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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.