Recipe

Evaluate a Legendre orthogonality or normalization integral

Use when: you need \int_{-1}^1P_n(x)P_k(x)\,dx for two specific degrees n and k

Goal: the integral's exact value, either 0 or 2/(2n+1), without re-deriving Rodrigues' formula from scratch

Taught in Orthogonality of Legendre polynomials, two ways

1. 1

   Write down which two Legendre polynomial degrees, n and k, actually appear in the product under the integral. Everything else in this recipe is decided by comparing these two numbers.

   Example

   For \int_{-1}^1P_1(x)P_2(x)\,dx, the two degrees are n=1 and k=2.

   Why: this fact states a different value, depending only on whether the two degrees match. So naming them is the first thing the rest of the procedure needs.

2. 2

   Check whether n equals k, or not. The theorem gives one value when they match, and a different value when they don't.

   Example

   Take n=1, k=3, both odd. Since n\neq k, the theorem already guarantees \int_{-1}^1P_1(x)P_3(x)\,dx=0. Here P_1(x)=x and P_3(x)=\frac{5x^3-3x}2.

   • n\neq k — the integral is 0 by orthogonality; skip straight to the check in the last step.
   • n=k — the integral is 2/(2n+1) by normalization; continue to the next step to pin down that value.

   ⚡ Shortcut

   When n and k have opposite parity (one even, one odd), you can confirm the zero instantly, without the theorem at all. An odd-degree P_n times an even-degree P_k is an odd function. Any odd function integrates to 0 over the symmetric interval [-1,1]. This shortcut only fires on opposite parity, though. The n=1,k=3 pair above is odd times odd -- an EVEN product. This trick gives no shortcut there. The zero still holds, but only the general theorem explains why.

   Why: this rule's two clauses are orthogonality and normalization. Only one clause ever applies to a given pair, and together they cover every possible pair of degrees.

3. 3

   When n=k, don't integrate by hand from scratch. Plug n into 2/(2n+1) directly. Or integrate as a check, if you want independent confirmation.

   Example

   For n=2, P_2(x)=\frac{3x^2-1}2, the formula gives 2/(2\cdot2+1)=2/5. Direct integration of P_2(x)^2 over [-1,1] confirms the same 2/5.

   Why: re-deriving the normalization by direct integration every time repeats work. That work is already settled, once and for all.

4. 4

   Suppose this integral is really one piece of pulling out a Fourier-Legendre coefficient, c_n=\frac{2n+1}2\int_{-1}^1f(x)P_n(x)\,dx. Then use the SAME n in the normalization factor \frac{2n+1}2 as the P_n actually used in that integral.

   Example

   Pulling out c_3 uses P_3, so the factor is \frac{2(3)+1}2=\frac72, not n=2's factor \frac52.

   Why: the normalization constant genuinely depends on n. A mismatched index scales the coefficient you pull out by the wrong factor entirely.