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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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    $\begingroup$ 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. $\endgroup$ May 23 '14 at 5:44
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    $\begingroup$ The problem is using antiderivatives (poorly called "indefinite integrals") as if they were integrals. $\endgroup$ May 23 '14 at 5:53
  • $\begingroup$ Related. $\endgroup$
    – Git Gud
    May 23 '14 at 9:01
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    $\begingroup$ 0 does equal 1 up to an additive constant $\endgroup$ May 23 '14 at 20:48
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    $\begingroup$ 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". $\endgroup$ May 24 '14 at 5:59
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Hint: Constant of integration.

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    $\begingroup$ About the most straight-to-the-point answer I've ever seen. +1. $\endgroup$ May 23 '14 at 13:28
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    $\begingroup$ @AidanF.Pierce Well.. math.stackexchange.com/a/74383/92774 $\endgroup$
    – MCT
    May 23 '14 at 22:09
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    $\begingroup$ Its too much of a hint. $\endgroup$
    – evil999man
    May 24 '14 at 6:04
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    $\begingroup$ @MichaelT Okay, mabye the second-most straight-to-the-point answer I've ever seen. $\endgroup$ May 24 '14 at 13:16
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    $\begingroup$ I would have written $C$ to match did $\endgroup$ Jun 8 '14 at 17:14
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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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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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    $\begingroup$ It appears that he is doing indefinite integration, for which this says nothing about. $\endgroup$
    – DanZimm
    May 23 '14 at 5:42
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    $\begingroup$ Indefinite integration is not integration, as the question shows. $\endgroup$ May 23 '14 at 5:54
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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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  • $\begingroup$ This isn't really an accurate analogy; every step of the proof in the OP is a reversible one. $\endgroup$
    – user14972
    May 24 '14 at 11:50
  • $\begingroup$ @Hurkyl Not perfectly accurate but trying to disprove it, he will find the word : arbitrary constant. $\endgroup$
    – evil999man
    May 24 '14 at 11:51
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The problem here is that when you applied the By Parts Formula $$uv-\int \frac{du}{dx}vdv $$

you took $u=\frac 1f$ and $dv=\frac {df}{dx}dx$ Now when you use by parts formula your first task is to get $v$ and for that you need to do this $\int dv$, right? You have done all righty, but the problem comes when you write $\int df =f$ Here you also need to add the constant C, Or your proof will get pretty much messed up. In the end where you got $0=1$ , you are lacking that $C$ in this equation here, when you do antiderivatives you can't just provide one answer.If you write $\int df=f$ only then you are only getting one of the infinite solutions because $\frac {d}{df}(f)=1$ but $\frac {d}{df}(f+2)$ is also 1, infact take the derivative of $f+c$ (w.r.t f) where $C$ is any constant, you will always end up with 1.So I guess that's the error.Next time be careful with constants.

Hope It answered your question.

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