Why don't we account for the constant in integration by parts?

In summary: I think it is clearer to write distinguishing integral parameter and up integral value, i.e.f(x)g(x) + C = \int^x (fg)'(y) \ dy +C_1= \int^x f(y)g'(y) \ dy + \int^x f'(y) g(y) \ dy+C_2Any down values for integrals are all right because of adjustment by a constant.
  • #1
Mayhem
302
194
As we all know, integration by parts can be defined as follows: $$\int u dv = uv - \int v du$$ And the usual strategy for solving problems of these types is to intelligently define ##u## and ##dv## such that the RHS integral can easily be evaluated. However, something that is never addressed is why the implicit constant that arises from integrating ##dv## is simply ignored on the RHS. And it is not as trivial as simply saying that "a constant plus a constant is just another constant" as the constant would sometimes become a function of the integrating variable upon integration. Or does the constant always disappear upon simplification for all problems that can be solved using IBP?
 
Physics news on Phys.org
  • #2
[tex]\int_a^b u dv = [uv]_a^b - \int_a^b v du[/tex]
Here you may find no ambiguity.
 
  • Like
Likes DaveE
  • #3
mitochan said:
[tex]\int_a^b u dv = [uv]_a^b - \int_a^b v du[/tex]
Here you may find no ambiguity.
Elaborate. AFAIK integration by parts works for indefinite integrals, no?
 
  • #4
More precisely
[tex]\int_{x=a}^{x=b} u(x) dv(x) = [u(x)v(x)]_{x=a}^{x=b} - \int_{x=a}^{x=b} v(x) du(x)[/tex]
or
[tex]\int_{a}^{b} u(x) v'(x)dx = [u(x)v(x)]_{a}^{b} - \int_{a}^{b} v(x) u'(x)dx[/tex]

The integrals are function of a and b. you may choose them constant or variables.

[tex]\int^{x} u(y) v'(y)dy = [u(y)v(y)]^{x} - \int^{x} v(y) u'(y)dy + C[/tex]
 
Last edited:
  • #5
Mayhem said:
AFAIK integration by parts works for indefinite integrals, no?
In an indefinite integral equation, the constant is implied, as each indefinite integral represents an equivalence class of functions. For example: $$\int \sin^2 x \ dx = \int (1 - \cos^2 x)dx = x - \int \cos^2 x \ dx$$ And note that, for example: $$x - \int \cos^2 x \ dx = x + C - \int \cos^2 x \ dx $$ As the equals sign here signifies an equality of equivalence classes of functions.
 
  • #6
Suppose you replace f with f+c in
$$\int g\,df = fg - \int f\,dg.$$ You get
$$\int g\,df = (f+c)g - \int (f+c)\,dg = fg + cg - \int f\,dg - c\int dg = fg - \int f\,dg.$$
 
  • Like
  • Love
Likes PeroK and Mayhem
  • #7
vela said:
Suppose you replace f with f+c in
$$\int g\,df = fg - \int f\,dg.$$ You get
$$\int g\,df = (f+c)g - \int (f+c)\,dg = fg + cg - \int f\,dg - c\int dg = fg - \int f\,dg.$$
This is the kind of explanation I was looking for. Thanks! I was doing some practice problems to see if I could find an example where the + C became a problem, but every single (correct) solution seemed to simplify nicely into the usual form.
 
  • #8
vela said:
Suppose you replace f with f+c in
$$\int g\,df = fg - \int f\,dg.$$ You get
$$\int g\,df = (f+c)g - \int (f+c)\,dg = fg + cg - \int f\,dg - c\int dg = fg - \int f\,dg.$$
I'd never thought of this before, but we have to use the same indefinite integral in both terms. We can't do:
$$\int g\,df = (f+c_1)g - \int (f+c_2)\,dg$$
 
  • #9
PeroK said:
I'd never thought of this before, but we have to use the same indefinite integral in both terms. We can't do:
$$\int g\,df = (f+c_1)g - \int (f+c_2)\,dg$$

Because in general [tex]
\frac{d}{dx}(f(x) + c_1)g(x)) \neq (f(x) + c_2) \frac{dg}{dx} + g(x)\frac{df}{dx}.[/tex]
 
  • #10
Mayhem said:
This is the kind of explanation I was looking for. Thanks! I was doing some practice problems to see if I could find an example where the + C became a problem, but every single (correct) solution seemed to simplify nicely into the usual form.
I recall seeing in one textbook an example where the author noted including the constant helped in solving the problem. Unfortunately, I don't remember the example, and I've never run across that situation since then.
 
