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Is the antiderivative of an odd function even?

The answer given by the book is yes.
However, I found a counterexample defined in $\mathbb{R}\setminus \{0\}$: $$f(x)=\begin{cases}\ln |x|+1& x<0\\\ln |x|&x>0\end{cases}$$ Its derivative is $\frac 1x$, which is an odd function.

Question: is my counterexample right?

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    $\begingroup$ I think the question implies that the odd function in question must contain $0$ in its domain. Otherwise, you can't integrate it across a symmetric interval $\endgroup$ – Dylan Dec 17 '18 at 6:56
  • $\begingroup$ You can actually do this for any odd function if you allow a piecewise function definition, as there are infinitely many anti-derivatives for a given function --- simply pick two that differ by a constant, then piece them together. E.g. consider $F(x) = x^4 + [x>0]$, where $[\cdot]$ is the Iverson bracket (equal to $1$ when the condition is true, otherwise $0$), which is an antiderivative of $f(x) = x^3$. $\endgroup$ – apnorton Dec 17 '18 at 7:19
  • $\begingroup$ @apnorton I don't think so. Your $f(x)$ is not differentiable at $x=0$, it contradicts with the definition of antiderivative. $\endgroup$ – Kemono Chen Dec 17 '18 at 7:23
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    $\begingroup$ $x^2 + C$ is even for any $C$. For the opposite case of whether the antiderivative of an even function is odd, this point would be valid. $\endgroup$ – badjohn Dec 17 '18 at 11:28
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    $\begingroup$ @1123581321 Constants are even. $\endgroup$ – Dylan Dec 17 '18 at 19:19
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I think that $f$ should be defined on an interval $I$ which contains $0$ and is symmetric to $0$. If $F$ is an antiderivative of $f$ on $I$, then there is a constant $c$ such that

$$F(x)=\int_0^x f(t) dt+c.$$

If you now calculate $F(-x)$ with the substitution $s=-t$ you will get $F(-x)=F(x).$

Try it !

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    $\begingroup$ I only get that when c is zero... $\endgroup$ – lalala Dec 17 '18 at 9:28
  • $\begingroup$ @lalala If $f(t)=\sin(t)$ then $\int_0^x f(t) \,dt = 1-\cos(t)$. Choose a $c$ such as $-1$ or if you prefer $1$ and do as Fred suggests to find $F(-x)$. What do you get? $\endgroup$ – Henry Dec 17 '18 at 13:13

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