Let's consider the indefinite integral $$\int \frac{dx}{x\ln x}.$$ We will compute it by integrating by parts: $$\int \frac{dx}{x\ln x}=\int (\ln x)'\frac{dx}{\ln x}=1+\int \frac{dx}{x\ln x}.$$ Hence, $0=1$. The question is : is there anything wrong in these computations?

Note: I came across this example while reading a book on counterexamples in real analysis. I thought that the people here would find this funny, especially because the error is not so obvious to everyone.


marked as duplicate by Eric Wofsey real-analysis Jun 26 at 21:48

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    $\begingroup$ Any two antiderivatives differ by only a constant. 1 is a perfect example of such a constant. $\endgroup$ – InterstellarProbe Jun 26 at 21:24
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    $\begingroup$ One of my favorites: $0 = (1 - 1) + (1 - 1) + (1-1) + \ldots = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + \ldots = 1$ $\endgroup$ – I.Chekhov Jun 26 at 21:34
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    $\begingroup$ Your Note suggests that you already know the answer to your question, so it's not really appropriate to post it here. $\endgroup$ – Blue Jun 26 at 21:37
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    $\begingroup$ @I.Chekhov I like this one too. Depending on how you manipulate it you can also show it equals $-1$ and $1/2$ since really it could take any value that satisfies $|S|\le 1$ $\endgroup$ – Henry Lee Jun 26 at 22:31
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    $\begingroup$ Thank you guys for your input! I thought that maybe there are more like this one, but I just wanted to show people here another example. @Blue the book only said "this is wrong because we cannot reduce indefinite integrals". I basically knew that this isn't allowed, but now other users added further explanations for this. $\endgroup$ – Alexdanut Jun 26 at 22:38

The issue here is what an indefinite integral means. An indefinite integral $\int f(x)dx$ is not a single function; rather, it's best thought of as a set of functions, namely, the set $$\{F: {d\over dx}F=f\}.$$ Elements of this set are called antiderivatives, or primitives, of $f$.

This set will always contain more than one element, since adding a constant won't affect the derivative. Notation like $$\int xdx={x^2\over 2}+C$$ is shorthand for "$\int xdx$ is the set of functions of the form $x\mapsto {x^2\over 2}+C$ for some $C\in\mathbb{R}$."

So no, your observation does not imply that zero equals one.

Note that it's not always the case that two antiderivatives must always differ by a constant: if the domain of the function is "not connected," we have more options. E.g. consider the function $f:x\mapsto {1\over x}$, defined on $\mathbb{R}-\{0\}$, has two "pieces" which can behave very differently: the function $F$ sending $x<0$ to $\ln(\vert x\vert) + 42$ and $x>0$ to $\ln( x) -17$ is an antiderivative of $f$, even though it doesn't differ from the "usual" antiderivative $x\mapsto\ln(\vert x\vert)$ by a constant.


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