## Mauro Fabrizio and Angelo Morro

Print publication date: 2003

Print ISBN-13: 9780198527008

Published to Oxford Scholarship Online: September 2007

DOI: 10.1093/acprof:oso/9780198527008.001.0001

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# A Some properties of Bessel functions

Source:
Electromagnetism of Continuous Media
Publisher:
Oxford University Press

The Bessel function of the first kind, of order n,

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is a solution of Bessel's equation of order n,
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The function Γ, also called Euler's function, is defined by

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for complex-valued z with ℜz > 0.

When n is an integer, positive or negative,

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To prove this property, first consider n > 0 and write

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As r = 0, 1, …, n - 1, the argument of Γ is a negative integer or zero, Γ is then infinite and its inverse vanishes. Hence, letting r = m + n, we have

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Comparison shows that it remains to prove the relation

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(p.614) this follows by the observation that (m + n)! = m!(m + 1) … (m + n) and that, by Γ(r + 1) = rΓ(r), we have Γ(n + m + 1) = (m + n) … (m + 1)Γ(m + 1). Now consider n < 0. Let n = -p, p > 0. We have to prove that
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or
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which is the result just proved.

For all values of n, the two independent solutions of Bessel's equation may be taken to be

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Yn(x) being called the Bessel function of the second kind (or the Neumann function) of order n. The Bessel functions are said to be generated by exp[½x(t - 1/t], in the sense that

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To prove this result, we expand exp[½x(t - 1/t)] in powers of t and show that the coefficient of tn is Jn(x):

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where the index n = r - s has been introduced and r, n are used instead of r, s. Examine the coefficient of tn; we have to prove that it is equal to Jn(x). Consider separately positive and negative values of n and begin with n ≥ 0. Letting p = r - n and observing that (p + n)! = Γ(p + n + 1), we have
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(p.615) Let now n < 0. The coefficient of tn is still the same as for n ≥ 0 but now the requirement s = r - n ≥ 0 is satisfied for all values of r (≥ 0). Then we consider the coefficient of tn and obtain

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which equals Jn(x) and the proof is complete.

If n is an integer then

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This property is easily proved by means of the generating function. For n integral, J-n(x) = (-1)n Jn(x), and hence

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By the change of variables t ↦ φ : t = exp(iφ), we have

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and hence
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If n is even,

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and, if n is odd,
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Accordingly,

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(p.616) whence, equating the real and imaginary parts, we have
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Multiply both sides of these equations by cos nφ, n ≥ 0, and sin nφ, n ≥ 1, respectively. Since

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integration from 0 to π yields
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Since the integrand equals cos(nφ - x sin φ), we have the desired result, for any positive integer n. If n is negative, let n = -p, with p positive. The desired result then takes the form

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The change of variable φ → θ = π - φ yields

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(p.617) This proves that the desired result holds also for negative n and the proof is complete.

The particular case for n = 0, namely,

(A.1)
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is invoked very often. Also, the integral of sin(x sin φ) over [0, 2π] vanishes and hence
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thus providing a further representation of J 0(x).

If a is a complex number, with ℜa > 0, and b ∈ ℝ, then

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The proof starts from the observation that, by (A.1),

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Substitution, interchange of the order of integration and some rearrangement lead to the result.

If a, b ∈ ℝ, a > b then

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To prove this result, substitute the expression for J 0(bx) and interchange the order of integration to have

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Per se, the value at infinity has no meaning. However, if a and/or b are the argument of distributions, we can regard the value as zero. In fact, we might replace ∞ by l and consider integrals of the form $∫ 0 ∞ exp ⁡ ( i a l ) Φ ( a ) d a$ with

(p.618) Φ test function; as l → ∞, the integral vanishes by the Riemann–Lebesgue lemma. Consequently, we have

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The integral on [0, π] is twice that on [0, π/2]. Letting u = cot φ, we have

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Substitution yields the result. If a < b, the integral can be shown to vanish.

Since

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if a > b and equal to zero if a < b, equating the real and imaginary parts of both sides gives
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More involved integral representations hold for the Bessel functions if the order n need not be integer. In this regard, we begin by considering the Euler function Γ and observing that it can be defined, or evaluated, by letting the variable t be complex. We write the integrand as the function

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which has a branch point at t = 0. Upon a cut along the positive real axis, the function In t is single valued; we let arg t = 0 as t approaches the cut from above. Consider the contour γ obtained by following closely the cut from t = ∞ to t = ε, then turning anticlockwise along a circle λε of radius ε around the origin and then following the cut from t = ε to t = ∞. Hence,
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(p.619) Since ℜ > 0, the integral on λε approaches 0 as ε → 0. Letting ε → 0, we have

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The integral on [0, ∞) is Γ (z) and hence we can write

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Now replace z by 1 - z so that

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Moreover, change the variable t → τ, t = -τ, to obtain

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γ′ being the specular contour of γ around the imaginary axis. Hence, we have

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Since

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we obtain
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Accordingly, we express 1/Γ(n + k + 1) by replacing z with n + k + 1 and hence represent Jn(z) in the form

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The summation over k yields exp(-z 2/4τ), whence

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(p.620) Upon the change of variable τ → t, τ = ½zt, we have

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the contour λ being the inverse image of γ′.

For simplicity, restrict attention to z = x ∈ ℝ+, in which case λ is equal to γ′ to within a magnification factor x/2. For convenience, consider a new variable w ∈ C, t = exp w. The contour l in the plane w consists of the line w = -iπ + ξ, ξ ∈ ℝ+, followed by the segment w = iη, η ∈ [-π, π], and the line w = iπ + ξ, ξ ∈ ℝ+. The representation of Jn(x) takes the form

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Upon substitution we have

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The function sin(x sin η - nη) (of η) is odd and hence the integral on [-π, π] vanishes while cos(x sin η - nη) is even. Accordingly, 1/2πi times the integral over l gives

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which is known as Schläfi's generalization of Bessel's integral. For, if n is integer then sin nπ = 0 and we have Bessel's integral
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(p.621) The definition of Yn gives

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Hence, we have the Hankel functions in the form

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and
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As a particular case, examine $H 0 ( 1 )$, namely,

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We prove that $H 0 ( 1 )$ may equivalently be represented as

(A.2)
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To this end, observe that the integral on ℝ is twice the integral on ℝ+ and consider the variable ν = u - iπ/2. It follows that

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(p.622) Now let ν = w + iφ and replace the line from (0 - iπ/2) to (∞ - iπ/2) by the segment w = 0, φ ∈ [-π/2, 0] and the line ℝ+; the ‘segment’ w = ∞, φ ∈ [0, -π/2] gives a vanishing contribution. Hence, we have

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Substitution yields (A.2).