Page 532 - Electromagnetics

a 2 2
p νm p νn a ν 2

J ν ρ J ν ρ ρ dρ = δ mn 1 − 2 J (p ), ν > −1 (E.23)
νm
ν
0 a a 2 p νm
∞
1

J ν (αx)J ν (βx)xdx = δ(α − β) (E.24)
0 α
a 3

α lm
α ln 2 a 2
j l r j l r r dr = δ mn j n+1 (α ln a) (E.25)
0 a a 2
π

∞
j m (x) j n (x) dx = δ mn , m, n ≥ 0 (E.26)
2n + 1
−∞
J m (p mn ) = 0 (E.27)

J (p ) = 0 (E.28)

m mn
j m (α mn ) = 0 (E.29)

j (α ) = 0 (E.30)
mn
m
Specific examples
sin z
j 0 (z) = (E.31)
z
cos z
n 0 (z) =− (E.32)
z
j
(1) jz
h (z) =− e (E.33)
0
z
j
(2) − jz
h (z) = e (E.34)
0
z
sin z cos z
j 1 (z) = − (E.35)
z 2 z
cos z sin z
n 1 (z) =− − (E.36)
z 2 z

3 1 3
j 2 (z) = − sin z − cos z (E.37)
z 3 z z 2
3 1 3

n 2 (z) = − + cos z − sin z (E.38)
z 3 z z 2
Functional relationships
n
J n (−z) = (−1) J n (z) (E.39)
n
I n (−z) = (−1) I n (z) (E.40)
n
j n (−z) = (−1) j n (z) (E.41)
n n (−z) = (−1) n+1 n n (z) (E.42)
n
J −n (z) = (−1) J n (z) (E.43)

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