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The question that follows is the final stage of the previous $3$ stages found here: Stage 1, Stage 2 and Stage 3 which are needed as part of a derivation of the Associated Legendre Functions Normalization Formula: $$\color{blue}{\displaystyle\int_{x=-1}^{1}[{P_{L}}^m(x)]^2\,\mathrm{d}x=\left(\frac{2}{2L+1}\right)\frac{(L+m)!}{(L-m)!}}\tag{1}$$ where for each $m$, the functions $${P_L}^m(x)=\frac{1}{2^LL!}\left(1-x^2\right)^{m/2}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\tag{2}$$ are a set of Associated Legendre functions on $[−1, 1]$.

The question in my textbook asks me to

Derive $(1)$ as follows: Multiply together the two formulas for ${P_{L}}^m(x)$ given by $(2)$ and $${P_L}^{m}(x)=(-1)^m\frac{(L+m)!}{(L-m)!}\frac{1}{2^LL!}\left(1-x^2\right)^{-m/2}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\quad\longleftarrow\text{(Stage 3)}$$ Then integrate by parts repeatedly lowering the $L+m$ derivative and raising the $L−m$ derivative until both are $L$ derivatives. Then use the regular Normalization formula for Legendre Functions: $$\displaystyle\int_{x=-1}^{1}[{P_{L}}(x)]^2\,\mathrm{d}x=\frac{2}{2L+1}\tag{3}$$ where ${P_{L}}(x)$ represents a Legendre function and ${P_{L}}^m(x)$ represents an associated Legendre function.


Start of attempt:

Multiplying together $(2)$ and the Stage $3$ formula yields: $$[{P_{L}}^m(x)]^2=\frac{(-1)^m}{(2^LL!)^2}\frac{(L+m)!}{(L-m)!}\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\tag{4}$$ Multiplying both sides of $(4)$ by $\mathrm{d}x$ and integrating gives: $$\int[{P_{L}}^m(x)]^2\,\mathrm{d}x=\frac{(-1)^m}{(2^LL!)^2}\frac{(L+m)!}{(L-m)!}\color{red}{\int\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\,\mathrm{d}x}\tag{5}$$

Focusing now on the part marked $\color{red}{\mathrm{red}}$ and integrating by parts: $$\int\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\,\mathrm{d}x$$ $$=\left.\frac{\mathrm{d}^{L-m}}{\mathrm{d}x^{L-m}}\left(x^2-1\right)^L\frac{\mathrm{d}^{L+m-1}}{\mathrm{d}x^{L+m-1}}\left(x^2-1\right)^L\right|_{-1}^1-\int\frac{\mathrm{d}^{L+m-1}}{\mathrm{d}x^{L+m-1}}\left(x^2-1\right)^L\,\frac{\mathrm{d}^{L-m+1}}{\mathrm{d}x^{L-m+1}}\left(x^2-1\right)^{L}\mathrm{d}x$$ $$=0-\int\frac{\mathrm{d}^{L+m-1}}{\mathrm{d}x^{L+m-1}}\left(x^2-1\right)^L\,\frac{\mathrm{d}^{L-m+1}}{\mathrm{d}x^{L-m+1}}\left(x^2-1\right)^{L}\mathrm{d}x$$

End of attempt.


I don't know how to take this calculation any further as I have no idea how to evaluate $$\color{#180}{\int\frac{\mathrm{d}^{L+m-1}}{\mathrm{d}x^{L+m-1}}\left(x^2-1\right)^L\,\frac{\mathrm{d}^{L-m+1}}{\mathrm{d}x^{L-m+1}}\left(x^2-1\right)^{L}\mathrm{d}x}$$

Could someone please help me reach equation $(1)$ and finally end this derivation of $$\color{blue}{\displaystyle\int_{x=-1}^{1}[{P_{L}}^m(x)]^2\,\mathrm{d}x=\left(\frac{2}{2L+1}\right)\frac{(L+m)!}{(L-m)!}}\tag{1}$$


EDIT:

The Latex didn't render correctly in the description below the bounty; so I type it here instead:

One user has already given a detailed answer to this question that uses mathematical induction. The problem is that I am finding it hard to understand this type of proof as it is beyond my current level of understanding. I am looking for an answer that doesn't use mathematical induction. Could someone please explain in simple English (where possible) why

