Bessel functions
The Bessel function Jν (z) of the first kind of order ν is defined by
(z/2)ν
Jν (z) =
0 F1
Γ(ν + 1)
−
z2
;−
ν+1
4
=
∞
z ν X
2
k=0
z 2k
(−1)k
.
Γ(ν + k + 1) k! 2
For ν ≥ 0 this is a solution of the Bessel differential equation
z 2 y 00 (z) + zy 0 (z) + z 2 − ν 2 y(z) = 0, ν ≥ 0.
(1)
(2)
For ν ∈
/ {0, 1, 2, . . .} we have that J−ν (z) is a second solution of the differential equation (2)
and the two solutions Jν (z) and J−ν (z) are clearly linearly independent.
For ν = n ∈ {0, 1, 2, . . .} we have
J−n (z) =
∞
z −n X
2
k=0
since
∞
z 2k z −n X
z 2k
(−1)k
(−1)k
=
,
Γ(−n + k + 1) k! 2
2
Γ(−n + k + 1) k! 2
k=n
1
=0
Γ(−n + k + 1)
for k = 0, 1, 2, . . . , n − 1.
This implies that
∞
z −n X
J−n (z) =
2
= (−1)n
k=n
∞
z 2k z −n X
z 2(n+k)
(−1)k
(−1)n+k
=
Γ(−n + k + 1) k! 2
2
Γ(k + 1) (n + k)! 2
k=0
∞
z n X
2
k=0
(−1)k
z 2k
Γ(n + k + 1) k! 2
= (−1)n Jn (z).
This implies that Jn (z) and J−n (z) are linearly dependent for n ∈ {0, 1, 2, . . .}.
A second linearly independent solution can be found as follows. Since (−1)n = cos nπ, we see
that Jν (z) cos νπ − J−ν (z) is a solution of the differential equation (2) which vanishes when
ν = n ∈ {0, 1, 2, . . .}. Now we define
Yν (z) :=
Jν (z) cos νπ − J−ν (z)
,
sin νπ
(3)
where the case that ν = n ∈ {0, 1, 2, . . .} should be regarded as a limit case. By l’Hopital’s
rule we have
(−1)n ∂J−ν (z)
1 ∂Jν (z)
Yn (z) = lim Yν (z) =
−
.
ν→n
π
∂ν
π
∂ν
ν=n
ν=n
This implies that Y−n (z) = (−1)n Yn (z) for n ∈ {0, 1, 2, . . .}.
The function Yν (z) is called the Bessel function of the second kind of order ν.
Using the definition (1) we find that
∂Jν (z)
∂ν
= Jn (z) ln
ν=n
z 2
−
∞
z n X
(−1)k ψ(n + k + 1) z 2k
2
1
k=0
(n + k)! k!
2
,
where
ψ(z) =
d
Γ0 (z)
ln Γ(z) =
.
dz
Γ(z)
For ν ∈
/ {0, 1, 2, . . .} we have
∞
z z −ν X (−1)k ψ(−ν + k + 1) z 2k
∂J−ν (z)
.
= −Jν (z) ln
+
∂ν
2
2
Γ(−ν + k + 1) k!
2
k=0
Now we use
lim (z + n)Γ(z) =
z→−n
and
(−1)n
n!
Γ(z) ∼
=⇒
(−1)n
(z + n) n!
for z → −n
n
(−1)
− (z+n)
2 n!
1
Γ0 (z)
∼ (−1)n
=−
ψ(z) =
Γ(z)
z+n
for z → −n.
(z+n) n!
This implies that
lim
z→−n
ψ(z)
= (−1)n+1 n!
Γ(z)
for
n = 0, 1, 2, . . . .
Hence
lim
ν→n
∞
X
(−1)k ψ(−ν + k + 1) z 2k
k=0
Γ(−ν + k + 1) k!
n
= (−1)
= (−1)n
n−1
X
k=0
n−1
X
2
∞
(n − k − 1)! z 2k X (−1)k ψ(−n + k + 1) z 2k
+
k!