  • #11
I'm not sure whether I saw examples in my calculus classes where we added the constant. The first time I recall seeing this was in Numerical methods for scientists and engineers by Hamming. He used it while deriving the integral form of the remainder for Taylor series. At the time I thought it was quite clever, which is certainly why I remember it. Start with
## f(x) = f(a) + \int_a^x f^\prime(t) dt ##.
In the standard integration by parts formula ##\int u dv = uv - \int v du##, we let ##u=f^\prime##, ##du = f^{\prime\prime} dt##, as expected. For the other term we have ##dv = dt##, but set ##v = t - x##. The constant addition makes our final expression
## f(x) = f(a) + (x-a) f^\prime(a) + \int_a^x f^{\prime\prime}(t) (x-t) dt ##.
The subsequent terms of the series follow from further integration by parts.

jason
 
Last edited:
  • #12
It's clear from the following what's happening. For functions ##f, g## we have: $$f(x)g(x) + C = \int (fg)'(x) \ dx = \int f(x)g'(x) \ dx + \int f'(x) g(x) \ dx$$ And, if we take one of the indefinite integrals to the LHS, then we can drop the constant: $$f(x)g(x) - \int f(x)g'(x) \ dx = \int f'(x) g(x) \ dx$$ And there is nothing to get confused about. Most of the confusion students have about calculus seems to stem from the shorthand differential notation. As soon as we use the full functional notation the ambiguities vanish.
 
  • Like
Likes jasonRF
  • #13
PeroK said:
It's clear from the following what's happening. For functions ##f, g## we have: $$f(x)g(x) + C = \int (fg)'(x) \ dx = \int f(x)g'(x) \ dx + \int f'(x) g(x) \ dx$$ And, if we take one of the indefinite integrals to the LHS, then we can drop the constant: $$f(x)g(x) - \int f(x)g'(x) \ dx = \int f'(x) g(x) \ dx$$ And there is nothing to get confused about. Most of the confusion students have about calculus seems to stem from the shorthand differential notation. As soon as we use the full functional notation the ambiguities vanish.

I think it is clearer to write distinguishing integral parameter and up integral value, i.e.
[tex]f(x)g(x) + C = \int^x (fg)'(y) \ dy +C_1= \int^x f(y)g'(y) \ dy + \int^x f'(y) g(y) \ dy+C_2[/tex]
Any down values for integrals are all right because of adjustment by ##C,C_1,C_2##.
 
  • Skeptical
Likes weirdoguy and PeroK
  • #14
mitochan said:
I think it is clearer to write distinguishing integral parameter and up integral value, i.e.
[tex]f(x)g(x) + C = \int^x (fg)'(y) \ dy +C_1= \int^x f(y)g'(y) \ dy + \int^x f'(y) g(y) \ dy+C_2[/tex]
Any down values for integrals are all right because of adjustment by ##C,C_1,C_2##.
In my view, that is very muddled.
 
  • #15
Let me tell the point again. For definite integral
[tex]f(x)g(x) -f(a)g(a) = \int^x_a (fg)'(y) \ dy = \int^x_a f(y)g'(y) \ dy + \int^x_a f'(y) g(y) \ dy[/tex]
by changing low integral parameters
[tex]f(x)g(x) -f(a)g(a) = \int^x_b (fg)'(y) \ dy + [f(b)g(b)-f(a)g(a)]= \int^x_d f(y)g'(y) \ dy + \int^x_e f'(y) g(y) \ dy + C[/tex]

So if we take indefinite integrals
[tex]f(x)g(x) = \int^x (fg)'(y) \ dy + C_1 = \int^x f(y)g'(y) \ dy + \int^x f'(y) g(y) \ dy + C_2[/tex]
two Cs are required.
 
  • #17
I thought indefinite integrals come from freedom of choice of lower integral period of definite integrals. I appreciate your teachings.
 

1. Why is the constant not included in integration by parts?

The constant is not included in integration by parts because it is a constant term that does not change when differentiated or integrated. Therefore, it does not affect the overall result of the integration by parts formula and can be added back in at the end if needed.

2. Can the constant be included in integration by parts?

Yes, the constant can be included in integration by parts if it is necessary for the specific problem being solved. However, it is typically not included because it does not affect the final result.

3. How does the constant affect the integration by parts formula?

The constant does not affect the integration by parts formula as it is a constant term that does not change when differentiated or integrated. It can be added back in at the end if needed.

4. Are there any cases where the constant must be included in integration by parts?

There may be cases where the constant must be included in integration by parts if it is necessary for the specific problem being solved. However, in most cases, it is not necessary and can be omitted.

5. How do I know when to include the constant in integration by parts?

The decision to include the constant in integration by parts depends on the specific problem being solved. If the constant is necessary for the final solution, then it should be included. Otherwise, it can be omitted.

Similar threads

Replies
2
Views
795
Replies
15
Views
2K
  • Calculus
Replies
2
Views
1K
Replies
10
Views
2K
Replies
5
Views
1K
Replies
5
Views
2K
Replies
1
Views
952
Back
Top