$\bbox[yellow]{\displaystyle-\int\dfrac{\mathrm{d}^{L+m-1}}{\mathrm{d}x^{L+m-1}}\left(x^2-1\right)^L\,\dfrac{\mathrm{d}^{L-m+1}}{\mathrm{d}x^{L-m+1}}\left(x^2-1\right)^{L}\mathrm{d}x} $

$\bbox[yellow]{\displaystyle=(-1)^m\int_{-1}^1\frac{\mathrm{d}^L}{\mathrm{d}x^L}(x^2-1)^{L}\frac{\mathrm{d}^L}{\mathrm{d}x^L}(x^2-1)^L\mathrm{d}x}$

I desperately need to understand this as this forms the final part of a $4$ step proof.

Thank you very much.

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    $\begingroup$ I removed the bounty text because it didn't render properly. See the last part of the question body for the details of the bounty. $\endgroup$ – Jyrki Lahtonen May 12 '16 at 5:35
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This answer has more explanatory character since most of the calculcations we need were already done by OP or provided in the answer of @user5713492. For convenience only we introduce the operator notation \begin{align*} D_x:=\frac{d}{dx} \end{align*}

First step: We focus at the $\color{red}{red}$ part of OPs calculation and obtain \begin{align*} \int&D_x^{L-m}\left(x^2-1\right)^LD_x^{L+m}\left(x^2-1\right)^L\,dx\tag{1}\\ &=\left.D_x^{L-m}\left(x^2-1\right)^LD_x^{L+m-1}\left(x^2-1\right)^L\right|_{-1}^1\\ &\qquad-\int D_x^{L-m+1}\left(x^2-1\right)^{L}\,D_x^{L+m-1}\left(x^2-1\right)^Ldx\\ &=0-\int D_x^{L-m+1}\left(x^2-1\right)^{L}\,D_x^{L+m-1}\left(x^2-1\right)^Ldx\\ &=-\int D_x^{L-m+1}\left(x^2-1\right)^{L}\,D_x^{L+m-1}\left(x^2-1\right)^Ldx\tag{2} \end{align*}

Comparing (1) with (2) we have successfully applied the integration by parts method and found \begin{align*} \int&D_x^{L-m}\left(x^2-1\right)^LD_x^{L+m}\left(x^2-1\right)^L\,dx\\ &\qquad=-\int D_x^{L-m+1}\left(x^2-1\right)^{L}\,D_x^{L+m-1}\left(x^2-1\right)^Ldx \end{align*} We see the RHS has a minus sign in front and we have a change of the pair \begin{align*} (L-m,L+m)\quad\rightarrow\quad(L-m+1,L+m-1)\tag{3} \end{align*}

We can read in the question of the textbook to apply integration by parts iteratively. This means to successively apply the method to the already calculated results and so we do integration by parts with (2) again.

Second step:

We obtain \begin{align*} -\int&D_x^{L-m+1}\left(x^2-1\right)^LD_x^{L+m-1}\left(x^2-1\right)^L\,dx\\ &=-\left.D_x^{L-m+1}\left(x^2-1\right)^LD_x^{L+m-2}\left(x^2-1\right)^L\right|_{-1}^1\\ &\qquad+\int D_x^{L-m+2}\left(x^2-1\right)^{L}\,D_x^{L+m-2}\left(x^2-1\right)^Ldx\\ &=0+\int D_x^{L-m+2}\left(x^2-1\right)^{L}\,D_x^{L+m-2}\left(x^2-1\right)^Ldx\\ &=\int D_x^{L-m+2}\left(x^2-1\right)^{L}\,D_x^{L+m-2}\left(x^2-2\right)^Ldx\tag{4} \end{align*} We see the RHS has now a plus sign in front and we have a change of the pair \begin{align*} (L-m+1,L+m-1)\quad\rightarrow\quad(L-m+2,L+m-2) \end{align*}