2
Γ(−n + k + 1) k!
2
k=n
(n − k − 1)! z 2k
k!
k=0
2
+ (−1)n
∞
z 2n X
(−1)k ψ(k + 1) z 2k
.
2
Γ(k + 1) (n + k)! 2
k=0
This implies that
∂J−ν (z)
∂ν
= −J−n (z) ln
ν=n
z 2
+ (−1)n
+ (−1)n
z −n n−1
X (n − k − 1)! z 2k
2
k=0
k!
∞
z n X
(−1)k ψ(k + 1) z 2k
2
k=0
(n + k)! k!
2
2
.
Finally we use the fact that J−n (z) = (−1)n Jn (z) to conclude that
Yn (z) =
z 1 z −n n−1
X (n − k − 1)! z 2k
2
Jn (z) ln
−
π
2
π 2
k!
2
k=0
∞
z 2k
1 z n X (−1)k
−
[ψ(n + k + 1) + ψ(k + 1)]
π 2
(n + k)! k!
2
k=0
for n ∈ {0, 1, 2, . . .}. Compare with the theory of Frobenius for linear second differential
equations.
2
In the theory of second order linear differential equations of the form
y 00 + p(z)y 0 + q(z)y = 0,
two solutions y1 and y2 are linearly independent if and only if
y1 (z) y2 (z) W (y1 , y2 )(z) := = y1 (z)y20 (z) − y10 (z)y2 (z) 6= 0.
y10 (z) y20 (z) This determinant is called the Wronskian of the solutions y1 and y2 .
It can (easily) be shown that this determinant of Wronski satisfies the differential equation
W 0 (z) + p(z)W (z) = 0.
This result is called Abel’s theorem or the theorem of Abel-Liouville. In the case of the Bessel
differential equation we have p(z) = 1/z, which implies that
1
W 0 (z) + W (z) = 0
z
=⇒
W (y1 , y2 )(z) =
c
z
for some constant c. Now we have
Theorem 1.
W (Jν , J−ν )(z) = −
2 sin νπ
πz
and
W (Jν , Yν )(z) =
2
.
πz
(4)
For ν ≥ 0 this implies that Jν (z) and J−ν (z) are linearly independent if ν ∈
/ {0, 1, 2, . . .} and
that Jν (z) and Yν (z) are linearly independent for all ν ≥ 0.
Proof. Note that
W (Jν , Yν )(z) = Jν (z)Yν0 (z) − Jν0 (z)Yν (z)
0 (z)
J 0 (z) cos νπ − J−ν
Jν (z) cos νπ − J−ν (z)
= Jν (z) ν
− Jν0 (z)
sin νπ
sin νπ
0 (z) − J 0 (z)J (z)
Jν (z)J−ν
W (Jν , J−ν )(z)
−ν
ν
= −
=−
.
sin νπ
sin νπ
Now we use the definition (1) to obtain
Jν (z) =
∞
X
k=0
(−1)k
z ν+2k
· ν+2k
Γ(ν + k + 1) k! 2
=⇒
Jν0 (z) =
∞
X
(−1)k (ν + 2k) z ν+2k−1
·
Γ(ν + k + 1) k! 2ν+2k
k=0
and
J−ν (z) =
∞
X
m=0
z −ν+2m
(−1)m
· −ν+2m
Γ(−ν + m + 1) m! 2
=⇒
0
J−ν
(z)
∞
X
(−1)m (−ν + 2m) z −ν+2m−1
=
·
.
Γ(−ν + m + 1) m! 2−ν+2m
m=0
3
Hence we have
0
z W (Jν , J−ν )(z) = z Jν (z)J−ν
(z) − Jν0 (z)J−ν (z)
∞ X
∞
X
z 2k+2m
(−1)k+m (−ν + 2m)
· 2k+2m
=
Γ(ν + k + 1)Γ(−ν + m + 1) k! m! 2
k=0 m=0
−
= −
∞ X
∞
X
k=0 m=0
∞
∞
XX
k=0 m=0
(−1)k+m (ν + 2k)
z 2k+2m
· 2k+2m
Γ(ν + k + 1)Γ(−ν + m + 1) k! m! 2
(−1)k+m (2ν + 2k − 2m)
z 2k+2m
· 2k+2m .
