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For an assignment I had, I had to prove that the derivative of an odd function is even. In the assignment we also had to prove that $F(x)=\int_0^x f(t)dt$ is odd given that $f$ is even, which I did do. Using that fact I stated the following:

Let us define $F(x)=\int_0^xf(t)dt$ such that $F(x)$ is odd. \begin{equation} F'(x)=\frac{d}{dx}\int_0^xf(t)dt=f(x) \nonumber \end{equation} Using 1.1 (the section where I proved that $F(x)=\int_0^x f(t)dt$ is odd given that $f$ is even) we know that $f(x)$ is even.

However, the teacher felt that the answer was not rigorous enough and that I was simply going backwards. Am I indeed simply moving backwards and not proving anything in which case: could someone point out places where I could make it more succinct and rigorous or alternatively supply better proof altogether.

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@AlexBecker We were supposed to prove that for any odd function $f$ then $f'$ is even. I chose to define an odd function $F$ as the integral of an even function $f$, which basically means that $F'(x)=f(x)$. I might as well have named them $g(x)=\int_0^xr(t)dt$, but that really is trivial. My intent all along was to prove that if $F$ is odd, then $f$ is even. – E.O. Feb 20 '12 at 10:32
You are quite right, I missed the word "also". – Alex Becker Feb 20 '12 at 10:33
@AlexBecker No harm done :) – E.O. Feb 20 '12 at 10:34
Differentiating $f(x)=-f(-x)$ using the chain rule will also give you the result. – David Mitra Feb 20 '12 at 15:32
@E. O. : I admit I haven't read every word of your question, but defining an odd function as the integral (a more precise term is "the primitive") of an even function is a bad idea, because then any odd function would have to be continuous. Also, it may trivialize the problem, and it differs from everyone else's definition of "odd". David Mitra's hint gives you a slick way to solve the problem. – Stefan Smith Oct 20 '13 at 16:55
up vote 7 down vote accepted

The proof is quite simple from the definition of the derivative: if $f$ is odd then $$ f'(-x) = \lim\limits_{h\to 0}\frac{f(-x+h)-f(-x)}{h} = -\lim\limits_{h\to 0}\frac{f(x-h)-f(x)}{h} = -f'(x). $$

W.r.t. your proof. You have showed that if $f$ is even, then $F = \int f$ is odd. You proved it - but you didn't prove that any odd function is an anti-derivative of the even function. That would be a reverse statement, as Alex has already told you.

Generally, you have $A\Rightarrow B$ where $A = \{f\text{ is even}\}$ and $B = \{F\text{ is odd}\}$ but to prove that the derivative of the odd function is even you need $B\Rightarrow A$ which you don't know at the moment.

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That is indeed a very nice proof! Would it be possible to comment on the validity of my proof as well? – E.O. Feb 20 '12 at 10:37
@Emile: sure, I've edited. Please tell me if it is clear – Ilya Feb 20 '12 at 11:08
You forgot a negative sign in the denominator, see Dr. Michael W. Ecker's comment/answer. – Ahmed S. Attaalla Dec 13 '15 at 17:48

I think your teacher is right that the argument is not quite rigorous. You've proved that the integral of an even function is odd. However, you haven't proved (or don't say specifically that you've proved) that the integral of a function that is not even is not odd. (Every dog has four legs, but it is not true that everything that is not a dog does not have four legs.)

Logically, there could be (a) even functions whose integrals are odd, (b) odd functions whose integrals are odd, and (c) functions that are neither even nor odd, whose integrals are odd. To complete your proof, you would have to show that cases b and c don't exist.

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The closest answer given was the direct proof from the limit definition. One major correction, however:

Technically, the last limit is NOT the claimed -1 times (the definition of) f '(x), owing to the negative h instead of the usual h. In fact, to make this work, the negative sign needs to be in the denominator so that h everywhere is replaced by -h. As a result, the right side is and should be just f '(x). So, now we get f '(-x) = f '(x) and NOT -f '(x)!

Besides, you want f '(-x) = f '(x) for the function f ' to be even! If you really had the negative one multiplier, then your derivative of your odd f would have also been odd!

(P.S. Sorry but I wrote those derivatives f ' because if I wrote f' it may look like just f. Please read very closely, as I am writing only about f prime, not original function f.)

Dr. Michael W. Ecker/ Associate Professor of Mathematics/ Pennsylvania State University/ Wilkes-Barre Campus/ Lehman, PA 18627

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