I'm doing a problem involving legendre polynomials and I got stuck in this integral:

$$I_k=\int_{-1}^{1} x P_{2k+1}(x)dx $$

Update: Note that the function in the integral is even

If $k=0$, then $I_0=2/3$, but for the other values I don't know how to proceed

For integer $k>0$ I computed the integral in Wolfram(here is for k=1) and I obtained:

$$\int_{-1}^{1} x P_{2k+1}(x)dx=0 $$

Is this an identity for legendre polynomials? and if so, How can I prove it?

  • $\begingroup$ take a look at this, it may help you: here! $\endgroup$ Mar 12, 2016 at 23:18
  • $\begingroup$ The legendre polynomials are designed such that the integral vanishes for polynomials with degree less than or equal to certain number. $\endgroup$
    – user220124
    Mar 12, 2016 at 23:57

1 Answer 1


The Legendre polynomials $P_n(x)$ are solutions of ODE of the form $$\frac{d}{dx}\left[ A(x) \frac{dP_n(x)}{dx} \right] = \lambda_n P_n(x) \quad\text{ with }\quad \begin{cases} A(x) &= 1-x^2,\\ \lambda_n &= -n(n+1) \end{cases}$$

For any two distinct $n, m \ge 0$, notice $$\begin{align} & \frac{d}{dx} \left[ A(x) \left(P_n(x) \frac{dP_m(x)}{dx} - P_m(x) \frac{dP_n(x)}{dx}\right)\right]\\ = & P_n(x) \frac{d}{dx}\left[ A(x) \frac{dP_m(x)}{dx} \right] - P_m(x) \frac{d}{dx}\left[ A(x) \frac{dP_n(x)}{dx} \right]\\ = & (\lambda_m - \lambda_n) P_m(x)P_n(x)\end{align} $$ We have $$\int_{-1}^1 P_m(x)P_n(x) dx = \frac{1}{\lambda_m - \lambda_n} \left[ A(x) \left(P_n(x) \frac{dP_m(x)}{dx} - P_m(x) \frac{dP_n(x)}{dx}\right)\right]_{-1}^1 = 0$$ because $A(x) = 0$ at $x = \pm 1$.

Since $P_1(x) = x$, this means for any $n \ne 1$, we have $$\int_{-1}^1 xP_n(x) dx = 0$$

  • 1
    $\begingroup$ I forgot that $P_1(x)=x$, the problem is just orthogonality of Legendre polynomials. $\endgroup$
    – Keith
    Mar 13, 2016 at 13:58

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