an answer that is easier for an instructor to grade? It is simplest. if the "correct" answer to finding $\int \cos(x^2+1) 2x dx$ is $\sin(u)$ instead of $\sin(u+1)$ or $\sin(u+14)$ etc. So, from the point of view of grading, we prefer that all students use the substitution $u = x^2 + 1$ instead of using $u = x^2$ or $u = x^2 - 13$ . That goal can be enforced by telling students that when they encounter a problem of the form "$\int f( <expression\ in\ x> ) ... dx$" that they should set $u$ equal to the entire expression within the parentheses.

The mathematical test of whether a substitution like $u = x^2 - 13$ gives a "correct" answer is whether
the definite integrals it implies give correct results. Different u-substitutions lead to different definite integrals with different limits of integration. So there can be many different u-substitutions that are correct. However, it is simplest for the grader if all students use the same u-substitution.

 

[benorin]

As a [former] reader (grader) I personally found it neat whenever a student put a non-trivially different answer that produced the correct result (though for an incorrect method of solution that miraculously happened to produce the correct result I would graciously award 1/10 points).

 

[archaic]

Some mathematicians also don't like the way I write the "differentials" in the front directly after the integration symbol, but that makes life so much easier that I don't care ;-)).

how so?

 

[fresh_42]

how so?

Because the notation $\int \ldots dx$ functions as parenthesis: open and end marks. If you write $\int dx \ldots$ then it is not clear where the integral ends and other equations start.

However, in physics we often have long expressions as integrands, so it is convenient to a) write the formal signs first and b) make sure from the start what the variable is and what the constants are. As long as one avoids the ambiguity at the end, it doesn't matter. Traditionally is $dx$ last.

There are a few things where mathematical and physical conventions differ. E.g. whether the Minkowski signature is written $(+,-,-,-,)$ or $(-,+,+,+)$, or the use of co- and contravariant. Sometimes it is just a way to make the statement: "Hey, I'm a physicist / mathematician!" In the case of integrals, it is because of convenience: If you write $dx$ at the end of a long formula, then you have to read it twice in order to concentrate on the variable and recognize the constants. If you write $dx$ at the beginning, you already know when you inspect the integrand. In cases like $f(x)=x$ then $\int x\,dx$ looks better than $\int dx \, x$. But if you have
$$
\int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty \dfrac{1}{\sqrt{2\pi}} e^{-i p'x} a^\dagger(p')\dfrac{1}{\sqrt{2\pi}} e^{i px} a(p) \,dpdp'dx
$$
it is somehow easier to read
$$
\int_{-\infty}^\infty dx \int_{-\infty}^\infty dp' \int_{-\infty}^\infty dp \;\dfrac{1}{\sqrt{2\pi}} e^{-i p'x} a^\dagger(p')\dfrac{1}{\sqrt{2\pi}} e^{i px} a(p)
$$

 

[vanhees71]

Well, in the latter case it's not so much of an advantage, but if the boundaries of the integral are different it's much more difficult to decipher, which integration variable belongs to which integral sign. Also writing the differential directly behind the integration symbol makes the integral more like an operator, which physicists love to handle in an informal way.