Γ(ν + k + 1)Γ(−ν + m + 1) k! m! 2
This implies that
lim z W (Jν , J−ν )(z) = −
z→0
2ν
2ν
2 sin νπ
=−
=−
.
Γ(ν + 1)Γ(−ν + 1)
νΓ(ν)Γ(1 − ν)
π
Using the definition (1) we obtain
d ν
[z Jν (z)] =
dz
=
∞
∞
X
d X
(−1)k
z 2ν+2k
(−1)k (2ν + 2k) z 2ν+2k−1
· ν+2k =
·
dz
Γ(ν + k + 1) k! 2
Γ(ν + k + 1) k!
2ν+2k
k=0
∞
X
k=0
(−1)k
k=0
Γ(ν + k) k!
·
z 2ν+2k−1
2ν+2k−1
= z ν Jν−1 (z).
Hence we have
d ν
[z Jν (z)] = z ν Jν−1 (z)
dz
⇐⇒
zJν0 (z) + νJν (z) = zJν−1 (z).
(5)
Similarly we have
d −ν
z Jν (z) =
dz
=
∞
∞
X
d X
(−1)k
z 2k
(−1)k 2k
z 2k−1
· ν+2k =
· ν+2k
dz
Γ(ν + k + 1) k! 2
Γ(ν + k + 1) k! 2
k=0
∞
X
k=0
k=0
(−1)k+1
z 2k+1
·
= −z −ν Jν+1 (z).
Γ(ν + k + 2) k! 2ν+2k+1
Hence we have
d −ν
z Jν (z) = −z −ν Jν+1 (z)
dz
⇐⇒
zJν0 (z) − νJν (z) = −zJν+1 (z).
Elimination of Jν0 (z) from (5) and (6) gives
Jν−1 (z) + Jν+1 (z) =
2ν
Jν (z)
z
and elimination of Jν (z) from (5) and (6) gives
Jν−1 (z) − Jν+1 (z) = 2Jν0 (z).
4
(6)
Special cases
For ν = 1/2 we have from the definition (1) by using Legendre’s duplication formula for the
gamma function
r
r
∞
∞
x 2k r 2x X
x X
2
(−1)k
(−1)k
2k
J1/2 (x) =
=
x =
sin x, x > 0
2
Γ(k + 3/2) k! 2
π
(2k + 1)!
πx
k=0
k=0
and for ν = −1/2 we have
r
r
∞
∞
x 2k r 2 X
2 X
2
(−1)k
(−1)k 2k
=
x =
cos x,
J−1/2 (x) =
x
Γ(k + 1/2) k! 2
πx
(2k)!
πx
k=0
x > 0.
k=0
Note that the definition (3) implies that
r
r
2
2
Y1/2 (x) = −J−1/2 (x) = −
cos x and Y−1/2 (x) = J1/2 (x) =
sin x,
πx
πx
x > 0.
Integral representations
First we will prove
Theorem 2.
z ν Z 1
1
√
eizt (1 − t2 )ν−1/2 dt,
Jν (z) =
Γ(ν + 1/2) π 2
−1
Proof. We start with
Z 1
izt
2 ν−1/2
e (1 − t )
dt =
−1
Z
∞
X
(iz)n
n=0
n!
Re ν > −1/2.
1
tn (1 − t2 )ν−1/2 dt.
−1
Note that the latter integral vanishes when n is odd. For n = 2k we obtain using t2 = u
Z 1
Z 1
Z 1
du
t2k (1 − t2 )ν−1/2 dt = 2
uk (1 − u)ν−1/2 √ =
uk−1/2 (1 − u)ν−1/2 du
2
u
0
−1
0
Γ(k + 1/2)Γ(ν + 1/2)
.
