Evaluation of some Integrals arising from Approximation on Spheres^{∗}

C. Odell and J. Levesley

Department of Mathematics, University of Leicester University Rd, Leicester LE1 7RH, UK

January 16, 2001

Abstract

Integrals arising in approximation on the sphere using radial basis functions in the ambient Euclidean space are computed.

1 Introduction

The ability to approximate both accurately and efficiently on the unit sphere S^{d−1} ⊂ IR^{d} is of interest to practitioners in a variety of scientific disciplines, astronomy, meteorology, and geology to name a few. The integrals we compute below arise from the desire to approximate on the sphere using functions defined on the ambient Euclidean space. We consider two families of radial basis functions on IR^{d}: for fixed y ∈ IR^{d}, ϕ(x) = ∥x − y∥^{2α} and ϕ(x) = ∥x − y∥^{2α} log ∥x − y∥, where ∥ · ∥ is the Euclidean norm in d–dimensional space. If we consider the restriction of these functions to the unit sphere we become interested in the following functions on S^{d−1}: for fixed y ∈ S^{d−1}, ϕ(x) = (1−xy)^{α} and ϕ(x) = (1 − xy)^{α} log(1 − xy), where xy denotes the inner product of vectors x, y ∈ S^{d−1}.

The purpose of this paper is to compute the Fourier–Gegenbauer series for the functions ϕ(t) = (1 − t)^{α} and ϕ(t) = (1 − t)^{α} log(1 − t); see Theorems 2 and 3. These expansions may be used to classify the space of functions which can be approximated using the above Euclidean radial basis functions for approximation on S^{d−1}. These integrals and others have been computed, using different techniques, by Baxter and Hubbert [2].

∗Technical Report No. 2001/02, Department of Mathematics & Computer Science, University of Leicester, University Road, Leicester LE1 7RH, England.

Let P_{n}^{λ}(x) be the Gegenbauer polynomial of degree n, orthogonal on [−1, 1] with respect to the weight (1−x^{2})^{λ−1/2}. These polynomials are normalised so that

h_{n}^{λ} := ∫_{−1}^{1} P_{n}^{λ}(x)(1 − x^{2})^{λ−1/2} dx
= \frac{π2^{1−2λ}Γ(n + 2λ)}{n!(n + λ)(Γ(λ))^{2}}  (1)

2 Results

Theorem 1 For α > 0 and λ > −1/2, let

I_{n}^{α,λ} = ∫_{−1}^{1}(1 − x^{2})^{α}P_{n}^{(λ)}(x) dx.

If n is even then

I_{n}^{α,λ} = \frac{Γ(λ + n/2)2^{−2α}πΓ(2α + 1)Γ(λ − α − 1/2 + n/2)}{(n/2)!Γ(λ)Γ(λ − α − 1/2)Γ(α + n/2 + 3/2)Γ(α + 1/2)}.q

However, if n is odd then I_{n}^{α,λ} = 0.

Proof The case of n odd is obvious.

From Szegö [3, p. 82], we have the following three term recurrence relation

nP_{n}^{(λ)}(x) = 2(n + λ − 1)xP_{n−1}^{(λ)}(x) − (n + 2λ − 2)P_{n−2}^{(λ)}(x).

Substituting this recurrence relation into the above integral gives:

I_{n}^{α,λ} = ∫_{−1}^{1}(1 − x^{2})^{α} \frac{2(n + λ − 1)xP_{n−1}^{(λ)}(x)}{n} dx

− ∫_{−1}^{1}(1 − x^{2})^{α} \frac{(n + 2λ − 2)}{n}P_{n−2}^{(λ)}(x) dx

= A − B,

with the obvious definitions of A and B. Then, using integration by parts,

A = ∫_{−1}^{1} \frac{(1 − x^{2})^{α+1}(n + λ − 1)2λP_{n−2}^{(λ+1)} dx}{(α + 1)n},

where we have used the fact that

\frac{d}{dx}P_{n−1}^{(λ)}(x) = 2λP_{n−2}^{(λ+1)}(x);

see [3, Page 82]. Thus

I_{n}^{α,λ} = \frac{2(n + λ − 1)λ}{n(α + 1)}I_{n−2}^{λ+1} − \frac{(n + 2λ − 2)}{n}I_{n−2}^{α,λ}

