Question

Show in two ways that \(\int_{-1}^{1} P_{2 n+1}(x) d x=0\).

Short answer

The integral of \(P_{2n+1}(x)\) over \([-1, 1]\) is zero because it is an odd function and due to the orthogonality property of Legendre polynomials.

Step-by-step solution

Define the Legendre Polynomial

Legendre polynomials, denoted as \(P_{n}(x)\), are a sequence of orthogonal polynomials. The first few polynomials are \(P_0(x) = 1\), \(P_1(x) = x\), and \(P_2(x) = \frac{1}{2}(3x^2 - 1)\).

Orthogonality Property

One key property of Legendre polynomials is that they are orthogonal over the interval \([-1, 1]\) with respect to the weight function \w(x) = 1\. Mathematically, this means: \int_{-1}^{1} P_{m}(x) P_{n}(x) d x = 0\ for \(m eq n\).

Apply the Orthogonality

Since \(P_{2n+1}(x)\) is an odd function for odd \(2n+1\), and its product with itself remains odd, the integral over \([-1, 1]\) will be zero.Given this orthogonality: \int_{-1}^{1} P_{2n+1}(x) \cdot P_0(x) d x = 0\, because \(P_0(x) = 1\) is an even function.

Use Symmetry of the Integrand

For an alternative approach, consider that \(P_{2n+1}(x)\) is an odd function. Specifically, \(P_{2n+1}(-x) = -P_{2n+1}(x)\). Integrating an odd function over a symmetric interval about the origin results in zero: \int_{-a}^{a} f(x) dx = 0,\ if \(f(x)\) is odd.

Consider the Interval

Since the interval \([-1, 1]\) is symmetric around 0, and \(P_{2n+1}(x)\) is an odd function: \int_{-1}^{1} P_{2n+1}(x) dx = 0\.

Key concepts

Orthogonal Polynomials

Legendre polynomials are a type of orthogonal polynomials. Orthogonal polynomials have the special property that the integral of their product over a specific interval is zero, provided the polynomials are different. This means that for any two different Legendre polynomials, say, \(P_{m}(x)\) and \(P_{n}(x)\), the integral of their product over the interval \([-1, 1]\) equals zero:
\[ \int_{-1}^{1} P_{m}(x) P_{n}(x) dx = 0, \quad \text{for} \; m e n. \]
This property is incredibly useful for solving problems involving Legendre polynomials because it simplifies the process of integrating functions that are composed of these polynomials. For example, \(P_0(x) = 1\), \(P_1(x) = x\), and \(P_2(x) = \frac{1}{2}(3x^2 - 1)\) are all Legendre polynomials. If you wanted to integrate \(P_0(x)\) times \(P_1(x)\) over \([-1, 1]\), using orthogonality, you know the result is zero without doing the actual math! This makes it easier to find solutions to many problems involving these polynomials.

Odd Function Symmetry

An odd function \(f(x)\) has the property \(f(-x) = -f(x)\). When an odd function is integrated over an interval symmetric around zero, the result is always zero. This is due to the areas above and below the x-axis canceling each other out. Let's take \(P_{2n+1}(x)\), which is an odd Legendre polynomial. This means:
\[ P_{2n+1}(-x) = -P_{2n+1}(x). \]
Now, consider the integral of this odd function over the symmetric interval \([-1, 1]\):
\[ \int_{-1}^{1} P_{2n+1}(x) dx. \]
Due to the symmetry of the interval and the odd nature of \(P_{2n+1}(x)\), each value of \(P_{2n+1}(x)\) on one side of the origin will have an equal and opposite value on the other side. Therefore, the integral of \(P_{2n+1}(x)\) over \([-1, 1]\) is zero. This property can significantly simplify many problems in mathematics and physics.

Integral Properties

Integrals involving polynomials can often be simplified by understanding certain properties of the integrand. One key property is orthogonality, as mentioned earlier, which helps simplify the integration of products of different polynomials. Another important property is parity, which classifies functions as even or odd.
An even function \(f(x)\) satisfies \(f(-x) = f(x)\), while an odd function satisfies \(f(-x) = -f(x)\). When integrating an odd function over a symmetric interval, the result will always be zero. Thus, for the odd Legendre polynomial \(P_{2n+1}(x)\), the integral
\[ \int_{-1}^{1} P_{2n+1}(x) dx = 0,\]
because \(P_{2n+1}(x)\) is odd, and \([-1, 1]\) is symmetric. Another useful trick involves decomposing the integral of a product of functions into simpler parts, especially when dealing with polynomials. Understanding these integral properties greatly helps in tackling exercises and problems involving Legendre polynomials effectively.