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Please don't be frightened by the length of this, I just wanted to provide ample detail. If you want, you can skip the derivation and go straight to the result at the bottom.

I have

$$ \begin{multline}\int_{-1}^{1}\left(\sum_{n_1-l_1=0}^{\infty}C_{n_1-l_1}^{l_1+1}(x)r^{n_1-l_1}\right)\left(\sum_{n_2-l_2=0}^{\infty}C_{n_2-l_2}^{l_2+1}(x)r^{n_1-l_1}\right)\times\\ \left(\sum_{n_3-l_3=0}^{\infty}C_{n_3-l_3}^{l_3+1}(x)r^{n_3-l_3}\right)(1-x^2)^{(l_1+l_2+l_3+1)/2}dx \end{multline} $$ Applying the identity $\left(\sum_{n=0}^\infty a_nr^n\right)\left(\sum_{n=0}^\infty b_nr^n\right)=\sum_{n=0}^\infty\sum_{k=0}^n a_kb_{n-k}r^n$ twice to the above gives: $$ =\sum_{i=0}^{\infty}\sum_{j=0}^{i}\sum_{k=0}^{i-j}r^{i}\int_{-1}^{1}C_{j}^{l_1+1}(x)C_{k}^{l_2+1}(x)C_{i-j-k}^{l_3+1}(x)(1-x^2)^{(l_1+l_2+l_3+1)/2}dx $$

Using the generating function for the Gegenbauer Polynomials: $$ (1-2xr+r^2)^{-\lambda}=\sum_{n=0}^{\infty}C_{n}^{\lambda}(x)r^n\ \ \ ,\left|r\right|<1 $$ And writing $L=l_1+l_2+l_3$, we get: $$ =\int_{-1}^{1}(1-2xr+r^2)^{-(L+3)}(1-x^2)^{(L+3)/2} $$ $$ =\sqrt{\pi}\frac{\Gamma(\frac{L+3}{2})}{\Gamma(\frac{L+4}{2})}(1-r^2)^{-(L+3)} $$ Rewriting $(1-r^2)^{-(L+3)}=(1-r^2)^{-(l_1+1)}(1-r^2)^{-(l_2+1)}(1-r^2)^{-(l_3+1)}$ and applying the binomial theorem gives: $$ \begin{multline} \sqrt{\pi}\frac{\Gamma\left(\frac{L+3}{2}\right)}{\Gamma\left(\frac{L+4}{2}\right)}\left(\sum_{n=0}^{\infty}\left(-1\right)^n\left(\begin{array}{c}-l_1-1\\n\end{array}\right)r^{2n}\right)\times\\ \left(\sum_{n=0}^{\infty}\left(-1\right)^n\left(\begin{array}{c}-l_2-1\\n\end{array}\right)r^{2n}\right)\left(\sum_{n=0}^{\infty}\left(-1\right)^n\left(\begin{array}{c}-l_3-1\\n\end{array}\right)r^{2n}\right) \end{multline} $$ Applying the product of series identity twice again gives: $$ =\sum_{n=0}^{\infty}\sum_{l=0}^{n}\sum_{m=0}^{n-l}r^{2n}(-1)^n\left(\begin{array}{c}-l_1-1\\l\end{array}\right)\left(\begin{array}{c}-l_2-1\\m\end{array}\right)\left(\begin{array}{c}-l_3-1\\n-l-m\end{array}\right)\sqrt{\pi}\frac{\Gamma\left(\frac{L+3}{2}\right)}{\Gamma\left(\frac{L+4}{2}\right)} $$