From this it is plausible how the iteration continues, namely \begin{align*} \int&D_x^{L-m}\left(x^2-1\right)^LD_x^{L+m}\left(x^2-1\right)^L\,dx\\ &=-\int D_x^{L-m+1}\left(x^2-1\right)^{L}\,D_x^{L+m-1}\left(x^2-1\right)^Ldx\\ &=\int D_x^{L-m+2}\left(x^2-1\right)^{L}\,D_x^{L+m-2}\left(x^2-1\right)^Ldx\\ &=-\int D_x^{L-m+3}\left(x^2-1\right)^{L}\,D_x^{L+m-3}\left(x^2-1\right)^Ldx\\ &=\cdots\\ &=(-1)^{m-1}\int D_x^{L-m+(m-1)}\left(x^2-1\right)^{L}\,D_x^{L+m-(m-1)}\left(x^2-1\right)^Ldx\\ &=(-1)^m\int D_x^{L-m+m}\left(x^2-1\right)^{L}\,D_x^{L+m-m}\left(x^2-1\right)^Ldx\\ &\qquad=(-1)^m\int D_x^{L}\left(x^2-1\right)^{L}\,D_x^{L}\left(x^2-1\right)^Ldx\\ \end{align*}

We observe, the sign in front of the integral alternates. It is obviously $(-1)^{\color{blue}{K}}$ corresponding with $D_x^{L-m+\color{blue}{K}}$. When we start with \begin{align*} (L-m,L+m)\quad&\rightarrow\quad(L-m+1,L+m-1)\\ \end{align*} we get after $m$ steps \begin{align*} (L-1,L+1)\quad&\rightarrow\quad(L,L)\\ \end{align*}

General step:

We are not content to argue based upon plausibility. We can rigorously show that everything claimed so far is valid. In fact this was done by the user @user5713492 right at the beginning of his answer (here with $K$ instead of $n$):

The following is valid for $0\leq K <m$ \begin{align*} \int& D_x^{L-m+K}\left(x^2-1\right)^{L}\,D_x^{L+m-K}\left(x^2-1\right)^Ldx\\ &=-\int D_x^{L-m+K+1}\left(x^2-1\right)^{L}\,D_x^{L+m-K-1}\left(x^2-1\right)^Ldx\\ \end{align*} The proof is precisely as we did it before in (2) and (4). Let $0\leq K <m$, then \begin{align*} \int& D_x^{L-m+K}\left(x^2-1\right)^{L}\,D_x^{L+m-K}\left(x^2-1\right)^Ldx\tag{5}\\ &=\left.D_x^{L-m+K}\left(x^2-1\right)^LD_x^{L+m-K-1}\left(x^2-1\right)^L\right|_{-1}^1\\ &\qquad-\int D_x^{L-m+K+1}\left(x^2-1\right)^{L}\,D_x^{L+m-K-1}\left(x^2-1\right)^Ldx\\ &=0-\int D_x^{L-m+K+1}\left(x^2-1\right)^{L}\,D_x^{L+m-K-1}\left(x^2-1\right)^Ldx\\ &=-\int D_x^{L-m+K+1}\left(x^2-1\right)^{L}\,D_x^{L+m-K-1}\left(x^2-1\right)^Ldx\tag{6}\\ \end{align*}

This proof of the general step from (5) to (6) is all we need. Since it is valid for all $0\leq K<m$ we can use it with $K=0$ and obtain \begin{align*} \int& D_x^{L-m}\left(x^2-1\right)^{L}\,D_x^{L+m}\left(x^2-1\right)^Ldx\qquad\quad&(K=0)\\ &=-\int D_x^{L-m+1}\left(x^2-1\right)^{L}\,D_x^{L+m-1}\left(x^2-1\right)^Ldx\qquad\quad&(K=1)\\ \end{align*} then we iteratively set $K=0,1,\ldots,m-1$ to finally obtain after $m$ steps \begin{align*} \int& D_x^{L-m}\left(x^2-1\right)^{L}\,D_x^{L+m}\left(x^2-1\right)^L\,dx\qquad\quad&(K=0)\\ &=-\int D_x^{L-m+1}\left(x^2-1\right)^{L}\,D_x^{L+m-1}\left(x^2-1\right)^L\,dx\qquad\quad&(K=1)\\ &=\cdots\\ &=(-1)^{m-1}\int D_x^{L-1}\left(x^2-1\right)^{L}\,D_x^{L+1}\left(x^2-1\right)^L\,dx\qquad\qquad&(K=m-1)\\ &=(-1)^m\int D_x^{L}\left(x^2-1\right)^{L}\,D_x^{L}\left(x^2-1\right)^L\,dx\\ \end{align*} Since each step alternates the sign of the expression, we have after $m$ steps the sign $(-1)^m$. There are $m+1$ lines altogether and each $K$ has a start line and a result line. The result line is the start line of the next $K$ value. So the result line of the first step is also the first line of the second step.

