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

[Mayhem]

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?

 

[mitochan]

\int_a^b u dv = [uv]_a^b - \int_a^b v du
Here you may find no ambiguity.

 

[Mayhem]

\int_a^b u dv = [uv]_a^b - \int_a^b v du
Here you may find no ambiguity.

Elaborate. AFAIK integration by parts works for indefinite integrals, no?

 

[mitochan]

More precisely
\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)
or
\int_{a}^{b} u(x) v'(x)dx = [u(x)v(x)]_{a}^{b} - \int_{a}^{b} v(x) u'(x)dx

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

\int^{x} u(y) v'(y)dy = [u(y)v(y)]^{x} - \int^{x} v(y) u'(y)dy + C

 

[PeroK]

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.

 

[vela]

Suppose you replace f with f+c in
$$\int g\,df = fg - \int f\,dg.$$ You get
$$\int g\,df = (f+c)g - \int (f+c)\,dg = fg + cg - \int f\,dg - c\int dg = fg - \int f\,dg.$$

 

[Mayhem]

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.

 

[PeroK]

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:
$$\int g\,df = (f+c_1)g - \int (f+c_2)\,dg$$

 

[pasmith]

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 <br /> \frac{d}{dx}(f(x) + c_1)g(x)) \neq (f(x) + c_2) \frac{dg}{dx} + g(x)\frac{df}{dx}.

 

[vela]

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.

 

[jasonRF]

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

 

[PeroK]

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.

 

[mitochan]

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.
f(x)g(x) + C = \int^x (fg)&#039;(y) \ dy +C_1= \int^x f(y)g&#039;(y) \ dy + \int^x f&#039;(y) g(y) \ dy+C_2
Any down values for integrals are all right because of adjustment by $C,C_1,C_2$.

 

[PeroK]

I think it is clearer to write distinguishing integral parameter and up integral value, i.e.
f(x)g(x) + C = \int^x (fg)&#039;(y) \ dy +C_1= \int^x f(y)g&#039;(y) \ dy + \int^x f&#039;(y) g(y) \ dy+C_2
Any down values for integrals are all right because of adjustment by $C,C_1,C_2$.

In my view, that is very muddled.

 

[mitochan]

Let me tell the point again. For definite integral
f(x)g(x) -f(a)g(a) = \int^x_a (fg)&#039;(y) \ dy = \int^x_a f(y)g&#039;(y) \ dy + \int^x_a f&#039;(y) g(y) \ dy
by changing low integral parameters
f(x)g(x) -f(a)g(a) = \int^x_b (fg)&#039;(y) \ dy + [f(b)g(b)-f(a)g(a)]= \int^x_d f(y)g&#039;(y) \ dy + \int^x_e f&#039;(y) g(y) \ dy + C

So if we take indefinite integrals
f(x)g(x) = \int^x (fg)&#039;(y) \ dy + C_1 = \int^x f(y)g&#039;(y) \ dy + \int^x f&#039;(y) g(y) \ dy + C_2
two Cs are required.

 

[PeroK]

That's way more complicated than post #12. Plus, we are dealing with indefinite integrals. See the definition here, for example:

https://tutorial.math.lamar.edu/classes/calci/indefiniteintegrals.aspx

 

[mitochan]

I thought indefinite integrals come from freedom of choice of lower integral period of definite integrals. I appreciate your teachings.