We now proceed by induction. Assume the truth of the induction hypothesis for the case n−2 for all λ and for all α. Using the recurrence relation just derived one obtains:

I_{n}^{α,λ} = \frac{2(n + λ − 1)λΓ(λ + n/2)2^{−2α}π}{4n(α + 1)Γ(λ + 1)Γ(α + 1/2)}

×\frac{(2α + 2)(2α + 1)Γ(2α + 1)Γ(λ − α − 3/2 + n/2)}{Γ(λ − α − 1/2)Γ(α + n/2 + 3/2)(n/2 − 1)!}

−\frac{(n + 2λ − 2)Γ(λ + n/2 − 1)2^{−2α}πΓ(2α + 1)Γ(λ − α − 3/2 + n/2)}{(n/2 − 1)!nΓ(λ − α − 1/2)Γ(α + n/2 + 1/2)Γ(λ)Γ(α + 1)Γ(α + 1/2)}

= \frac{Γ(λ + n/2 − 1)2^{−2α}πΓ(2α + 1)Γ(λ − α − 3/2 + n/2)}{(n/2 − 1)!Γ(λ)Γ(α + 1/2)Γ(λ − α − 1/2)Γ(α + n/2 + 1/2)}

× [\frac{(n + λ − 1)2λ(λ + n/2 − 1)(2α + 2)(2α + 1)}{4n(α + 1)λ(α + 1/2)(α + n/2 + 1/2)} − \frac{(n + 2λ − 2)}{n}] .

However,

\frac{(n + 2λ − 2)}{n} (\frac{2(n + λ − 1)}{2α + n + 1} − 1) = \frac{n + 2λ − 2}{n/2(2α + n + 1)}(λ − α + n/2 − 3/2).

Therefore

I_{n}^{α,λ} = \frac{Γ(λ + n/2 − 1)2^{−2α}πΓ(2α + 1)Γ(λ − α − 3/2 + n/2)}{(n/2 − 1)!Γ(λ)Γ(α + 1/2)Γ(λ − α − 1/2)Γ(α + n/2 + 1/2)}

×\frac{(λ − α + n/2 − 3/2)(λ + n/2 − 1)}{n/2(α + n/2 − 1)}

= \frac{Γ(λ + n/2)2^{−2α}πΓ(2α + 1)Γ(λ − α − 1/2 + n/2)}{(n/2)!Γ(λ)Γ(α + 1/2)Γ(λ − α − 1/2)Γ(α + n/2 + 3/2)},

as required.

We complete the proof by noting that, from (1) with λ = α + 1/2 and n = 0,

I_{0}^{α,λ} = \frac{2^{−2α}πΓ(2α + 1)}{(Γ(α + 1/2))^{2}(α + 1/2)}.

Theorem 2 Let α > 0 and λ > −1/2. Then

J_{n}^{α,λ} := ∫_{−1}^{1}(1 − x)^{α}(1 − x^{2})^{λ−1/2}P_{n}^{(λ)}(x)
= \frac{(−1)^{n}2^{α+2λ+2n}Γ(α + λ + 1/2)Γ(λ + n + 1/2)Γ(2λ + n)}{n!Γ(2λ + 2n)Γ(α + 2λ + n + 1)Γ(α − n + 1)Γ(λ)}
×Γ(α + 1)Γ(λ + n).

Proof Using integration by parts

J_{n}^{α,λ} = ∫_{−1}^{1} \frac{(1 − x)^{α+1}}{α + 1} \frac{d}{dx} ((1 − x^{2})^{λ−1/2}P_{n}^{(λ)}(x)) dx.

