Closed form for an orthogonal polynomial integral?

It turns out the identities I needed for resolving

$int_{-1}^1P_n(x)T_j(x)mathrm{d}x$

into a closed form was well-hidden in Abramowitz and Stegun and Gradshteyn and Ryzhik.

As I had mentioned in the edit to my original question, the integral is 0 if $j<n$ by virtue of the orthogonality property of the Legendre polynomials.

I now considered the following integral:

$int_{-1}^1P_n(x)T_{n+k}(x)mathrm{d}xquad kgeq0$

To dispose of an elementary case first, I noted that $P_n(x)T_{n+k}(x)$ is an odd function iff $k$ is odd and even iff $k$ is even; the integral is therefore 0 for odd $k$.

The even $k$ case I had solved by making use of two identities: this series representation for $T_{n}(x)$ (also in Abramowitz and Stegun as 22.3.6), and an integral I derived from a more general form in Gradshteyn and Ryzhik:

$int_0^1x^{n+2rho}P_n(x)mathrm{d}x=frac{left(2rho+1right)_n}{2^{n+1}left(rho+frac1{2}right)_{n+1}}$

where $left(aright)_n$ is the Pochhammer symbol. (The identity actually listed in G&R was an integral for a Gegenbauer (ultraspherical) polynomial, of which the Legendre polynomial is a special case.)

I only needed to retain terms in the series greater than or equal to $n$, again due to orthogonality of the Legendre polynomial. Applying the integral formula to each term (with an additional factor of 2 because the integrand is even), and feeding the resulting sum to Mathematica netted the following closed form:

$int_{-1}^1P_n(x)T_{n+2k}(x)mathrm{d}x=-frac1{4}frac{left(n+2kright)Gammaleft(n+kright)Gammaleft(k-frac1{2}right)}{Gammaleft(k+1right)Gammaleft(n+k+frac{3}{2}right)}$

(The original result returned by Mathematica 5.2 had nasty cosecant factors, which I disposed of using the reflection formula for the gamma function).

This can then be applied to the original integral with the three polynomials by exploiting the product-sum identity for the Chebyshev polynomial.