Note: We do not need any induction based proof. The proof of the general step is sufficient as it was done here corresponding to the first part of @user5713492's answer.

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  • $\begingroup$ @BLAZE: Good to see the answer is helpful and good review. :-) Typos corrected. Please, check again. $\endgroup$ – Markus Scheuer May 13 '16 at 7:17
  • $\begingroup$ Hi, I made a few small changes to your answer for clarity. Please comment if you agree/disagree with it. Congratulations on your bounty, why no one else besides me upvoted your answer is a mystery to me! It should have like $5$ votes or more in my opinion. Thank you :-) $\endgroup$ – BLAZE May 17 '16 at 22:30
  • $\begingroup$ @BLAZE: Many thanks for accepting the answer and granting the bounty. I agree with your changes, everything is fine! :-) Since this answer addresses a very specific aspect, I suppose the interest of the users (and so the number of upvotes) is rather small. Regards, $\endgroup$ – Markus Scheuer May 18 '16 at 9:42
  • $\begingroup$ In my studies of series solutions to ODE's I am now teaching myself Bessel's differential equation and it's solution(s). I was wondering if you would take a look at the following question? Kindest regards :-) $\endgroup$ – BLAZE May 21 '16 at 11:13
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We are trying to evaluate $$\int_{-1}^1\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}dx$$ For $0\le n<m$. Let $$u=\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}$$ Then $$du=\frac{d^{\ell+1-m+n}}{dx^{\ell+1-m+n}}(x^2-1)^{\ell}dx$$ And $$dv=\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}dx$$ So $$v=\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}$$ Integrating by parts, $$\begin{align}\int_{-1}^1u\,dv&=\left.u\,v\right|_{-1}^1-\int_{-1}^1v\,du\\ &=\int_{-1}^1\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}dx\\ &=\left.\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\right|_{-1}^1\\ &-\int_{-1}^1\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+1+n}}{dx^{\ell-m+1+n}}(x^2-1)^{\ell}dx\\ &=-\int_{-1}^1\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+1+n}}{dx^{\ell-m+1+n}}(x^2-1)^{\ell}dx\end{align}$$ Let's see... we were tasked with establishing that for any polynomial $p(x)$ and any $0\le n<k$ there is a polynomial $q(x)$ such that $$\frac{d^n}{dx^n}\left((x-x_0)^kp(x)\right)=(x-x_0)^{k-n}q(x)$$ Well, it's true for $n=0$, and if true for some $0\le n<k-1$, $$\begin{align}\frac{d^{n+1}}{dx^{n+1}}\left((x-x_0)^kp(x)\right)&=\frac d{dx}\frac{d^n}{dx^n}\left((x-x_0)^kp(x)\right)\\ &=\frac d{dx}\left((x-x_0)^{k-n}q(x)\right)\\ &=(k-n)(x-x_0)^{k-n-1}q(x)+(x-x_0)^{k-n}q^{\prime}(x)\\ &=(x-x_0)^{k-n-1}\left((k-n)q(x)+(x-x_0)q^{\prime}(x)\right)\\ &=(x-x_0)^{k-n-1}r(x)\end{align}$$ Where $r(x)$ is a polynomial. Having established the proposition for $n=0$ and having demonstrated its validity for $n+1$ given that it is true for $n<k-1$, it follows by mathematical induction for $0\le n<k$. From this we may conclude that $$\left.\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\right|_{-1}^1=0$$ For $n<m$ because $\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}$ still has at least one of its $\ell$ factors of $(x^2-1)$ left in it.

