I just saw the following fake proof.

$$\int \frac1x dx =\int 1\cdot \frac1x dx=x\frac1x+\int x \frac1{x^2} dx = 1+ \int \frac1x dx$$

Which would imply $1=0$, hence the fake proof tag.

The explanation given was that there was a discontinuity in the function that we were in essence "integrating over", however the integral is indefinite so we are not explicitly integrating over anything (or maybe integrating over everything?).

Will any function with a discontinuity be exploitable in this way? Could we actually solve the problem by converting the integral to a definite integral in a region there is no discontinuity? I think the trick used would still work, or am not sure why it would provide a result that make sense.

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    $\begingroup$ "constant of integration" $\endgroup$ Jun 20, 2014 at 19:18
  • $\begingroup$ What you showed is that if $F$ is a primitive of $x\mapsto\frac1x$ (on some interval where the latter is defined), then so is $x\mapsto1+F(x)$; this is true. I'll make that an answer. $\endgroup$ Jun 21, 2014 at 10:21
  • $\begingroup$ To add even further to the confusion, letting $u=\dfrac1x$ we have $\displaystyle\int\frac1xdx=-\int\frac1udu$ :-) $\endgroup$
    – Lucian
    Jun 21, 2014 at 12:20

6 Answers 6


I disagree with alexqwx. His answer is misleading. There is no constant of integration generated in integration by parts until integration actually occurs. The above answer also does not mention that the fallacy occurs when subtracting indefinite integrals not because of a lack of a constant of integration.

There is no reason to write $\int \frac{1}{x}dx=1+\int \frac{1}{x}dx +C$. The error is in assuming $\int \! f\, dx - \int\! f\, dx = 0$ when, in reality, $\int \! f\, dx - \int\! f\, dx = \int \! 0 \, dx = C.$

There is nothing wrong with the statement of $$\int \frac1x dx =1+ \int \frac1x dx,$$

only the conclusion obtained from it. It is just a proof that $1$ is constant. If you decide at this point to add bounds to your integral you find

$$\int_a^b \frac1x dx =1_a^b+ \int_a^b \frac1x dx$$

$$\int_a^b \frac1x dx = \int_a^b \frac1x dx$$

Because $1_a^b = 1 - 1 = 0$.


$$\int \frac1x dx =1+ \int \frac1x dx$$

$$\int \frac1x dx - \int \frac1x dx =1$$

$$\int \frac1x - \frac1x dx = 1$$

$$\int 0\, dx = 1 $$

$$C = 1$$

The fallacy occurs when subtracting the indefinite integrals, not with anything mentioned by alexqwx.

  • 2
    $\begingroup$ My answer isn't misleading at all. The $+C$ at the end, whilst not strictly necessary, is to emphasise that there must be an arbitrary constant (or two, if you like, merged into 1) when integrating, as the OP is assuming that $\int \frac{1}{x}dx=\ln|x|$, rather than $\ln|x|+c_1$. There's nothing wrong with my answer. $\endgroup$
    – beep-boop
    Jun 20, 2014 at 21:39
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    $\begingroup$ @alexqwx Your answer is wrong because you imply that the error is that there should be a constant when using integration by parts. Like I said, the fallacy occurs when subtracting the indefinite integrals. You say that the $+C$ is there to emphasize the arbitrary constant when integrating but fail to realize that we haven't done any integrating yet. You also come to the conclusion that the arbitrary constant equals $-1$ which does not make sense. See my answer for more details. $\endgroup$
    – Brad
    Jun 20, 2014 at 22:43
  • $\begingroup$ @alexqwx there is also no reason to downvote my previous questions because you do not like my response to your answer $\endgroup$
    – Brad
    Jun 20, 2014 at 22:51
  • $\begingroup$ I agree with the fact that $\int f(x) dx-\int f(x) dx$ is equal to a constant (and not necessarily zero), but to say that we shouldn't introduce arbitrary constants when doing indefinite integration is wrong. And I have no idea what you're talking about. Do you have any evidence to support this claim? $\endgroup$
    – beep-boop
    Jun 20, 2014 at 23:29
  • $\begingroup$ There should be two arbitrary constants (one on each side, for each integral, which can then become another single constant), as indefinite integration is defined up to an additive constant; it is by ignoring the existence of such a constant that we 'deduce' that 1=0. $\endgroup$
    – beep-boop
    Jun 20, 2014 at 23:34

Indefinite integral of f is a set of all such functions F that $F'=f$. They all different by additive constant. And indeed sets $\{F: F'=\frac{1}{x}\}$ and $\{F + 1: F'=\frac{1}{x}\}$ are same.


It just occurred to me that none of the answers here, although they critique the “proof” and the "reasoning" given for why the proof is wrong, really answer the question as stated in the title: which is can we get a flawed “proof” from a function with integration no matter what the function? The “discontinuity” requirement has been shown to be superfluous, but the question I mention remains and is itself legitimate. So for a general function, can we rewrite its integral in such a way as to “prove” something false, using valid transformations of integrals but then a final fallacious (but seemingly "sensible") step?

