Question about substitution method in integration

kelvin490

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It is common that we replace [tex]\int u(x)v'(x)dx[/tex] by [tex]\int udv[/tex] where both u and v are continuous functions of x. My question is, must we ensure that u can be written as a function of v before applying this? The above substitution method is involved in the proof of integration by parts but I cannot find textbooks that addressed this point.

I find it easier to understand for definite integrals because I can think it as summation. But for indefinite integrals it is just anti-derivative and I am not sure how this kind of replacement is valid.
 
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I think the point is that there is definitely a v if u conforms to the definition of an integral function in x and it stems from the diffentiation rule for a function of the form u(v(x)).

I don't know how you prove that v exists unless it's some fundamental rule that any independent variable can be replaced by an expression based on another independent variable which I've always accepted as true.
 

SteamKing

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It is common that we replace [tex]\int u(x)v'(x)dx[/tex] by [tex]\int udv[/tex] where both u and v are continuous functions of x. My question is, must we ensure that u can be written as a function of v before applying this? The above substitution method is involved in the proof of integration by parts but I cannot find textbooks that addressed this point.
It's not clear what you mean by "My question is, must we ensure that u can be written as a function of v before applying this?". The functions u(x) and v(x) are functions of x, not each other.

I find it easier to understand for definite integrals because I can think it as summation. But for indefinite integrals it is just anti-derivative and I am not sure how this kind of replacement is valid.
The idea of integration by parts stems from applying the product rule to finding the derivative of the product of two functions.

See this article for a derivation:

http://en.wikipedia.org/wiki/Integration_by_parts
 

kelvin490

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[/QUOTE]

The functions u(x) and v(x) are functions of x, not each other.
[/QUOTE]

What I mean is if u(x) cannot be expressed as function of v(x) then we cannot do [tex]\int udv[/tex]. How do we know in general u(v(x)) exist?

In justifying the method of substitution, many textbooks says [tex]\int F[u(x)]u'(x)dx = \int F(u)du[/tex], in this case the F must be able to be expressed as function of u.
 

kelvin490

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In some simple cases it is easy to express u(x) as u(v(x)) so that [tex]\int u(x)v'(x)dx[/tex] = [tex]\int u(v(x))v'(x)dx[/tex] , which is simple substitution method commonly mentioned in textbook. What I concern is how do we know in general u(x) can be expressed as u(v(x)) so that [tex]\int u(v(x))v'(x)dx = \int u(v)dv[/tex] can be applied?
 

SteamKing

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What I mean is if u(x) cannot be expressed as function of v(x) then we cannot do [tex]\int udv[/tex]. How do we know in general u(v(x)) exist?

In justifying the method of substitution, many textbooks says [tex]\int F[u(x)]u'(x)dx = \int F(u)du[/tex], in this case the F must be able to be expressed as function of u.
I think you are getting hung up on the notation here. There are some shortcuts taken with the usual integration by parts formula, and you should not assume that because

[tex]\int udv[/tex]

that u must be written as a function of v or vice versa for this integration procedure to apply.

After all, if we want to find the derivative of u(x)*v(x) w.r.t x, we just write d(uv)/dx, with the understanding that u and v are functions of x.

Since d(u*v)/dx = u * dv/dx + v * du/dx, then we can integrate both sides and maintain the equal sign:

[tex]\int d(u*v)/dx = \int [u * dv/dx + v * du/dx] = \int u*dv/dx + \int v * du/dx [/tex]

The Fundamental Theorem of Calculus states that [tex]\int dF(x)/dx = F(x) + C [/tex]

so we can write the equation above as

[tex]\int d(u*v)/dx = u*v = \int [u * dv/dx + v * du/dx] = \int u*dv/dx + \int v * du/dx [/tex]

And doing a little rearrangement we have:

[tex]\int u*dv/dx = u*v - \int v * du/dx [/tex]

which is abbreviated as

[tex]\int u dv = u v - \int v du [/tex]

In the latter form of the integration by parts formula, there is no implication that u is a function of v or v is a function of u in order for the formula to apply.
 