Now, using Rodriguez’ formula (see e.g. [3, Page 82]),

(1 − x^{2})^{λ−1/2}P_{n}^{(λ)}(x) = γ_{n}^{λ} \frac{d^{n}}{dx^{n}} (1 − x^{2})^{n+λ−1/2},  (2)

where

γ_{n}^{λ} = \frac{(−2)^{n}Γ(n + λ)Γ(n + 2λ)}{n!Γ(λ)Γ(2n + 2λ)}.  (3)

Thus

\frac{d}{dx}(1 − x^{2})^{λ−1/2}P_{n}^{(λ)}(x) = γ_{n}^{λ} \frac{d^{n+1}}{dx^{n+1}} (1 − x^{2})^{n+λ−1/2}
= γ_{n}^{λ} \frac{d^{n+1}}{dx^{n+1}} (1 − x^{2})^{n+1+(λ−1)−1/2}.

Using (2) again we have

\frac{d^{n+1}}{dx^{n+1}} (1 − x^{2})^{n+λ−1/2} = \frac{(1 − x^{2})^{λ−3/2}P_{n+1}^{(λ−1)}}{γ_{n−1}^{λ+1}} .

Thus

\frac{d}{dx}(1 − x^{2})^{λ−1/2}P_{n}^{(λ)}(x) = \frac{γ_{n}^{λ}}{γ_{n+1}^{λ−1}} (1 − x^{2})^{λ−3/2}P_{n+1}^{(λ−1)}(x).

Putting this back into the integral gives

J_{n}^{α,λ} = \frac{γ_{n}^{λ}}{γ_{n+1}^{λ−1}} ∫_{−1}^{1} \frac{(1 − x)^{α+1}}{α + 1} (1 − x^{2})^{λ−3/2}P_{n+1}^{(λ−1)}(x)dx.

Therefore

J_{n}^{α,λ} = \frac{γ_{n}^{λ}J_{n+1}^{α+1,λ−1}}{γ_{n−1}^{λ+1}α + 1} ,

so that

J_{n}^{α,λ} = \frac{αγ_{n}^{λ}J_{n−1}^{α−1,λ+1}}{γ_{n−1}^{λ+1}} .

Now, from (3),

\frac{γ_{n}^{λ}}{γ_{n−1}^{λ+1}} = \frac{−2λα}{n(n + 2λ)},

with the result that

J_{n}^{α,λ} = \frac{α(−2λ)}{n(n + 2λ)}J_{n−1}^{α−1,λ+1}

We can now proceed by induction. Assume the result is true for n − 1, and all α > 0 and λ > −1/2:

J_{n−1}^{α−1,λ+1} = \frac{(−1)^{n−1}2^{α−1+2λ+2n}Γ(α + λ + 1/2)Γ(λ + n + 1/2)}{(n − 1)!Γ(2λ + 2n)Γ(α + 2λ + n + 1)}

×\frac{Γ(2λ + n + 1)Γ(α)Γ(λ + n − 1)}{Γ(α − n + 1)Γ(λ − 1)}.

Substituting this into the recurrence relation gives

J_{n}^{α,λ} = \frac{α(−2λ)(−1)^{n−1}2^{α+2λ+2n−1}}{n(n + 2λ)(n − 1)!Γ(2λ + 2n)}

×\frac{Γ(α + λ + 1/2)Γ(λ + n + 1/2)Γ(2λ + n + 1)Γ(α)Γ(λ + n)}{Γ(α + 2λ + n + 1)Γ(α − n + 1)Γ(λ + 1)}

= \frac{(−1)^{n}2^{α+2λ+2n}Γ(α + λ + 1/2)Γ(λ + n + 1/2)Γ(2λ + n)}{n!Γ(2λ + 2n)Γ(α + 2λ + n + 1)Γ(α − n + 1)Γ(λ)}
×Γ(α + 1)Γ(λ + n)

We conclude the induction by noting from [3, Page 68], that for all α > 0 and λ > −1/2:

J_{0}^{α,λ} = \frac{2^{α+2λ}Γ(α + λ + 1/2)Γ(λ + 1/2)}{Γ(α + 2λ + 1)}.