While a thorough investigation of all integration techniques would seem to be difficult, we can investigate integration by parts. The integration by parts formula is

$$\int u\ dv = uv\ – \int v\ du$$

To get a fallacy of the form shown in the original post, we need that

$$\int u\ dv = -\int v\ du$$

which implies

$$u\ dv = -v\ du$$


$$u = -v \frac{du}{dv}$$

which is a differential equation for $u$ with solution $u = K/v$, where $K$ is a constant. Note that in the given example, $u = 1/x$, $dv = dx$, $v = x$, $du = -\frac{1}{x^2} dx$. Note that $1/v = 1/x = u$, so with $K = 1$ the condition is satisfied. Also, note that $uv = K$.

So for the case of integration by parts, any integrand $u dv$ where $u$ is a constant times the reciprocal of $v$ will work. We can take $u = x^2 + 1$, so that $v = \frac{1}{x^2 + 1}$, then $dv = -\frac{2x}{(x^2 + 1)^2} dx$, so the function to integrate is $-\frac{2x}{x^2 + 1}$. If you integrate that with $u = x^2 + 1$ and $dv = -\frac{2x}{(x^2 + 1)^2}$, you can get another fallacy of the form shown in the post.

Indeed, this shows that the answer to the question is in fact yes: Just write the function (“times” $dx$) as a product $u dv$ where $uv = K$. The condition $uv = K$ implies $u'v + uv' = 0$, or $v' = -\frac{u'}{u} v = -\frac{u'}{u} \frac{K}{u} = -K\frac{u'}{u^2}$. Thus $uv' = u \frac{dv}{dx} = -K \frac{u'}{u}$. Setting this equal to an arbitrary function $f$ gives $u(x) = Le^{\int -f(x)/K\ dx}$, $L$ another arbitrary constant. Now you have $u$ and $dv$ for any function you want and can "prove" something false from it.


alexqwx's answer is wrong. Brad's is notationally confusing because it pretends the "$1$" is a nonzero constant when it's not.

There is no actual "$1$" after the integration by parts. It's easy to see if you pretend there are integration bounds:

$$\int_a^b \frac1x dx =\int_a^b 1\cdot \frac1x dx=\left[x\frac1x\right]_a^b+\int_a^b x \frac1{x^2} dx = \left[1\right]_a^b + \int_a^b \frac1x dx$$

No matter what bounds you pick, $\left[1\right]_a^b = 0$. So integration by parts just gives

$$\int \frac1x dx = \int \frac1x dx$$

We don't need to worry about whether subtracting indefinite integrals is okay.

  • $\begingroup$ "it pretends the "$1$" is a nonzero constant when it's not" - I'm not sure what you're trying to say there. Do you think $1$ is actually zero? Or that its value is allowed to vary? Or did you screw something up when writing that? $\endgroup$ Jun 21, 2014 at 1:45
  • $\begingroup$ @Pringoooals Well, I was about to go all-guns-blazing with Tom, but, since you've asked me to moderate my attitude, I think I'll turn the other cheek. Thanks for the sage advice, oh, Wise One! $\endgroup$
    – beep-boop
    Jun 21, 2014 at 3:14
  • $\begingroup$ Not to fan the flame war here, but @alexqwx -- Brad already said why your answer was wrong: integration by parts doesn't generate an additive constant. That would only come later during actual integration. $\endgroup$
    – tom
    Jun 21, 2014 at 3:19
  • $\begingroup$ @user2357112 let me try to clarify. When you do indefinite integration by parts, the first term you get is like an antiderivative, in the sense that it still has implicit integration bounds. It's confusing to write a "1" there as Brad did, because what you really have is the function F(x) = 1, which wants to be evaluated at two bounds [a,b]. To actually evaluate this equation, you have to plug in specific integration bounds and you will always get F(b) - F(a) = 1 - 1 = 0. That's why I say there's not really a 1 there. (Note esp. that for some bounds, the integral is not even well-defined.) $\endgroup$
    – tom
    Jun 21, 2014 at 3:41

What you showed is that if $F$ is a primitive of $x\mapsto\frac1x$ (on some interval where the latter is defined, in other words that does not contain$~0$), then so is $x\mapsto1+F(x)$. This is true, since the derivative of $x\mapsto 1$ is $x\mapsto 0$.


$\int \frac{1}{x}dx=1+\int \frac{1}{x}dx \color{red}{+C}$ (you forgot the constant of integration).

Then we have that $C=-1$, which is fine.

  • $\begingroup$ @alexqwk if I understand correctly if you were to finish the integration in the "normal way" you would get a second independent constant. The first one (shown above) gets you back to the starting position and the second constant would be used for the particular solution, right? $\endgroup$
    – kleineg
    Jun 20, 2014 at 19:35
  • $\begingroup$ So, something like $1 + ln(x) + C_1 + C_2$ with $C_1 = -1$ $\endgroup$
    – kleineg
    Jun 20, 2014 at 19:37
  • $\begingroup$ Well, we'd have $ \require{cancel}\cancel{\ln |x|}= 1+\cancel{\ln|x|}+C$. Alternatively, we can do what you suggested and write $ \require{cancel}\cancel{\ln |x|}+c_1= 1+\cancel{\ln|x|}+c_2$, but the we can just re-introduce another arbitrary constant, $C=c_2-c_1$, so it doesn't matter, as long as, in the end, we've only got one arbitrary constant. $\endgroup$
    – beep-boop
    Jun 20, 2014 at 19:55

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