# Is this integral correct?

I used substitution and got that: $$\int_0^\pi \sin x \cdot P_n(\cos x ) \, dx=0$$ where $P_n$ is the $n$-th Legendre polynomial.

• What integral you were calculating? – Mhenni Benghorbal Jul 23 '13 at 11:02
• Are you calculating $\int_{-1}^{1}P_0(t)P_n(t)dt$? – Mhenni Benghorbal Jul 23 '13 at 11:13

Recall that Legendre polynomials are defined as orthogonal polynomials on $[-1,1]$ with weight function $w(x)=1$. In other words, we have by definition $$(P_m,P_n)=\int_{-1}^1P_m(x)P_n(x)dx\sim \delta_{mn}.$$ But, since $P_0(x)=1$, your integral is equal to $(P_0,P_n)$ and therefore it vanishes whenever $n>0$. For $n=0$, however, it is equal to $2$.
• Thank you, but why is it equal to 2, I mean if my integral is $(P_0,P_n)$ then it should be equal to 1, as $(P_0,P_0)=1$ is the only one that survives? – user66906 Jul 23 '13 at 11:02
• @Lipschitz You are welcome! In fact, in front of the Kronecker symbol there is a factor $2/(2n+1)$. But here you don't even need to know it, just integrate $1$ from $-1$ to $1$. Of course, the result will reproduce $2/(2\cdot 0+1)$. – Start wearing purple Jul 23 '13 at 11:07
If you look at the Power reduction formulas for the cosine, you see that all monomials in your Legendre polynomial involve only cosines of integer multiples of $x$. These are odd functions on the interval $[0,\pi]$ while the sine function is even, thus their product must be odd and the integral is zero.
• I think the oddness is not quite true: e.g. $\cos2x=1-2\sin^2x$. – Start wearing purple Jul 23 '13 at 11:16