Symmetry in Definite Vector Integrals

Why Symmetry Simplifies Vector Integrals

A definite vector integral is computed component by component. This means a symmetry in even one component can simplify that component's integral without affecting the others. On a symmetric interval, odd components cancel between matching points on opposite sides of the center, while even components have matching values and can be integrated over just half the interval.

The key is to check the parity of each component separately. A vector-valued function need not be entirely odd or entirely even: its components may have different symmetries, or a component may contain both odd and even terms.

Central rule: On an interval symmetric about zero, an integrable odd component contributes zero. An integrable even component contributes twice its integral over the positive half of the interval.

Odd and Even Vector-Valued Functions

For a scalar-valued function \(g\), defined on an interval symmetric about zero, \(g\) is even if \(g(-t)=g(t)\) for every \(t\) in the interval. It is odd if \(g(-t)=-g(t)\) for every such \(t\). The interval must include both \(t\) and \(-t\) for these comparisons to make sense.

For a vector-valued function \(\mathbf{F}(t)=\langle f_1(t),\ldots,f_n(t)\rangle\), parity is determined componentwise. The vector function is even if every component is even, and odd if every component is odd. In general, its components can have different parities.

Definition: A vector-valued function \(\mathbf{F}\) is even on a symmetric domain if \(\mathbf{F}(-t)=\mathbf{F}(t)\) for every \(t\) in the domain. It is odd if \(\mathbf{F}(-t)=-\mathbf{F}(t)\). These vector equalities hold exactly when the corresponding equality holds in every component.

For example, \(\langle t,t^2\rangle\) is neither an even nor an odd vector function: its first component is odd and its second is even. That does not prevent using symmetry. The integral of the first component vanishes, and the second can be doubled from the positive half-interval.

The Symmetry Rules

Assume that each component of \(\mathbf{F}\) is continuous on \([-a,a]\), where \(a>0\). (Continuity is a sufficient condition for the component integrals to exist.) The vector integral is defined by integrating each component on the same interval.

\int_{-a}^{a}\mathbf{F}(t)\,dt = \left\langle \int_{-a}^{a} f_1(t)\,dt,\ldots, \int_{-a}^{a} f_n(t)\,dt \right\rangle.

If a component is odd, the contributions at \(t\) and \(-t\) cancel. If it is even, those contributions agree. Consequently, an odd vector function has zero integral, and an even vector function has an integral equal to twice its integral from \(0\) to \(a\).

Theorem: If \(\mathbf{F}\) is continuous on \([-a,a]\), then

if \(\mathbf{F}\) is odd, \(\displaystyle\int_{-a}^{a}\mathbf{F}(t)\,dt=\mathbf{0}\);

if \(\mathbf{F}\) is even, \(\displaystyle\int_{-a}^{a}\mathbf{F}(t)\,dt=2\int_0^a\mathbf{F}(t)\,dt\).

The same rules apply to individual odd or even components when the vector function has mixed parity.

Why the Rules Work

It is enough to justify the scalar rules, because a vector integral is computed componentwise. Split a scalar integral at zero. In the negative half, substitute \(u=-t\), so \(dt=-du\). As \(t\) runs from \(-a\) to \(0\), \(u\) runs from \(a\) to \(0\). Therefore, for a continuous function \(g\),

\int_{-a}^{0}g(t)\,dt = \int_a^0 g(-u)(-du) = \int_0^a g(-u)\,du.

If \(g\) is odd, then \(g(-u)=-g(u)\). Adding the integral over \([0,a]\) gives

\int_{-a}^{a}g(t)\,dt = \int_0^a g(-u)\,du+\int_0^a g(u)\,du = \int_0^a[-g(u)+g(u)]\,du = 0.

If \(g\) is even, then \(g(-u)=g(u)\), and the same split gives

\int_{-a}^{a}g(t)\,dt = \int_0^a g(u)\,du+\int_0^a g(u)\,du = 2\int_0^a g(u)\,du.

