The following integral came up in one of my applications:


$\int\_{-1}^1P\_n(x)T\_j(x)T\_k(x)\mathrm{d}x$


where $P\_n(x)$ is a Legendre polynomial, $T\_k(x)$ is a Chebyshev polynomial, and $j$, $k$, and $n$ are nonnegative integers.


I want to ask if there might be a closed-form representation for this integral. I have a feeling it will involve gamma functions and Pochhammer symbols, but I seem to be unable to figure out how to proceed.


Alternatively, since I am aware that Legendre polynomials can be expressed as a linear combination of Chebyshev polynomials, it might be easier to instead simplify the integral


$\int\_{-1}^1T\_n(x)T\_j(x)T\_k(x)\mathrm{d}x$


or in trigonometric form


$\int\_{0}^{\pi}\cos(n\theta)\cos(j\theta)\cos(k\theta)\sin(\theta)\mathrm{d}\theta$


but I do not know of any closed form for this either.


I have already tried looking in Abramowitz and Stegun, the DLMF, Gradshteyn and Ryzhik, and the Wolfram functions site to no avail.


(edit:


I had neglected to exploit the identity


$T\_j(x)T\_k(x)=\frac1{2}\left(T\_{j+k}(x)+T\_{j-k}(x)\right)$


when I first formulated my question. I now amend my question to asking for a closed form for


$\int\_{-1}^1P\_n(x)T\_j(x)\mathrm{d}x$


of which the only fact I know about it is that it is 0 if $j<n$ by virtue of the orthogonality of the Legendre polynomial.)



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:


$\int\_{-1}^1P\_n(x)T\_{n+k}(x)\mathrm{d}x\quad k\geq0$


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](http://functions.wolfram.com/Polynomials/ChebyshevT/06/01/02/01/0009/) for $T\_{n}(x)$ (also in Abramowitz and Stegun as [22.3.6](http://people.math.sfu.ca/%7Ecbm/aands/page_775.htm)), and an integral I derived from a more general form in Gradshteyn and Ryzhik:


$\int\_0^1x^{n+2\rho}P\_n(x)\mathrm{d}x=\frac{\left(2\rho+1\right)\_n}{2^{n+1}\left(\rho+\frac1{2}\right)\_{n+1}}$


where $\left(a\right)\_n$ 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:


$\int\_{-1}^1P\_n(x)T\_{n+2k}(x)\mathrm{d}x=-\frac1{4}\frac{\left(n+2k\right)\Gamma\left(n+k\right)\Gamma\left(k-\frac1{2}\right)}{\Gamma\left(k+1\right)\Gamma\left(n+k+\frac{3}{2}\right)}$


(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.