Comparing the series in $i,j,k$ to this one in $n,l,m$ I find that $i=2n$ (so that the original series is zero unless i is an even integer). Then comparing the coefficients of $r$ I get: $$ \begin{multline} \sum_{j=0}^{i}\sum_{k=0}^{i-j}\int_{-1}^{1}C_{j}^{l_1+1}(x)C_{k}^{l_2+1}(x)C_{i-j-k}^{l_3+1}(x)(1-x^2)^{(l_1+l_2+l_3+1)/2}dx\\=\sum_{l=0}^{n}\sum_{m=0}^{n-l}(-1)^n\left(\begin{array}{c}-l_1-1\\l\end{array}\right)\left(\begin{array}{c}-l_2-1\\m\end{array}\right)\left(\begin{array}{c}-l_3-1\\n-l-m\end{array}\right)\sqrt{\pi}\frac{\Gamma\left(\frac{L+3}{2}\right)}{\Gamma\left(\frac{L+4}{2}\right)} \end{multline} $$ I then set $$ \begin{align} i=2n\\j=l=n_1-l_1\\k=m=n_2-l_2\\i-j-k=n_3-l_3 \end{align} $$ Whence I get the $\bf{FINAL\ RESULT}$: $$ \int_{-1}^{1}C_{n_1-l_1}^{l_1+1}(x)C_{n_2-l_2}^{l_2+1}(x)C_{n_3-l_3}^{l_3+1}(x)(1-x^2)^{(l_1+l_2+l_3+1)/2}dx= $$ $$ (-1)^{(N-L)/2}\left(\begin{array}{c}-l_1-1\\ n_1-l_1\end{array}\right)\left(\begin{array}{c}-l_2-1\\n_2-l_2\end{array}\right)\left(\begin{array}{c}-l_3-1\\-\frac{1}{2}((n_1-l_1)+(n_2-l_2)-(n_3-l_3))\end{array}\right)\sqrt{\pi}\frac{\Gamma\left(\frac{L+3}{2}\right)}{\Gamma\left(\frac{L+4}{2}\right)} $$

So, $\bf{HERE's}$ my question:
Why is this incorrect? I've tested it and it is not true for all values of $n_1,l_1,n_2,l_2,n_3,l_3$.

I've now spent the better part of a week trying to figure out why this doesn't work, but I can't figure it out. It seems like maybe it has something to do with the way I equate the indices, or perhaps a theorem about the convergence of the product of power series that would make something I did illegal, but I can't find anything so far that would tell me that there's anything wrong with it. In case it's helpful the answer seems to be off by a multiplicative factor that is a function of the indices (a gamma function maybe?).

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this is indeed very scary. –  Patrick Da Silva Jan 20 '12 at 17:43
    
I have done some tests (using Mathematica) on the final equality of double summations and it appears that everything is correct to that point, so the problem is something with my very last step of setting $$\begin{align}j=l=n_1-l_1\\k=m=n_2-l_2\\i-j-k=n_3-l_3\end{align}$$ But I don't see why this assignment is invalid. Any ideas? –  okj Jan 20 '12 at 19:09

1 Answer 1

up vote 2 down vote accepted

The problem is that I set the summands equal to each other $$ \begin{multline*} \sum_{j=0}^{i}\sum_{k=0}^{i-j}\int_{-1}^{1}C_{j}^{l_1+1}(x)C_{k}^{l_2+1}(x)C_{i-j-k}^{l_3+1}(x)(1-x^2)^{(l_1+l_2+l_3+1)/2}dx\\=\sum_{l=0}^{i/2}\sum_{m=0}^{i/2-l}(-1)^{i/2}\left(\begin{array}{c}-l_1-1\\l\end{array}\right)\left(\begin{array}{c}-l_2-1\\m\end{array}\right)\left(\begin{array}{c}-l_3-1\\i/2-l-m\end{array}\right)\sqrt{\pi}\frac{\Gamma\left(\frac{L+3}{2}\right)}{\Gamma\left(\frac{L+4}{2}\right)} \end{multline*} $$ but there is no guarantee that they are equal term by term, and in fact the upper limits don't match since $i=2n$.

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Lol. I'm glad you found it yourself because I don't think someone would've. –  Patrick Da Silva Jan 21 '12 at 18:31

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