= B(k + 1/2, ν + 1/2) =
Γ(ν + k + 1)
Now we use Legendre’s duplication formula to find that
√
√
√
π Γ(2k)
π Γ(2k + 1)
π (2k)!
Γ(k + 1/2) = 2k−1 ·
= 2k ·
= 2k ·
.
Γ(k)
Γ(k + 1)
k!
2
2
2
Hence we have
Z 1
eizt (1 − t2 )ν−1/2 dt =
−1
∞
X
(iz)2k
k=0
(2k)!
·
Γ(k + 1/2)Γ(ν + 1/2)
Γ(ν + k + 1)
∞
√ X
= Γ(ν + 1/2) π
k=0
This proves the theorem.
5
z 2k
(−1)k
.
Γ(ν + k + 1) k! 2
(7)
We also have Poisson’s integral representations
Theorem 3.
Jν (z) =
=
z ν Z π
1
√
eiz cos θ (sin θ)2ν dθ
Γ(ν + 1/2) π 2
0
z ν Z π
1
√
cos(z cos θ) (sin θ)2ν dθ,
Γ(ν + 1/2) π 2
0
Re ν > −1/2.
Proof. Use the substitution t = cos θ to obtain
Z
Z 0
Z 1
iz cos θ
2
ν−1/2
izt
2 ν−1/2
e
(1 − cos θ)
· (− sin θ) dθ =
e (1 − t )
dt =
−1
π
eiz cos θ (sin θ)2ν dθ.
0
π
Further we have
eiz cos θ = cos(z cos θ) + i sin(z cos θ)
and
π
Z
sin(z cos θ)(sin θ)2ν dθ = 0.
0
This shows that Poisson’s integral representations follow from the integral representation (7).
Remarks:
1. The Fourier transform is defined by
1
F (z) := √
2π
Z
∞
e−izt f (t) dt
−∞
with inversion formula
Z ∞
1
√
f (t) =
eizt F (z) dz.
2π −∞
This implies that the Fourier transform of the function
(
(1 − t2 )ν−1/2 , |t| ≤ 1
f (t) =
0,
|t| > 1
is
F (z) =
Γ(ν + 1/2) z −ν
√
Jν (z).
2
2
2. Instead of the substitution t = cos θ in (7) one can also use the substitution t = sin θ,
which leads to slightly different forms of Poisson’s integral representations. In fact we
have
z ν Z π/2
1
√
Jν (z) =
eiz sin θ (cos θ)2ν dθ
Γ(ν + 1/2) π 2
−π/2
z ν Z π/2
1
√
=
cos(z sin θ) (cos θ)2ν dθ
Γ(ν + 1/2) π 2
−π/2
Z π/2
ν
2
z
√
=
cos(z sin θ) (cos θ)2ν dθ, Re ν > −1/2.
Γ(ν + 1/2) π 2
0
6
Integrals of Bessel functions
The Hankel transform of a function f is defined by
Z ∞
tf (t)Jν (st) dt
F (s) =
0
for functions f for which the integral converges. The inversion formula is given by
Z ∞
sF (s)Jν (st) ds.
f (t) =
0
This pair of integrals is called a Hankel pair of order ν.
An example of such an integral is
Z ∞
Γ µ+ν
(µ + ν)/2
s2
sν
µ−1 −ρ2 t2
2
t
e
Jν (st) dt = ν+1
;− 2 ,
· µ+ν · 1 F1
ν+1
2 Γ(ν + 1) ρ
4ρ
0
Re(µ + ν) > 0.
It can be shown that the integral converges for Re(µ + ν) > 0. Now we use the definition (1)
to obtain
Z ∞
Z ∞
∞
X
(−1)k
sν+2k
2 2
µ−1 −ρ2 t2
t
e
Jν (st) dt =
· ν+2k ·
tµ+ν+2k−1 e−ρ t dt.