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I think the big thing here is just the notation used. There are probably numerous definitions used for how to properly describe substitution. Therefore, I don't think the issue of whether we view this as an anti-derivative or a summation matters here.
The way I see it:
##v = v(x)## (i.e., ##v## is a function of ##x##)
##\frac{dv}{dx} = v'(x)##
##dv = v'(x) \ dx##
So, to me, ##dv## and ##v'(x) \ dx## are the same, so I think that little replacement is perfectly allowable.
So, to answer your question, no; I don't believe ##u## has to be written as a function of ##v##, as the shorthand ##dv## is taking the place of ##v'(x) \ dx##. Just because there's a ##u## and the differential operator is in terms of ##v## doesn't mean that rewriting it that way suddenly makes ##u## a function of ##v##.
 

kelvin490

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I think the big thing here is just the notation used. There are probably numerous definitions used for how to properly describe substitution. Therefore, I don't think the issue of whether we view this as an anti-derivative or a summation matters here.
The way I see it:
##v = v(x)## (i.e., ##v## is a function of ##x##)
##\frac{dv}{dx} = v'(x)##
##dv = v'(x) \ dx##
So, to me, ##dv## and ##v'(x) \ dx## are the same, so I think that little replacement is perfectly allowable.
So, to answer your question, no; I don't believe ##u## has to be written as a function of ##v##, as the shorthand ##dv## is taking the place of ##v'(x) \ dx##. Just because there's a ##u## and the differential operator is in terms of ##v## doesn't mean that rewriting it that way suddenly makes ##u## a function of ##v##.
In terms of differential, of course ##dv## and ##v'(x) \ dx## are the same. But [tex]\int [/tex] and [tex]dx [/tex] together forms one single mathematical sign meaning anti-derivative for indefinite integral, [tex]\int f(x) dx [/tex] means finding the primitive function F(x) which is a function of x. So to me [tex]\int u dv [/tex] implies u is function of v.

I just wonder why in applications substitution is done in this way without considering the differentiability of u with respect to v.
 
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SteamKing

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In terms of differential, of course ##dv## and ##v'(x) \ dx## are the same. But [tex]\int [/tex] and [tex]dx [/tex] together forms one single mathematical sign meaning anti-derivative for indefinite integral, [tex]\int f(x) dx [/tex] means finding the primitive function F(x) which is a function of x. So to me [tex]\int u dv [/tex] implies u is function of v.

I just wonder why in applications substitution is done in this way without considering the differentiability of u with respect to v.
I can't address why you interpret the formulas different from most everyone else. This notation, although it may be confusing to you, is pretty much standard. You'll just have to retrain yourself on how you interpret the notation.

I've never had the notion that just because you saw u dv written down that this juxtaposition implied u was a function of v. Other integral expressions make liberal use of dummy variables of integration, so a certain amount of flexibility in interpreting mathematical notation is desirable.

In any event, if you take the derivative of any result obtained from integration by parts, you should obtain the original function which was integrated.
 

kelvin490

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I think in practice if u can simply be written as function of v then [tex]\int udv[/tex] can be done in usual way. Otherwise we still need to express it as [tex]\int u(x)v'(x)dx[/tex] in order to do the integration. So [tex]\int udv[/tex] is just a notation without the information about the integrability of u with respect to v.
 

Stephen Tashi

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how do we know in general u(x) can be expressed as u(v(x)) so that [tex]\int u(v(x))v'(x)dx = \int u(v)dv[/tex] can be applied?
That's a good question.

To answer it, we'd have to state the "rule" as actual mathematical theorem. That would mean putting it in the form "If...... then.....". We'd have to bring everything out in the open instead of hiding it notation. For example, the notation suggests that v(x) is function of x and that v'(x) exists. But, stated as a theorem, more is required of v(x) than that..

