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I'm trying to prove that the primitive function of an odd function is even without using integrals. It goes down to proving that $$g(x) = F(-x) - F(x) = 0$$

(Such that $F'(x) = f(x)$ and $f$ is odd)

I calculated $g'(x)$ but I keep getting $$g'(x) = -2F(x)$$ And the same goes for trying to prove that the primitive of an even function is odd.

Can you please point out where's my mistake ?

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  • $\begingroup$ Note that $g'(x)=-F'(-x)-F'(x)=0$ since $F'=f$ assumed to be odd. Since $g'(x)=0$ and $g(0)=0$ with $g$ continuous it follows that $g \equiv 0$ as required. $\endgroup$ – Kal S. Oct 26 '17 at 11:49
  • $\begingroup$ Note that you have $(F(-x))'=- F'(-x)$. See en.wikipedia.org/wiki/Chain_rule $\endgroup$ – Kal S. Oct 26 '17 at 12:00
  • $\begingroup$ But F is not a composition of two functions. $\endgroup$ – Abbkey Oct 26 '17 at 12:04
  • $\begingroup$ $F(-x)= F\circ h (x)$ with $h(x)=-x$. You need to review the chain rule for differentiation. $\endgroup$ – Kal S. Oct 26 '17 at 12:06
  • $\begingroup$ Ok thank you. This is what I was missing then. $\endgroup$ – Abbkey Oct 26 '17 at 12:08
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Its just chain rule. Let g be the primitive of $f$. Then $$ (g(-x))'=-g'(-x)=-f(-x)=f(x)=(g(x))' $$

so that $g(-x)$ and $g(x)$ differ by a constant. Plugging $x=0$ this constant is $0$ too so proved.

[Another great proof by Great Hands :v]

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    $\begingroup$ "[Another great proof by Great Hands :v]" lmaof $\endgroup$ – noctusraid Oct 26 '17 at 12:24
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First of, there is no 'the' primitive of a function. We are going to show however that for an odd function, there is 'a' primitive of $f$ which is even.

Write: $f = f_+ + f_-$ where $f_+(x) = \frac{f(x)+f(-x)}{2}$ and $f_-(x) = \frac{f(x)-f(-x)}{2}$ are the even/odd part of $f$. Do the same with $F =F_+ + F_-$.

By assumption, $F_-' = \frac{d}{dx}\frac{F(x) - F(-x)}{2} = \frac{f(x)+f(-x)}{2} = f_+ = 0$, hence $F_-$ is piece-wise constant (on each connected component of the domain of $f$). Thus $F_+$ is also a primitive of $f$ which is even.

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