Γ(ν + k + 1) k! 2
0
0
k=0
Using the substitution ρ2 t2 = u we find that
Z ∞
Z ∞
du
2 2
u(µ+ν+2k−1)/2 e−u √
tµ+ν+2k−1 e−ρ t dt = ρ−µ−ν−2k+1
2ρ
u
0
Z 0∞
1 −µ−ν−2k
=
ρ
u(µ+ν+2k−2)/2 e−u du
2
0
1 −µ−ν−2k
µ + ν + 2k
=
ρ
Γ
.
2
2
Hence we have
Z ∞
2 2
tµ−1 e−ρ t Jν (st) dt =
0
=
=
∞
µ+ν
1 −µ−ν X (−1)k Γ 2 + k
sν+2k
ρ
· ν+2k 2k
2
Γ(ν + k + 1) k!
2
ρ
k=0
∞
Γ µ+ν
sν X (−1)k ((µ + ν)/2)k
s2k
2
·
·
· 2k 2k
µ+ν
ν
2ρ
Γ(ν + 1) 2
(ν + 1)k k!
2 ρ
k=0
Γ µ+ν
sν
(µ + ν)/2
s2
2
·
·
F
;
−
.
1 1
2ν+1 Γ(ν + 1) ρµ+ν
ν+1
4ρ2
The special case µ = ν + 2 is of special interest: in that case we have (µ + ν)/2 = ν + 1. This
implies that the 1 F1 reduces to a 0 F0 which is an exponential function. The result is
Z ∞
sν
2 2
2
2
tν+1 e−ρ t Jν (st) dt =
· e−s /4ρ , Re ν > −1.
2 )ν+1
(2ρ
0
7
Hankel functions
(1)
(2)
The functions Hν (z) and Hν (z) are defined by
Hν(1) (z) := Jν (z) + iYν (z)
and
Hν(2) (z) := Jν (z) − iYν (z).
These functions are called Hankel functions or Bessel functions of the third kind.
Note that these definitions imply that
(1)
Jν (z) =
(2)
(1)
Hν (z) + Hν (z)
2
and Yν (z) =
(2)
Hν (z) − Hν (z)
.
2i
Further we have
(1)
H−1/2 (x)
r
= J−1/2 (x) + iY−1/2 (x) =
2
(cos x + i sin x) =
πx
r
2 ix
e ,
πx
x>0
and
(2)
H−1/2 (x)
r
= J−1/2 (x) − iY−1/2 (x) =
2
(cos x − i sin x) =
πx
r
2 −ix
e ,
πx
x > 0.
Similarly we have
(1)
H1/2 (x)
r
= J1/2 (x) + iY1/2 (x) =
and
(2)
H1/2 (x)
r
= J1/2 (x) − iY1/2 (x) =
r
2
2 ix
(sin x − i cos x) = −i
e ,
πx
πx
x>0
r
2
2 −ix
(sin x + i cos x) = i
e ,
πx
πx
x > 0.
Modified Bessel functions
The modified Bessel function Iν (z) of the first kind of order ν is defined by
X
∞
z 2k
(z/2)ν
−
z2
z ν
1
Iν (z) :=
;
=
.
0 F1
Γ(ν + 1)
ν+1 4
2
Γ(ν + k + 1) k! 2
k=0
For ν ≥ 0 this is a solution of the modified Bessel differential equation
z 2 y 00 (z) + zy 0 (z) − z 2 + ν 2 y(z) = 0, ν ≥ 0.
For ν ∈
/ {0, 1, 2, . . .} we have that I−ν (z) is a second solution of this differential equation and
the two solutions Iν (z) and I−ν (z) are linearly independent.
For ν = n ∈ {0, 1, 2, . . .} we have I−n (z) = In (z).
The modified Bessel function Kν (z) of the second kind of order ν is defined by
π
Kν (z) :=
[I−ν (z) − Iν (z)] for ν ∈
/ {0, 1, 2, . . .}
2 sin νπ
and
Kn (z) = lim Kν (z) for n ∈ {0, 1, 2, . . .}.
