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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.)

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The integral of a product of sines and cosines can always be evaluated; one just writes each factor as a sum of complex exponentials and expands. Alternately, repeatedly apply the product-to-sum formula. –  Qiaochu Yuan Jul 27 '10 at 5:32
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Well, playing around with it in Mathematica, there seems to be no trouble evaluating it exactly as long as n, j, and k are given specific values. What I'm curious about is if there is a simpler closed-form. Even better I suppose would be nice papers/references. There also seems to be nothing on this in Szego's or Chihara's books. –  J. M. Jul 27 '10 at 5:41
    
What I described will get you a very simple closed form, which maybe splits into a few cases. Have you tried it? –  Qiaochu Yuan Jul 27 '10 at 5:53
    
@JM: Qiaochu says that your final, purely trigonometric integral, obviously has a simple closed evaluation; see product-to-sum identities on en.wikipedia.org/wiki/List_of_trigonometric_identities. What I can't inderstand is your reduction from the integral involving $P_n$ to Chebyshev's triple: even if the latter has a closed form, this is not sufficient for having a closed form for the former. In fact, I doubt that your starting integral, say $I(n,j,k)$, has a closed form evaluation (and this is not hard by constructing a recursion for $I(n,n,n)$, say, using gfun in MAPLE). –  Wadim Zudilin Jul 27 '10 at 7:18
    
@Qiaochu: Your hint actually became useful to me, but probably not in the way you intended: I had neglected to exploit the following property of the Chebyshev polynomials $T_j(x)T_k(x)=\frac1{2}\left(T_{j+k}(x)+T_{j-k}(x)\right)$ which of course follows from the product to sum identities for the cosine. I will have to amend my question to asking about the closed form of the integral of the product of a Chebyshev and Legendre polynomial. –  J. M. Jul 27 '10 at 9:07

1 Answer 1

up vote 4 down vote accepted

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 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:

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

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.

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"The identities I needed were well-hidden in Abramowitz and Stegun and Gradshteyn and Ryzhik" - as is so often the case. –  Emilio Pisanty Sep 14 '12 at 8:55

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