Let

Ψ(x) = \frac{d}{dx} log Γ(x) = \frac{Γ′(x)}{Γ(x)}

be the digamma function (see [1, Chapter 6]). Theorem 3 Let α > 0, λ > −1/2 and

K_{n}^{α,λ} = ∫_{−1}^{1}(1 − x)^{α} log(1 − x)(1 − x^{2})^{λ−1/2}P_{n}^{(λ)}(x) dx.

Then, if α ∈ IN and n > α, K_{n}^{α,λ}

= \frac{(−1)^{α}2^{α+2λ+2n}Γ(α + λ + 1/2)Γ(λ + n + 1/2)Γ(2λ + n)Γ(α + 1)Γ(λ + n)}{n!Γ(2λ + 2n)Γ(α + 2λ + n + 1)(n − α − 1)!Γ(λ)}.

Otherwise, K_{n}^{α,λ}

= J_{n}^{α,λ}(log 2 + Ψ(α + λ + 1/2) + Ψ(α + 1) − Ψ(α + 2λ + n) − Ψ(α − n + 1)).

Proof The key observation in this proof is that

\frac{d}{dα}(1 − x)^{α} = (1 − x)^{α} log(1 − x).

Thus,

K_{n}^{α,λ} = \frac{d}{dα}J_{n}^{α,λ}.

If we write

A_{n}^{λ} = \frac{(−1)^{n}2^{2λ+2}Γ(λ + n + 1/2)Γ(2λ + n)Γ(λ + n)}{n!Γ(2λ + 2n)Γ(λ)}

then

J_{n}^{α,λ} = A_{n}^{λ} \frac{2^{α}Γ(α + λ + 1/2)Γ(α + 1)}{Γ(α + 2λ + n)Γ(α − n + 1)}.

Thus, using the facts that Γ′(x) = Γ(x)Ψ(x) and that

\frac{d}{dx} \frac{1}{Γ(x)} = −\frac{Ψ(x)}{Γ(x)},

we see that

K_{n}^{α,λ}

= A_{n}^{λ} \frac{∂}{∂α} \frac{2^{α}Γ(α + λ + 1/2)Γ(α + 1)}{Γ(α + 2λ + n)Γ(α − n + 1)}
= A_{n}^{λ} \frac{2^{α}Γ(α + λ + 1/2)Γ(α + 1)}{Γ(α + 2λ + n)Γ(α − n + 1)}

(log 2 + Ψ(α + λ + 1/2) + Ψ(α + 1) − Ψ(α + 2λ + n) − Ψ(α − n + 1))),

and the result follows.

Now, when α ∈ IN and n > α, Γ(α − n + 1) is undefined, as is Ψ(α − n + 1). Thus in this case we need to compute the limit

lim_{x→−k} \frac{Ψ(x)}{Γ(x)}

for k ∈ IN. Since Γ possesses a simple pole at −k with residue (−1)^{k}/k!, (see [1, Page 255]), Ψ also possesses a simple pole with residue −1. Thus, for k ∈ IN,

lim_{x→−k} \frac{Ψ(x)}{Γ(x)} = (−1)^{k+1}/k!.

Therefore, for α ∈ IN and n ≥ α,

K_{n}^{α,λ} = (−1)^{n−α}A_{n}^{λ} \frac{2^{α}Γ(α + λ + 1/2)Γ(α + 1)}{Γ(α + 2λ + n)(n − α − 1)!}

Acknowledgements: We would like to thank Leon Greenberg for directing us towards the complex analytic approach to the calculation of the limit in Theorem 3.

References

[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, National Bureau of Standards, Dover Publications, 1964.

[2] B. J. C. Baxter and S. Hub, Radial basis functions for the sphere, preprint 2000.

[3] G. Szegö, Orthogonal Polynomials, American Mathematical Society, 1958.