ν→n
Now we have for x > 0
r
2
sinh x,
I1/2 (x) =
πx
r
I−1/2 (x) =
2
cosh x and K1/2 (x) = K−1/2 (x) =
πx
8
r
π −x
e .
2x
A generating function
The Bessel function Jn (z) of the first kind of integer order n ∈ Z can also be defined by means
of the generating function
∞
X
1
−1
=
Jn (z)tn .
(8)
exp
z t−t
2
n=−∞
In fact, the series on the right-hand side is a so-called Laurent series at t = 0 for the function
at the left-hand side. Using the Taylor series for the exponential function we obtain
∞
∞
∞
z X
X
zt
1
1 zt j X (−1)k z k
−1
= exp
· exp −
=
exp
z t−t
=
·
an tn .
2
2
2t
j! 2
k!
2t
n=−∞
j=0
k=0
For n ∈ {0, 1, 2, . . .} we have
∞
∞
z n+k (−1)k z k z n X
X
(−1)k z 2k
1
·
=
= Jn (z)
an =
(n + k)! 2
k!
2
2
(n + k)! k! 2
k=0
k=0
and
a−n =
∞
X
1 z j (−1)j+n z n+j
·
= (−1)n Jn (z) = J−n (z).
j! 2
(n + j)! 2
j=0
This proves (8).
If t = eiθ , then we have
eiθ − e−iθ
1
t − t−1 =
= i sin θ
2
2
1
−1
=⇒ exp
x t−t
= eix sin θ = cos(x sin θ) + i sin(x sin θ).
2
Hence we have
cos(x sin θ) + i sin(x sin θ) = eix sin θ =
∞
X
∞
X
Jn (x)einθ =
n=−∞
Jn (x) [cos(nθ) + i sin(nθ)] .
n=−∞
Since J−n (x) = (−1)n Jn (x) this implies that
∞
X
cos(x sin θ) =
Jn (x) cos(nθ) = J0 (x) + 2
n=−∞
and
J2k (x) cos(2kθ)
k=1
∞
X
sin(x sin θ) =
∞
X
Jn (x) sin(nθ) = 2
n=−∞
∞
X
J2k+1 (x) sin(2k + 1)θ.
k=0
For θ = π/2 this implies that
cos x = J0 (x) + 2
∞
X
k
(−1) J2k (x)
and
sin x = 2
k=1
∞
X
k=0
For θ = 0 we also have
1 = J0 (x) + 2
∞
X
k=1
9
J2k (x).
(−1)k J2k+1 (x).
The generating function (8) can be used to prove that
∞
X
Jn (x + y) =
Jk (x)Jn−k (y).
k=−∞
The proof is as follows:
∞
X
1
(x + y) t − t−1
2
1
1
−1
−1
= exp
x t−t
· exp
y t−t
2
2
!
∞
∞
∞
∞
X
X
X
X
k
m
=
Jk (x)t ·
Jm (y)t =
Jk (x)Jn−k (y) tn .
Jn (x + y)tn = exp
n=−∞
m=−∞
k=−∞
n=−∞
k=−∞
We can also find an integral representation for the Bessel function Jn (x) of the first kind of
integral order n starting from the generating function
∞
X
eix sin θ =
Jn (x)einθ .
n=−∞
We use the orthogonality property of the exponential function, id est
(
Z π
0,
k ∈ Z, k 6= 0
eikθ dθ =
2π, k = 0.
−π
Hence we have
Z
π
ix sin θ
e
−inθ
·e
dθ =
−π
∞
X
Z
π
eikθ · e−inθ dθ = 2π · Jn (x).
Jk (x)
−π
k=−∞
This implies that
1
Jn (x) =
2π
Z
π
i(x sin θ−nθ)
e
−π
1
dθ =
π
Z
π
cos (x sin θ − nθ) dθ.
0
The special case n = 0 reads
1
J0 (x) =
π
Z
0
π
2
cos (x sin θ) dθ =
π
10
Z
0
1
cos(xt)
√
dt.
1 − t2