Odd / Even integrals

Question

My textbook doesn't really have an explanation for this so could someone explain this too me.

If f(x) is even, then what can we say about: $$\int_{-2}^{2} f(x)dx$$

If f(x) is odd, then what can we say about $$\int_{-2}^{2} f(x)dx$$

I guessed they both are zero? For the first one if its even wouldn't this be the same as $$\int_{a}^{a} f(x)dx = 0$$

Now if its odd f(-x) = -f(x). Would FTOC make this zero as well?

Try some examples. Constant functions are even, $x^2$ is even. $x$ is odd. — lulu, Feb 13 '17 at 18:49
Hint #1: Try drawing out graphs of $f(x) = x$ and $f(x) = x^2$, and consider the visual interpretation of your integrals. — FraGrechi, Feb 13 '17 at 18:50
Also see http://math.stackexchange.com/questions/1986770/definite-integral-of-even-and-odd-functions-proof — David K, Feb 13 '17 at 19:10
I think more information is needed about $f(x)$ in terms of continuity — imranfat, Feb 13 '17 at 19:22
I assumed being even/odd meant continuous on all domain of x because they are even in that domain. If not, then apologies — user349557, Feb 13 '17 at 19:24
@imranfat Judging from nature of the question, I think we can safely assume that $f$ is integrable over $[-2, 2]$ — FraGrechi, Feb 13 '17 at 19:25
@user349557 Consider the function which is identically $+1$ on $[0, 2]$ and identically $-1$ on $[-2, 0)$ — FraGrechi, Feb 13 '17 at 19:31
@GrancescoFrechi: That function is neither even nor odd (although it only differs from an odd function on a set of zero measure). — celtschk, Feb 13 '17 at 23:58

score 8 · Accepted Answer · answered Feb 13 '17 at 18:50

8

If $f(x)$ is even then $f(-x) = f(x)$. So $$\int_{-2}^2 f(x) \, \mathrm{d}x = \int_{-2}^0 f(x)\, \mathrm{d}x + \int_0^2 f(x) \, \mathrm{d}x = \int_0^2 f(-x) \, \mathrm{d}x + \int_0^2 f(x) \, \mathrm{d}x$$

But then $f(-x) = f(x)$ so that simplifies to $2\int_0^2 f$.

Similarly, if $f$ is odd - that is: $f(-x) = -f(x)$ we get $$\int_{-2}^2 f(x) \, \mathrm{d}x = \int_{-2}^0 f(x)\, \mathrm{d}x + \int_0^2 f(x) \, \mathrm{d}x = \int_0^2 f(-x) \, \mathrm{d}x + \int_0^2 f(x) \, \mathrm{d}x = 0$$

answered Feb 13 '17 at 18:50

Zain Patel

16,802

1

Note that $f(x)=\frac{1}{x}$ is also odd. Caution! – imranfat Feb 13 '17 at 19:21
Yes, user has explained in my other comment just now... – imranfat Feb 13 '17 at 19:27

score 3 · Answer 2 · answered Feb 13 '17 at 19:02

Not exactly: $$\begin{cases}\displaystyle\int_{-a}^a f(x)\,\mathrm d\mkern1mu x=2\int_{0}^a f(x)\,\mathrm d\mkern1mu x &\text{if }\;f\;\text{ is even,}\\ \displaystyle\int_{-a}^a f(x)\,\mathrm d\mkern1mu x=0&\text{if }\;f\;\text{ is odd.}\end{cases}$$ To see it, make the substitution $\;t=-x$, $\;\mathrm d\mkern1mu x=-\mathrm d\mkern1mu t$: $$\int_{-a}^0 f(x)\,\mathrm d\mkern1mu x=-\int_{a}^0 f(-t)\,\mathrm d\mkern1mu t=\int_{0}^a f(-t)\,\mathrm d\mkern1mu t=\begin{cases}\displaystyle\int_{0}^a f(-t)\,\mathrm d\mkern1mu t&(f\;\text{even}),\\\displaystyle-\int_{0}^a f(-t)\,\mathrm d\mkern1mu t&(f\;\text{odd}),\end{cases}$$ then use Chasles relation.

How about the 2 dimensions? I mean if $\int_{R^2} f=0$, can we say $f$ is odd? — Hermi, Apr 09 '20 at 08:11
I don't think one can apply the same arguments, because the change of variables formula in multiple integrals involves the absolute value of the Jacobian. — Bernard, Apr 09 '20 at 08:58

score 2 · Answer 3 · answered Feb 13 '17 at 18:50

Start by splitting the integral into two pieces, the part over negatives values of $x$ and the part over positive values.

$$ \int_{-2}^{2} f(x)\,dx = \int_{-2}^{0} f(x)\,dx + \int_{0}^{2} f(x)\,dx$$

From here you can apply the definition of an even or odd function

score 1 · Answer 4 · answered Feb 13 '17 at 18:57

$$ \int_a^b f(x) dx = \int_a^0 f(x)dx + \int_0^bf(x)dx $$ if and only if $0 \in (a,b)$ i.e. $0$ is in the interval of your integral. So $a=-2$ and $b=2$ satisfies this easily. $$ \int_{a}^0 f(x)dx = -\int_{-a}^0 f(-x)dx = \int_0^{-a}f(-x)dx $$ now since we have the requirement that $0$ is in the interval then we must have $a < 0$ and $b>0$ this implies $-a > 0$ (easy to see)

putting this together we have $$ \int_a^b f(x) dx = \int_a^0 f(x)dx + \int_0^bf(x)dx = \int_0^{-a}f(-x)dx + \int_0^bf(x)dx $$ if we have symmetric bounds i.e. $|a| = |b|$ or $a = - b$ then we have $$ \int_{-b}^{b} f(x) dx = \int_0^{b}f(-x)dx + \int_0^bf(x)dx = \int_0^b f(-x) + f(x) dx $$ The final part is what is the parity of a function, the example we have here is odd/even in this sense $$ f(-x) = -f(x)\;\;\text{odd}\\ f(-x) = f(x)\;\;\text{even} $$ so we can replace this in the integral. $$ \int_{-b}^{b} f(x) dx = \int_0^b -f(x) + f(x) dx = \int_0^b 0 dx\;\;\text{odd}\\ \int_{-b}^{b} f(x) dx = \int_0^b f(x) + f(x) dx = \int_0^b 2f(x) dx\;\;\text{even} $$

score 0 · Answer 5 · answered Feb 13 '17 at 18:50

The intuition for this comes from the pictures (although it could also be proved rigorously). If a function is even then it is symmetrical with respect to the y-axis. Therefore when you integrate it you only need to integrate half of it (greater than zero part or less than zero part) and double your answer.

If the function is odd, it is also symmetric with respect to the y-axis expect this time one side is the negative of the other. This means that when add the integral of the two halves together they will cancel out and you will get zero as your answer.

Odd / Even integrals

5 Answers5