I know that $\int_{-a}^a f(x) \, dx = 0$ if $f$ is an odd function, but I'm having difficulty understanding the proof behind it.

Every website I find is telling me basically the same proof in some form or another: here's one of them (scroll down to p. 7)

As you can see in the link, they have this:

$$\int_{-a}^a f(x)\,dx=\int_{-a}^0f(x)\,dx+\int_0^af(x)\,dx$$

$$\int_{-a}^a f(x)\,dx=-\int_0^{-a}f(x)\,dx+\int_0^af(x)\,dx$$

From which they do a $u$-substitution to get:

$$\int_{-a}^a f(x)\,dx=\int_0^{a}f(-u)\,du+\int_0^af(x)\,dx$$

I understand up to this point. What I don't understand is that in this proof and in every proof I've found, they somehow pull this off in the next step:

$$\int_{-a}^a f(x)\,dx=\int_0^{a}f(-x)\,dx+\int_0^af(x)\,dx$$

How is this possible? Didn't we already define $u$ to be $-x$? How can we just turn around and claim $x=u$ again?

  • $\begingroup$ Are you asking why $\int_0^a f(-u)\, du = \int_0^a f(-x)\, dx$? The $u$ and $x$ there are just a dummy variables. $\endgroup$ – anomaly Mar 10 '15 at 22:03
  • $\begingroup$ But how is changing the dummy variable legal in this situation? Even if they're dummy variables, they still hold significance in respect to each other ($u=-x$), and the dummy variable is used in a certain context (other functions in the integrals are in terms of $x$). $\endgroup$ – user3932000 Mar 10 '15 at 22:30
  • $\begingroup$ Writing $u$ in terms of $x$ is just an artifact of the way such substitutions are often presented in a calculus class. There's no inherent significance to $u$; it's just a dummy variable. $\endgroup$ – anomaly Mar 10 '15 at 22:53

At this point $u$ is a dummy variable, which means the definite integral does not depend on $u$ at all. You can see as it is a definite integral, the answer will be a number depending on $a$, which has nothing to do with $u$. So it can be represented by any variable, such as $x$, $t$, $y$, etc.

  • $\begingroup$ But surely you can't "mix up" the dummy variables in the same problem? (I mean, of course you can, since the proof is correct, but still) Even if they are dummy variables, is it actually possible to be inconsistent with them and claim $u=-x=x$? $\endgroup$ – user3932000 Mar 10 '15 at 22:08
  • $\begingroup$ As long as you're never doing it in the same step, yes you can. It's just a marker to let you know how the integral is operating, in the grander scheme of the equation nothing is changing. $\endgroup$ – ConMan Mar 10 '15 at 22:17
  • $\begingroup$ It is in one integral instead of "one step". If they are in one integral, you cannot mix them up. But outside of that integral, the $x$ has nothing to do with the $u$. $\endgroup$ – KittyL Mar 10 '15 at 22:19

Remember that in any integral, the letter used as the integration variable is arbitrary, so $$ \int_{-a}^a f(x) \, dx = \int_{-a}^{a} f(y) \, dy = \int_{-a}^a f(\lambda) \, d\lambda $$ and so on.

  • $\begingroup$ Any definite integral (which these are, so that's OK). $\endgroup$ – David K Mar 10 '15 at 23:12

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