It's an exercise just say what the equation [itex]\int u(v(x))v'(x)dx = \int u(v) dv[/itex] claims. If the left hand size is an antiderivative then it is a function of [itex] x [/itex]. The right hand size is a function of [itex] v [/itex].

Is the rule saying two functions are the same function? Suppose someone asked you "Are [itex] G(x) = sin^2(x) [/itex]and [itex] H(v) = v^2 [/itex], the same function?" you'd probably say "No" or "Not necessarily" or "Not unless there is a definite relation between [itex] x [/itex] and [itex] v [/itex].

Suppose there is the relation [itex] v = sin(x) [/itex] and suppose the the two functions are the same function. What function is it? A real valued function of a real variable is a set of ordered pairs of real numbers. So does this "same" function contain the ordered pair [itex] ( 1.2, sin^2(1.2)) [/itex] or does it contain the ordered pair [itex] (1.2, (1.2)^2) [/itex] ? It can't contain both. You can say "That depends on whether we're talking about [itex] x [/itex] or [itex] v [/itex] as the variable. But you can't reconcile that kind of terminology with the mathematical definition of a function.

In the Wikipedia , integration by substitution, as a mathematical theorm, deals with the equality of definite integrals., not with the equality of two functions. http://en.wikipedia.org/wiki/Integration_by_substitution . The Wikipedia shows integration by substitution used as a "technique" for finding antiderivatives, but it doesn't state it as a theorem.

Examining rigorous basis for calculus techniques falls under the heading of "Analysis", so the people who frequent the "Topology And Analysis" section might be the experts on your question.
 

kelvin490

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That's a good question.

To answer it, we'd have to state the "rule" as actual mathematical theorem. That would mean putting it in the form "If...... then.....". We'd have to bring everything out in the open instead of hiding it notation. For example, the notation suggests that v(x) is function of x and that v'(x) exists. But, stated as a theorem, more is required of v(x) than that..

It's an exercise just say what the equation [itex]\int u(v(x))v'(x)dx = \int u(v) dv[/itex] claims. If the left hand size is an antiderivative then it is a function of [itex] x [/itex]. The right hand size is a function of [itex] v [/itex].

Is the rule saying two functions are the same function? Suppose someone asked you "Are [itex] G(x) = sin^2(x) [/itex]and [itex] H(v) = v^2 [/itex], the same function?" you'd probably say "No" or "Not necessarily" or "Not unless there is a definite relation between [itex] x [/itex] and [itex] v [/itex].

Suppose there is the relation [itex] v = sin(x) [/itex] and suppose the the two functions are the same function. What function is it? A real valued function of a real variable is a set of ordered pairs of real numbers. So does this "same" function contain the ordered pair [itex] ( 1.2, sin^2(1.2)) [/itex] or does it contain the ordered pair [itex] (1.2, (1.2)^2) [/itex] ? It can't contain both. You can say "That depends on whether we're talking about [itex] x [/itex] or [itex] v [/itex] as the variable. But you can't reconcile that kind of terminology with the mathematical definition of a function.

In the Wikipedia , integration by substitution, as a mathematical theorm, deals with the equality of definite integrals., not with the equality of two functions. http://en.wikipedia.org/wiki/Integration_by_substitution . The Wikipedia shows integration by substitution used as a "technique" for finding antiderivatives, but it doesn't state it as a theorem.

Examining rigorous basis for calculus techniques falls under the heading of "Analysis", so the people who frequent the "Topology And Analysis" section might be the experts on your question.
Absolutely agree. The equality [itex]\int u(x)v'(x)dx = \int u dv[/itex] and its application in proving the formula of integration by parts should be backed up by a separate theorem. I am not sure about that yet but I guess there should be a theorem like "If both u(x) and v(x) are continuous functions of x then u can always be expressed as function of v". Otherwise "u is function of v" is actually a hidden assumption that for [itex]\int u(x)v'(x)dx = \int u dv[/itex] to be true.
 

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