Prove that $\int_{0}^{1} P_{l}(x)dx = \dfrac{P_{l-1} (0)}{l + 1}$

Question

I've tried to prove this through the recurrence relation $$ lP_{l}(x) + P'_{l-1}(x) - xP'_{l}(x) = 0 \rightarrow l\int_{0}^{1} P_{l}(x) dx + \int_{0}^{1} P'_{l-1}(x) dx - x \int_{0}^{1} P'_{l}(x)dx = 0 $$ $$ l\int_{0}^{1} P_{l}(x) dx = P_{l-1}(0) - P_{l-1}(1) + x [P_{l}(1) - P_{l}(0)]$$ But I'm stuck here, I'm not sure how to manipulate the legendre polynomials from here.

Answer 1

You integrated your recurrence incorrectly. So: $$0 = l \int_{0}^{1} P_l(x) \ dx + \int_{0}^{1} P'_{l-1}(x) \ dx - \int_{0}^{1} xP'_l(x) \ dx$$ You can't just take the $x$ out of the last integral. You can simplify it by integrating by parts: $$\int_{0}^{1} x P'_l(x) \ dx = [xP_l(x)]_{0}^{1} - \int_{0}^{1} P_l(x) \ dx = P_l(1) - \int_{0}^{1} P_l(x) \ dx$$ Finally, you get that: $$0 = (l+1) \int_{0}^{1} P_l(x) \ dx + P_{l-1}(1)-P_{l-1}(0)-P_l(1)$$

It is the case that $P_l(1) = 1$ for every $l$. Hence, it follows that: $$\int_{0}^{1} P_l(x) \ dx = \frac{P_{l-1}(0)}{l+1}$$ as was desired.