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Use Rodrigues’ formula to show that $I_l = \int_{-1}^1 P_l(x)P_l(x)dx = \frac{2}{2l +1}$.

Using Rodrigue's formula, we have $$I_l = \frac{1}{2^{2l}(l!)^2} \int_{-1}^1 \left[\frac{d^l(x^2-1)^l}{dx^l}\right]\left[\frac{d^l(x^2-1)^l}{dx^l}\right]dx$$

Now, it is in these below steps I'm having trouble with. Repeated integration by parts, with all boundary terms vanishing, reduces this to

$$I_l = \frac{(-1)^l}{2^{2l}(l!)^2} \int_{-1}^1 (x^2-1)^l \frac{d^{2l}(x^2-1)^l}{dx^{2l}}dx$$ or, $$I_l = \frac{(2l)!}{2^{2l}(l!)^2} \int_{-1}^1 (1 - x^2)^l dx$$

I could not find any literature that explains the above steps. Thank You.

P.S. - This was my first time writing in Mathjax.

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  • $\begingroup$ You should explain exactly where you're having difficulties. $\endgroup$ – Cameron Williams Jun 19 '17 at 3:38
  • $\begingroup$ I could not understand, how we got to the second last step using repeated integration by parts $\endgroup$ – AgentRock Jun 19 '17 at 3:39
  • $\begingroup$ one way is to put $x=\sin(t)$ and remember the beta/gamma function. Sorry, i read wrong, your problem was upper. $\endgroup$ – Veridian Dynamics Jun 19 '17 at 4:41
  • $\begingroup$ Crossposted to physics.stackexchange.com/q/340403/2451 $\endgroup$ – Qmechanic Jun 20 '17 at 10:36
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The proof works by mathematical induction, which is a basic mathematical technique.

Start with the integral you gave, and integrate by parts once: \begin{align} I_l & = \frac{1}{2^{2l}(l!)^2} \int_{-1}^1 \left[\frac{d^l(x^2-1)^l}{dx^l}\right]\left[\frac{d^l(x^2-1)^l}{dx^l}\right]dx \\ & = \frac{1}{2^{2l}(l!)^2} \left[ \left. \left[\frac{d^{l-1}(x^2-1)^l}{dx^{l-1}}\right]\left[\frac{d^{l}(x^2-1)^l}{dx^{l}}\right] \right|_{-1}^1 - \int_{-1}^1 \left[\frac{d^{l-1}(x^2-1)^l}{dx^{l-1}}\right]\left[\frac{d^{l+1}(x^2-1)^l}{dx^{l+1}}\right]dx \right] \\ & = \frac{(-1)}{2^{2l}(l!)^2} \int_{-1}^1 \left[\frac{d^{l-1}(x^2-1)^l}{dx^{l-1}}\right]\left[\frac{d^{l+1}(x^2-1)^l}{dx^{l+1}}\right]dx , \end{align} where the boundary terms vanish, because $(x^2-1)^l = (x+1)^l (x-1)^l$ has $l$-fold zeros at both endpoints, and the $(l-1)$-fold derivative leaves one zero on each end.

This suggests the intermediate result for the inductive step: the claim that \begin{align} I_l & = \frac{(-1)^k}{2^{2l}(l!)^2} \int_{-1}^1 \left[\frac{d^{l-k}(x^2-1)^l}{dx^{l-k}}\right]\left[\frac{d^{l+k}(x^2-1)^l}{dx^{l+k}}\right]dx , \end{align} for all $k=0,1,\ldots, l$. To prove this claim via induction, we assume that it's true for some $k$, and then we integrate by parts again: \begin{align} I_l & = \frac{(-1)^k}{2^{2l}(l!)^2} \int_{-1}^1 \left[\frac{d^{l-k}(x^2-1)^l}{dx^{l-k}}\right]\left[\frac{d^{l+k}(x^2-1)^l}{dx^{l+k}}\right]dx \\ & = \frac{(-1)^{k}}{2^{2l}(l!)^2} \left[ \left. \left[\frac{d^{l-k-1}(x^2-1)^l}{dx^{l-k-1}}\right]\left[\frac{d^{l+k}(x^2-1)^l}{dx^{l+k}}\right] \right|_{-1}^1 - \int_{-1}^1 \left[\frac{d^{l-k-1}(x^2-1)^l}{dx^{l-k-1}}\right]\left[\frac{d^{l+k+1}(x^2-1)^l}{dx^{l+k+1}}\right]dx \right] \\ & = \frac{(-1)^{k+1}}{2^{2l}(l!)^2} \int_{-1}^1 \left[\frac{d^{l-k-1}(x^2-1)^l}{dx^{l-k-1}}\right]\left[\frac{d^{l+k+1}(x^2-1)^l}{dx^{l+k+1}}\right]dx , \end{align} where the boundary terms vanish for the same reason as above. This is exactly the same claim with $k$ replaced by $k+1$, which completes the proof by induction.

The desired result, $$I_l = \frac{(-1)^l}{2^{2l}(l!)^2} \int_{-1}^1 (x^2-1)^l \frac{d^{2l}(x^2-1)^l}{dx^{2l}}dx,$$ then follows as the special case $k=l$. Finally, to get to $$ I_l = \frac{(2l)!}{2^{2l}(l!)^2} \int_{-1}^1 (1 - x^2)^l dx $$ you simply need to realize that $(x^2-1)^l$ is a polynomial of degree $2l$ with leading coefficient $1$, and that under a $2l$-fold derivative the only thing that will survive is $(2l)!$ times that leading coefficient.

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  • $\begingroup$ Woww. Thank you so much. Everything got cleared $\endgroup$ – AgentRock Jun 20 '17 at 16:49
  • $\begingroup$ Shouldn't the first term when you integrated by parts be $d^(l+1)$, $\endgroup$ – AgentRock Jun 20 '17 at 16:55
  • $\begingroup$ No, the combination $d^{l-1}(·)d^l(·)$ is correct. $\endgroup$ – E.P. Jun 20 '17 at 17:05
  • $\begingroup$ Sorry for noob question, but Is it that when we integrate a l fold derivative, we reduce its order by 1? $\endgroup$ – AgentRock Jun 20 '17 at 17:10
  • $\begingroup$ This is standard integration by parts, of the form $\int u\mathrm dv = uv -\int v\mathrm du$, where $u$ is an $l$-fold derivative ($l+k$ in the induction step) and $v$ is an $(l-1)$-fold one ($l-k-1$ in the induction step). For more details, see your favourite calculus textbook. $\endgroup$ – E.P. Jun 20 '17 at 17:13

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