Applying these equalities to each component proves the vector statements. In particular, the zero vector conclusion for an odd vector function means every component integral is zero.

Separating Mixed Components

A component need not be purely odd or purely even. For any scalar function \(g\) defined on a symmetric domain, its even and odd parts are

g_{\mathrm{even}}(t)=\frac{g(t)+g(-t)}{2}, \qquad g_{\mathrm{odd}}(t)=\frac{g(t)-g(-t)}{2}.

Adding these expressions gives \(g(t)=g_{\mathrm{even}}(t)+g_{\mathrm{odd}}(t)\). Substitution of \(-t\) shows that the first part is even and the second is odd. Thus, on \([-a,a]\), the odd part contributes zero and only the even part needs to be integrated. For polynomials, this separation is often immediate: terms with even powers are even, and terms with odd powers are odd.

Worked Example: A Vector Function with Different Component Parities

Evaluate \(\displaystyle\int_{-2}^{2}\langle t^3-2t,\;3t^2+1,\;t^4\rangle\,dt\).

The first component is odd, since both \(t^3\) and \(t\) are odd. Its integral is therefore \(0\). The second and third components are even, so each can be doubled from \(0\) to \(2\):

\int_{-2}^{2}\langle t^3-2t,\;3t^2+1,\;t^4\rangle\,dt = \left\langle 0,\; 2\int_0^2(3t^2+1)\,dt,\; 2\int_0^2t^4\,dt \right\rangle.

For the second component,

2\int_0^2(3t^2+1)\,dt = 2\left[t^3+t\right]_0^2 = 2\big((8+2)-(0+0)\big) = 20.

For the third component,

2\int_0^2t^4\,dt = 2\left[\frac{t^5}{5}\right]_0^2 = 2\left(\frac{32}{5}-0\right) = \frac{64}{5}.

Thus the integral is \(\langle 0,20,64/5\rangle\). To check, antiderivatives of the three components are \(t^4/4-t^2\), \(t^3+t\), and \(t^5/5\). Evaluating each from \(-2\) to \(2\) gives \(0\), \(20\), and \(64/5\), respectively, as found.

Worked Example: Removing the Odd Part of Each Component

Evaluate \(\displaystyle\int_{-2}^{2}\langle t^5+2t^2,\;4t^3-t,\;6-t^4\rangle\,dt\).

The first component has an odd part \(t^5\) and an even part \(2t^2\). The second component is entirely odd. The third is entirely even. Therefore,

\int_{-2}^{2}\langle t^5+2t^2,\;4t^3-t,\;6-t^4\rangle\,dt = \left\langle 2\int_0^2 2t^2\,dt,\; 0,\; 2\int_0^2(6-t^4)\,dt \right\rangle.

The first coordinate is

2\int_0^2 2t^2\,dt = 2\left[\frac{2t^3}{3}\right]_0^2 = 2\left(\frac{16}{3}-0\right) = \frac{32}{3}.

The third coordinate is

2\int_0^2(6-t^4)\,dt = 2\left[6t-\frac{t^5}{5}\right]_0^2 = 2\left(12-\frac{32}{5}\right) = 2\left(\frac{60-32}{5}\right) = \frac{56}{5}.

The answer is \(\langle 32/3,0,56/5\rangle\). For a check, antiderivatives are \(t^6/6+2t^3/3\), \(t^4-t^2/2\), and \(6t-t^5/5\). Their endpoint differences on \([-2,2]\) are \(32/3\), \(0\), and \(56/5\), confirming the result.

Products and Trigonometric Parity

Parity also helps with expressions involving products. The product of two even functions is even, the product of two odd functions is even, and the product of an even and an odd function is odd. For instance, if \(p\) is odd and \(q\) is even, then \((pq)(-t)=p(-t)q(-t)=(-p(t))q(t)=-(pq)(t)\), so \(pq\) is odd.

