# How to get an expression for this integral(Numerically/Analytically)

I have the following problem:

I need to evaluate the integral $$\int_{\cos(\alpha)}^{1} P_l(t)P_{l'}(t) dt$$ for $\alpha \in [0,\pi]$ and each combination of $l$ and $l'$, where $P_l$ is the l-th Legendre polynomial. The thing is that this integral occurs in a double series with truly messy functions $f(l)$ and $g(l')$(unfortunately they are more than two lines long, so no chance to evaluate anything here), such that:

$$\sum_{l=0}^{\infty} \sum_{l'=0}^{\infty} f(l)g(l') \int_{\cos(\alpha)}^{1} P_l(t)P_{l'}(t) dt$$

So my whole calculation depends on this one integral.

Is there any chance to get an expression (analytically or numerically for this integral in the double series)? The problem that numerical integration faces, is that there is no chance to actually evaluate the double series for single values) A foolish approximation would be to consider just a few terms here of this whole double sum, but has somebody a better idea?

$$f(l)=\left(\frac{c_1}{2l+1} \left(P_{l+1}(\cos(\alpha))-P_{l-1}(cos(\alpha))\right)+c_2 \delta_{1l}\right)\int_{R}^{\infty} k_{l}(c_3r) r^2 r^{-(l+1)} dr$$ I should add that I am sure that one can solve this integral analytically, although I have not done it yet, so this might get some sum that also depends on l($k_l$ is the modified spherical bessel function of the second kind)

$$g(l')=c_4 P_{l'}(\cos(\alpha_2))$$

Constant terms are given by: $c_1,...,c_4$

• If $f, g$ are polynomial, a closed form can possibly be given. – Subhajit Jana Jul 30 '13 at 11:27
• they are not(they also contain a lot of legendre functions) but still I would be interested in what idea you had? – user37929 Jul 30 '13 at 11:29
• @Kunnysan: Integral of product is product of integrals because...? – Gerald Edgar Jul 30 '13 at 11:42
• Oh I am sorry, that was a typo. I meant,$$\int_{\cos\alpha}^1\left(\sum_{n=0}^\infty f(n)P_n(t)\right)\left(\sum_{n=0}^\infty g(n)P_n(t)\right).$$ – Subhajit Jana Jul 30 '13 at 11:50
• no problem, I appreciate every effort! Thank you a lot – user37929 Jul 30 '13 at 11:52