I know that such question has already been answered but what I am trying to figure out is that whether theres a more elementary way of doing so. I know the proof where we take the help of Legendre differential equation ie $$(1-x^2)y''-2xy'+n(n+1)y=0.$$ But I do not find it intuitive as that's not the proof that will strike you when you have just started dealing with these polynomials.

My Attempt

So my first approach towards this problem of proving $$\int _{-1}^{1} P_m (x) P_n (x) dx=0$$ where $P_i(x)$ is the $i$th degree Legendre polynomial is as follows:

Starting with $$P_m (x) = a \frac{d^m}{dx^m}(x^2-1)^m=a (x-1)^m(x+1)^m,$$ doing similar thing for $n$ and then somehow using Leibniz rule and saying that at $x=-1,1$ since $(x+1),(x-1)$ are $0$ respectively that's why the whole integral sums up to zero.

  • $\begingroup$ Rodrigues formula implies orthogonality by integration by parts, it is pretty straightforward. The non-trivial part is to deduce Rodrigues formula from the differential equation! $\endgroup$ Commented Feb 24, 2018 at 18:29
  • $\begingroup$ So with some effort I can make this proof work right? $\endgroup$ Commented Feb 24, 2018 at 18:30
  • $\begingroup$ And I have just started this course so our professor has omitted the derivation of Rodrigue's formula $\endgroup$ Commented Feb 24, 2018 at 18:31
  • $\begingroup$ Your approach works, sure. In order to prove Rodrigues formula, you may just plug the polynomial defined by $\frac{d^m}{dx^m}$ into the differential equation and check it works. $\endgroup$ Commented Feb 24, 2018 at 18:33

1 Answer 1


So preceding with your idea of using Rodrigue's formula for the Legendre polynomials $P_n (x)$, namely $$P_n (x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2 - 1)^n],$$ I will start by considering the integral $$\int_{-1}^1 f(x) P_n (x) \, dx.$$

Substituting Rodrigues formula for $P_n (x)$ into the above integral followed by integrating by parts $n$ times, recoginising the term $(x^2 - 1)$ repeatedly cancels at either of the end-points, one will be left with $$\int_{-1}^1 f(x) P_n (x) \, dx = \frac{(-1)^n}{2^n n!} \int_{-1}^1 f^{(n)}(x) (x^2 - 1)^n \, dx.\tag1$$

If $f(x) = P_m (x)$ where $m$ is an integer such that $0 \leqslant m < n$, since $P_m (x)$ is a polynomial of degree $m$ which is less than $n$, we have $$f^{(n)}(x) = \frac{d^n}{dx^n} [P_m(x)] = 0,$$ and from (1) we see that $$\int_{-1}^1 P_m (x) P_n (x) \, dx = 0, \quad m \neq n,$$ as required.

  • $\begingroup$ Do we know what happens when $m=n$? Using your neat method. $\endgroup$
    – blahblah
    Commented Dec 13, 2020 at 9:15
  • 1
    $\begingroup$ @Никита Васильев If $m = n$ we cannot find $f^{(m)}(x)$ as it is no longer equal to zero. In this case you can always check out my answer here. $\endgroup$
    – omegadot
    Commented Dec 13, 2020 at 9:39

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