Question

Find the value of the integral
\[\int_{-1}^{1} x\,P_n(x)\,P_{n-1}(x)\,dx\]

• A. Zero
• B. \(\dfrac{2}{2n+1}\)
• C. \(\dfrac{1}{4n^2-1}\)
• D. \(\dfrac{2n}{4n^2-1}\)

Hint:

Write \(x\,P_{n-1}\) using the Legendre recurrence, then use orthogonality and \(\int_{-1}^{1}P_n^2\,dx = \tfrac{2}{2n+1}\).

Answer 1

The Correct Option is D

Step 1: Use the Legendre recurrence relation to express \(x\,P_{n-1}(x)\) in terms of neighbouring polynomials:
\[(2n-1)\,x\,P_{n-1}(x) = n\,P_n(x) + (n-1)\,P_{n-2}(x)\]
so that
\[x\,P_{n-1}(x) = \frac{n\,P_n(x) + (n-1)\,P_{n-2}(x)}{2n-1}\]
Step 2: Multiply by \(P_n(x)\) and integrate over \([-1,1]\):
\[\int_{-1}^{1} x\,P_n P_{n-1}\,dx = \frac{1}{2n-1}\left[n\int_{-1}^{1}P_n^2\,dx + (n-1)\int_{-1}^{1}P_n P_{n-2}\,dx\right]\]
Step 3: Apply orthogonality: \(\int_{-1}^{1}P_n P_{n-2}\,dx = 0\) and \(\int_{-1}^{1}P_n^2\,dx = \dfrac{2}{2n+1}\).
\[\int_{-1}^{1} x\,P_n P_{n-1}\,dx = \frac{n}{2n-1}\cdot\frac{2}{2n+1} = \frac{2n}{(2n-1)(2n+1)}\]
Step 4: Since \((2n-1)(2n+1) = 4n^2-1\),
\[\boxed{\int_{-1}^{1} x\,P_n P_{n-1}\,dx = \frac{2n}{4n^2-1}}\]