Worked Example: Using Parity in a Trigonometric Vector Integral

Evaluate \(\displaystyle\int_{-\pi}^{\pi}\langle\sin t,\;\cos t+t\sin t,\;t^3\cos t\rangle\,dt\).

The first component \(\sin t\) is odd, so its integral is zero. In the second component, \(\cos t\) is even. Also, \(t\) and \(\sin t\) are both odd, so their product \(t\sin t\) is even. The third component is odd because \(t^3\) is odd and \(\cos t\) is even. Thus only the second coordinate remains:

\int_{-\pi}^{\pi}\langle\sin t,\;\cos t+t\sin t,\;t^3\cos t\rangle\,dt = \left\langle 0,\;2\int_0^\pi(\cos t+t\sin t)\,dt,\;0\right\rangle.

For that coordinate,

2\int_0^\pi(\cos t+t\sin t)\,dt = 2\left[\sin t-t\cos t+\sin t\right]_0^\pi = 2\big((0+\pi+0)-(0-0+0)\big) = 2\pi.

The integral is \(\langle 0,2\pi,0\rangle\). To check directly, antiderivatives of the components are \(-\cos t\), \(2\sin t-t\cos t\), and \(t^3\sin t+3t^2\cos t-6t\sin t-6\cos t\), respectively. Evaluating each antiderivative at \(\pi\) and \(-\pi\) gives \(0\), \(2\pi\), and \(0\), which verifies all three coordinates.

Intervals Symmetric About Another Center

Symmetry need not be centered at zero. The interval \([c-a,c+a]\) is symmetric about \(c\). To use the same rules there, set \(s=t-c\). A component is even about \(c\) when its values at \(c+s\) and \(c-s\) agree; it is odd about \(c\) when those values are negatives. Equivalently, apply the usual parity rules to the function written in terms of \(s\). The center, rather than the number zero itself, is what matters.

Common Pitfalls

Misclassifying a product: It is tempting to call \(t\sin t\) odd because it contains \(t\). That is incorrect: both factors are odd, so their product is even. The direct check is \((-t)\sin(-t)=(-t)(-\sin t)=t\sin t\). Therefore, its integral on a symmetric interval generally does not vanish.

Using symmetry on an interval that is not symmetric: The function \(g(t)=t\) is odd, but it does not follow that its integral over \([-1,2]\) is zero. In fact,

\int_{-1}^{2}t\,dt = \left[\frac{t^2}{2}\right]_{-1}^{2} = 2-\frac{1}{2} = \frac{3}{2}.

The odd-function cancellation rule requires a symmetric interval about the relevant center. Always check the endpoints and the parity of the function on that interval before discarding a component.

Why This Method Matters

Symmetry can reduce the amount of calculation, reveal which components must vanish, and provide a useful check on a result. For example, if an odd component produces a nonzero integral over a symmetric interval, then either its parity was classified incorrectly or an arithmetic step needs review.

The method does not replace componentwise integration. It identifies which component integrals are zero and which can be simplified using symmetry; the remaining integrals still have to be evaluated. The same principle applies whether the components are polynomials, trigonometric expressions, or other integrable functions.

Check Your Understanding

For each question, consider the parity of every component and whether the interval is symmetric about the relevant center.

On \([-3,3]\), which components of \(\langle t^4,t^3,t^2-t\rangle\) have zero integral by symmetry?
Is \(t^2\sin t\) odd or even? What does its parity imply about its integral on \([-a,a]\)?
For a continuous even vector function \(\mathbf{F}\), how can \(\int_{-a}^{a}\mathbf{F}(t)\,dt\) be written using an integral over \([0,a]\)?
Explain why \(\langle t,t^2\rangle\) is neither an odd nor an even vector function, and state what symmetry still tells you about its integral on \([-a,a]\).
Does oddness alone imply that \(\int_{-1}^{2}g(t)\,dt=0\)? Explain the interval condition required for the cancellation rule.