# How does knowing a function as even or odd help in integration ??

So, I am learning Fourier Series and it involves integration. I am not too good at integration.
Now, the resource I use is videos by Dr. Chris Tisdell. In the opening video he says that knowing whether the function as even or odd will greatly simplify the integration process.

So, I have two questions:
1. What is even/odd function?
2. How will that simplify integration ?

## Odd and even functions

Even and odd are terms used to describe particularly well-behaved functions.

An even function is symmetric about the $y$-axis. That is, if we reflect the graph of the function in the $y$-axis, then it doesn’t change. Formally, we say that $\,f$ is even if, for all $x$ and $-x$ in the domain of $\,f$, we have $$f(-x)=f(x)$$ Examples of even functions are $\,f(x)=x^2$ and $\,f(x)=\cos x$.

An odd function has rotational symmetry of order two about the origin. That is, if we rotate the graph of the function $180^\circ$ about the origin, then it doesn’t change. Formally, we say that $\,f$ is odd if, for all $x$ and $-x$ in the domain of $\,f$, we have $$f(-x)=-f(x)$$ Examples of odd functions are $\,f(x)=x^3$ and $\,f(x)=\sin x$.

## Integration

When calculating Fourier series, you often consider integrals of the form $$I=\int_{-a}^a f(x)\,\mathrm{d}x$$ If $\,f$ is odd or even, then sometimes you can make this simpler. We can rewrite that integral in the following way: \begin{align*} I=\int_{-a}^a f(x)\,\mathrm{d}x &= \int_{-a}^0 f(x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \\ &= \int_0^a f(-x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \end{align*} For an even function, we have $f(-x)=f(x)$, whence $$I = 2\int_0^a f(x)\,\mathrm{d}x$$ For an odd function, we have $f(-x)=-f(x)$, whence $$I = \int_0^a (-f(x)+f(x))\,\mathrm{d}x = 0$$ That’s what it means to simplify the integration: the integral of an odd or even function over the interval $[-a,a]$ can be put into a nicer form (and sometimes we can see that it vanishes without ever computing an integral).

• Well most of the $f(x)$ I have seen in Fourier Series so far have been even. Can you gimme an example of $f(x)$ that is not even ?? – An SO User Jul 8 '13 at 8:15
• @LittleChild: Sure. $f(x)=x+1$ isn’t even. Nor are the odd functions. In general, most functions are neither odd nor even; we just happen to deal with particularly nice ones in Fourier series. – alexwlchan Jul 8 '13 at 8:32
• So, integration of all odd functions from $[-L,+L]$ will be zero -- no matter what. Right ? :) – An SO User Jul 8 '13 at 9:12
• @LittleChild: Some oddities happen if the function isn’t defined in $[-L,L]$ or there are asymptotes, but in the vast majority of cases, yes. :) – alexwlchan Jul 8 '13 at 9:29

An even function is the one which satisfies the folloving conditions: 1) for any $x$ from the domain of $f$ the point $-x$ is also in the domain of $f$ 2) $f(-x) = f(x)$ for any $x$ from the domain of $f$.

An odd function is the one that satisfies the conditions 1) for any $x$ from the domain of $f$ the point $-x$ is also in the domain of $f$ 2) $f(-x) = -f(x)$ for any $x$ from the domain of $f$.

The graph of an even function is symmetric with respect to the Y-axis, and the graph of an odd function is symmetric with respect to the origin (0,0). Hope that'll be helpful...

The "how does it help in integration" part is easy: If your integration interval is symmetric around $0$ (which especially includes integration over all of $\mathbb R$), then the integral over any integrable odd function is zero, no exception. Therefore as soon as you've found that your integrand is odd and your integration interval is symmetric, you're done. Also, for general functions, if you can easily split them into even and odd parts, you only have to consider the integral over the even part for symmetric integration intervals.

Another important property is that the product of two even or of two odd functions is even, and the product of an even and an odd function is odd. For example, if $f$ is even, $x\mapsto f(x)\sin(x)$ is odd, and therefore the integral over it is zero (provided it is well defined).

• How about the 2 dimensions? I mean if $\int_{R^2} f=0$, can we say $f$ is odd? – Bob Apr 9 '20 at 8:12
• @BobOakley: You have the direction wrong: In $1$ dimension, if an integrable function is odd, it has integral $0$. The reverse is not true. For example, the function $$f(x)=\begin{cases}1 & -2<x<0\\ -2 & 0<x<1\\ 0 & \text{otherwise}\end{cases}$$ has integral $0$, but clearly is not odd. Now the actual definition of odd functions, namely $f(-x)=-f(x)$, may be carried over to two-dimensional functions, and if you do so, then yes, if a thusly defined odd function is integrable over $\mathbb R^2$, then its integral over $\mathbb R^2$ is $0$. – celtschk Apr 9 '20 at 11:01
• Thanks for your nice example! – Bob Apr 9 '20 at 11:18

An even function $f(x)$ satisfies $f(-x)=f(x)$ for all $x$ in the domain. An odd function satisfies $f(-x) = -f(x)$.

You can also think of these properties as symmetry conditions at the origin. These imply certain things about the function's Fourier series. In particular a periodic even function's Fourier series contains only cosines, and a periodic odd function's Fourier series contains only sines.