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I've run into a strange situation while trying to apply Integration By Parts, and I can't seem to come up with an explanation. I start with the following equation:

$$\int \frac{1}{f} \frac{df}{dx} dx$$

I let:

$$u = \frac{1}{f} \text{ and } dv = \frac{df}{dx} dx$$

Then I find:

$$du = -\frac{1}{f^2} \frac{df}{dx} dx \text{ and } v = f$$

I can then substitute into the usual IBP formula:

$$\int udv = uv - \int v du$$

$$\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$

$$\int \frac{1}{f} \frac{df}{dx} dx = 1 + \int \frac{1}{f} \frac{df}{dx} dx$$

Then subtracting the integral from both sides, I've now shown that:

$$0 = 1$$

Obviously there must be a problem in my derivation here... What wrong assumption have I made, or what error have I made? I'm baffled.

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39  
There is no mistake, and it does not prove 0 = 1. It proves only that 0 = 1 + $C$ for some constant $C$. An indefinite integral always has a constant of integration to be taken into account. You may have noticed in Calculus 2 that you come up with the exact same kind of result when you integrate by parts the expression $\ln (x+1)$ directly without intermediate $u$-substition vs. doing it with intermediate $u$-substition, as another example. –  Joseph Myers May 23 at 5:44
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The problem is using antiderivatives (poorly called "indefinite integrals") as if they were integrals. –  Martin Argerami May 23 at 5:53
    
Related. –  Git Gud May 23 at 9:01
7  
0 does equal 1 up to an additive constant –  Tim Seguine May 23 at 20:48
12  
I'd call on several students, one after another, and demand that they tell me what is $\displaystyle\int\dfrac{dx}{x}$, $\displaystyle\int\dfrac{du}{u}$, $\displaystyle\int\dfrac{dz}{z}$, $\displaystyle\int\dfrac{da}{a}$, and then, as the clincher, I'd ask about $\displaystyle\int\dfrac{d(\text{cabin})}{\text{cabin}}$. Some of them would grin amiably and shout out "log cabin", and they were surprised when I told them that I didn't agree. The right answer (as I learned when I was learning calculus) is "house-boat", "log cabin plus sea". –  Awal Garg May 24 at 5:59

4 Answers 4

up vote 58 down vote accepted

Hint: Constant of integration.

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4  
About the most straight-to-the-point answer I've ever seen. +1. –  Aidan F. Pierce May 23 at 13:28
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@AidanF.Pierce Well.. math.stackexchange.com/a/74383/92774 –  Soke May 23 at 22:09
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Its too much of a hint. –  Awesome May 24 at 6:04
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@MichaelT Okay, mabye the second-most straight-to-the-point answer I've ever seen. –  Aidan F. Pierce May 24 at 13:16
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I would have written $C$ to match did –  Awal Garg Jun 8 at 17:14

This line $$\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$ should be $$\int_a^b \frac{1}{f} \frac{df}{dx} dx = \left[\frac{1}{f} f\right]_a^b - \int_a^b f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$ so $$\int_a^b \frac{1}{f} \frac{df}{dx} dx = \left[1\right]_a^b - \int_a^b f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$ and $\left[1\right]_a^b=0$

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4  
It appears that he is doing indefinite integration, for which this says nothing about. –  DanZimm May 23 at 5:42
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Indefinite integration is not integration, as the question shows. –  Martin Argerami May 23 at 5:54

You have correctly derived that $0 = 1$... modulo constants.

Antiderivatives are only well-defined modulo constants*; e.g. both $x$ and $x+1$ are antiderivatives (with respect to $x$) of $1$. The equation you wrote is implicitly only meant to be an equation modulo constants; that is, the two sides of the equation don't have to be equal: they're allowed to differ by a constant.

This is traditionally worked around by adding a "constant of integration" in an ad-hoc manner rather than trying to introduce modular arithmetic. This ad-hoc fix can be tricky to get right in a nontrivial algebraic calculation if you don't fully understand what's going on, as your calculation shows.

When you cancel out the two copies of $\int\frac{1}{f} \frac{df}{dx} \, dx$, that doesn't get rid of the fact that the equation is still only meant to hold modulo constants: you've merely eliminated your mental cue (the presence of an antiderivative) to remind you of that fact.

*: Technically, I should say "locally constant functions in the integration variable" rather than "constants"

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As an exercise, look at this proof of $0=1$

Differentiate both sides $wrt $ $x$

$0=0$

Which is true, hence proved.

If you can find the error here, so you can in your above question.

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This isn't really an accurate analogy; every step of the proof in the OP is a reversible one. –  Hurkyl May 24 at 11:50
    
@Hurkyl Not perfectly accurate but trying to disprove it, he will find the word : arbitrary constant. –  Awesome May 24 at 11:51

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