## Monday Math 2: Symmetry and Integrals

Symmetry and integration:

Consider a one-dimensional definite integral of a real function: $\int_{a}^{b}f(x)\,dx$.
Now, let us make a simple transformation: reflection about the y axis. This may be done by means of the substitution $x'=-x$. Then $dx'=-dx$, the limits become –a and –b, and we get $\int_{a}^{b}f(x)\,dx=-\int_{-a}^{-b}f(-x')\,dx'$.
If b>a, then –a>-b, and we should reverse the limits of integration in the transformed coordinates:
$\int_{a}^{b}f(x)\,dx=\int_{-b}^{-a}f(-x')\,dx'$ (eq 1).

Now, let us consider the case of a symmetric interval about the origin, say (-a,a). Then we have
$\int_{-a}^{a}f(x)\,dx=\int_{-a}^{a}f(-x')\,dx'$ (eq. 2).

Now suppose f(x) is an odd function: $f(-x)=-f(x)$. Then
$\int_{-a}^{a}f(x)\,dx=\int_{-a}^{a}f(-x')\,dx'=-\int_{-a}^{a}f(x')\,dx'$
Thus the integral is equal to it’s opposite, and must then be zero; any odd function integrated about an interval symetric about the origin is zero.
[Note that for an even function $f(-x)=f(x)$, equation two is trivially true.]
Here, however we can make an important point: any function whose domain is symmetric about the origin can be considered the sum of an odd function and an even function: $f(x)=o(x)+e(x)$. It can be shown rather simply that these functions are given by
$o(x)=\frac{1}{2}(f(x)-f(-x))$ and
$e(x)=\frac{1}{2}(f(x)+f(-x))$

This means that for f(x),

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

$=\int_{-a}^{a}o(x)\,dx+\int_{-a}^{a}e(x)\,dx$

$=\int_{-a}^{a}e(x)\,dx$

and only the even component contributes to the integral if the limits are symmetric. If the integrand can be easily broken into odd and even terms, then this can be used to simplify some integrals. For example, if $f(x)=x^9-10x^7+3x^5-6x^2+\frac{x}{\sqrt{25-x^2}}+\sin(x)$, then $-6x^2$ is the only even term (all others are odd), and:
$\int_{-2}^{2}f(x)\,dx=-6\int_{-2}^{2}x^2\,dx=-2x^3\|_{x=-2}^2=-32$

Or consider $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(ax+by)\,dx$.
We use the sum identity
$\cos(ax+by)=\cos(ax)\cos(by)-\sin(ax)\sin(by)$; the first term is even in x, the second odd; thus the integral becomes:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(ax+by)\,dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(ax)\cos(by)\,dx$

$\,=\cos(by)\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(ax)\,dx$

$\,=\cos(by)[-\sin(ax)]\|_{x=-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\,=-2\sin(\frac{a\pi}{2})\cos(by)$

Now let us again consider equation 1, but with lower limit 0 and upper limit a>0. Then it becomes $\int_{0}^{a}f(x)\,dx=\int_{-a}^{0}f(-x')\,dx'$.

Now, let us consider when f is an even function: then we have
$\int_{0}^{a}f(x)\,dx=\int_{-a}^{0}f(x)\,dx$. Thus:

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

$\int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx$ (eq. 3)

when f is even.

Combining with our earlier result, we thus have for any function f(x) whose domain is symmetric about x=0,
$\int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}e(x)\,dx$, where $e(x)=\frac{1}{2}(f(x)+f(-x))$ is the even component of f(x).

This result is useful in the early stages of solving the definite integral from Friday’s physics problem. There we had:

$R_{eq}=\frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{\left(1-\cos[2x+y]\right)}{2-\cos(x)-\cos(y)}\,dx\,dy$

The denominator is even in both x and y, and thus we break the numerator into odd and even components as in the second example, and apply our result to the x integral to get

$R_{eq}=\frac{1}{2\pi^2}\int_{-\pi}^{\pi}\int_{0}^{\pi}\frac{\left(1-\cos(2x)\cos(y)\right)}{2-\cos(x)-\cos(y)}\,dx\,dy$
and as the integrand is even in y, we can use equation 3 to get:
$R_{eq}=\frac{1}{\pi^2}\int_{0}^{\pi}\int_{0}^{\pi}\frac{\left(1-\cos(2x)\cos(y)\right)}{2-\cos(x)-\cos(y)}\,dx\,dy$

From there, we can reverse the order of double integration (integrate first in y, then in x), split the integral into two integrals by the two terms in the numerator, and integrate in y using a table of integrals with $2-\cos(x)$ treated as a constant: one can find an integral of the form $\int\frac{du}{p+q\cos(au)}$, which can be used to integrate both
$\frac{1}{[2-\cos(x)]-\cos(y)}$
and
$\frac{\cos(y)}{[2-\cos(x)]-\cos(y)}=[2-\cos(x)]\frac{1}{[2-\cos(x)]-\cos(y)}-1$
with respect to y