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558 CHAPTER 15 Special Functions and Eigenfunction Expansions

These Fourier coefficients are
1 π
A k = sin(x sin(θ))cos(k(θ))dθ = 0 (15.35)
π −π
for k = 0,1,2,···, and

1 π 0 if k is even
B k = sin(x sin(θ))sin(kθ)dθ = (15.36)
π −π 2J 2n−1 (x) if k = 2n − 1 is odd.
Upon adding equations (15.35) and (15.36), we obtain

1 π 1 π
cos(x sin(θ))cos(kθ)dθ + sin(x sin(θ))sin(kθ)dθ
π −π π −π
1 π
= cos(kθ − x sin(θ))dθ
π −π

2J k (x) if k is even
=
2J k (x) if k is odd.

In summary,
1 π
J k (x) = cos(kθ − x sin(θ))dθ
2π −π
for k = 0,1,2,···. Finally, observe that cos(kθ − x sin(θ)) is an even function, so the
integral over [−π,π] is twice the integral over [0,π]. Then
1 π
J n (x) = cos(nθ − x sin(θ))dθ
π 0
for n = 0,1,2,···. These integrals are called Bessel’s integrals.

We will apply Bessel’s integrals to the solution of Kepler’s problem in astronomy. A planet
moves along its elliptical orbit with the sun at one focus. In Figure 15.12, the center of the orbit

Q

P *
P

B A
C S

Elliptical orbit

FIGURE 15.12 Kepler’s problem.

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October 14, 2010 15:20 THM/NEIL Page-558 27410_15_ch15_p505-562