Using this result, we now are requested to show that for $0\le n\le m$, $$\int_{-1}^1\frac{d^{\ell-m}}{dx^{\ell-m}}(x^2-1)^{\ell}\frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^{\ell}dx\\=(-1)^n\int_{-1}^1\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}dx$$ Clearly it's true for $n=0$, and way back up at the top, we showed that if it's true for $0\le n<m$, then $$\begin{align}&\int_{-1}^1\frac{d^{\ell-m}}{dx^{\ell-m}}(x^2-1)^{\ell}\frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^{\ell}dx\\ &=(-1)^n\int_{-1}^1\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-n}}{dx^{\ell+m-n}}(x^2-1)^{\ell}dx\\ &=(-1)^n\left\{-\int_{-1}^1\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+1+n}}{dx^{\ell-m+1+n}}(x^2-1)^{\ell}dx\right\}\\ &=(-1)^{n+1}\int_{-1}^1\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\frac{d^{\ell-m+1+n}}{dx^{\ell-m+1+n}}(x^2-1)^{\ell}dx\end{align}$$ So it's true for $n+1\le m$, thus the result follows by mathematical induction and in particular for $n=m$, $$\int_{-1}^1\frac{d^{\ell-m}}{dx^{\ell-m}}(x^2-1)^{\ell}\frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^{\ell}dx=(-1)^m\int_{-1}^1\frac{d^{\ell}}{dx^{\ell}}(x^2-1)^{\ell}\frac{d^{\ell}}{dx^{\ell}}(x^2-1)^{\ell}dx$$ Thus we can see that $$\int_{-1}^1\frac{d^{\ell+m}}{dx^{\ell+m}}(x^2-1)^{\ell}\frac{d^{\ell-m}}{dx^{\ell-m}}(x^2-1)^{\ell}dx=(-1)^m\int_{-1}^1\left(2^{\ell}\ell!P_{\ell}(x)\right)^2dx=\frac{2\cdot2^{2\ell}(\ell!)^2}{2\ell+1}$$ Where we have used the Rodrigues formula for the Legendre polynomials $$P_{\ell}(x)=\frac1{2^{\ell}\ell!}\frac{d^{\ell}}{dx^{\ell}}(x^2-1)^{\ell}$$

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  • $\begingroup$ Thank you very much for a great answer. There is one step in your calculation (marked red) for which I do not understand. Could you please elaborate and show intermediate steps for the part marked red? I know this is not always easy to do but I just need to know where the $(-1)^m$ factor comes from, as I don't know what you mean by "it takes at least $L$ derivatives to create a polynomial that is nonzero at either of these endpoints, and we know that , so $L−m$ derivatives is not enough. Then we had such success the first time, we repeated $m−1$ more times". Best Regards, thanks. $\endgroup$ – BLAZE May 9 '16 at 13:25
  • $\begingroup$ Sorry, just noticed one more thing; Could you please show how $\color{red}{(-1)^m\int_{-1}^1\dfrac{d^{L}}{dx^{L}}(x^2-1)^L\dfrac{d^{L}}{dx^{L}}(x^2-1)^Ldx}=(-1)^m\int_{-1}^1\left(2^LL!P_L(x)\right)^2dx$? Can't figure that part out either :( Thanks again! $\endgroup$ – BLAZE May 9 '16 at 14:12
  • $\begingroup$ Hi again, I don't follow your logic for the yellow highlighted part as it was my understanding that $\bbox[#AFA]{\left.\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}\frac{d^{\ell+m-1-n}}{dx^{\ell+m-1-n}}(x^2-1)^{\ell}\right|_{-1}^1=0}$ no matter what the derivatives are, since we just substitute the limits into the expression and it is equal to zero since $([-1]^2-1)^L=0,([1]^2-1)^L=0$. What am I missing here? $\endgroup$ – BLAZE May 10 '16 at 5:42
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    $\begingroup$ I painstakingly proved just above that part that $$\frac{d^{\ell-m+n}}{dx^{\ell-m+n}}(x^2-1)^{\ell}=(x^2-1)^{m-n}q(x)$$ for some polynomial $q(x)$. That is $0$ for $x\in\{-1,1\}$ if $n< m$. $\endgroup$ – user5713492 May 10 '16 at 6:03

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