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Niels Walet 5/14/2020 10.5.1 CC- BY-NC- SA https://math.libretexts.org/link?8329
10.5: Properties of Bessel functions
Bessel functions have many interesting properties:
Let me prove a few of these. First notice from the definition that is even or odd if is even or odd,
Substituting in the definition of the Bessel function gives if , since in that case we have the sum of positive
powers of , which are all equally zero.
Let’s look at :
Here we have used the fact that since , [this can also be proven by defining a recurrence relation
for ]. Furthermore we changed summation variables to .
The next one:
(0)
J
0
(
x
)
J
ν
(
x
)
J
n
[
(
x
)
]
d
dx
x
ν
J
ν
[ (
x
)]
d
dx
x
ν
J
ν
[ (
x
)]
d
dx
J
ν
x
(
x
)
J
ν
+1
(
x
)
dx
x
ν
J
ν
+1
(
x
)
dx
x
ν
J
ν
1
= 1,
= 0 (if
ν
> 0),
= (1 (
x
),)
n
J
n
= (
x
),
x
ν
J
ν
+1
= (
x
),
x
ν
J
ν
1
= [ (
x
) (
x
)] ,
1
2
J
ν
1
J
ν
+1
= 2
ν
(
x
)
x
(
x
),
J
ν
J
ν
1
= (
x
)+
C
,
x
ν
J
ν
= (
x
)+
C
.
x
ν
J
ν
(
x
)
J
n
n
(
x
) = .
J
n
k
=0
(1
)
k
k
!(
n
+
k
)!
( )
x
2
n
+2
k
(10.5.1)
x
= 0 0
ν
> 0
0
J
n
(
x
)
J
n
=
k
=0
(1
)
k
k
!Γ(
n
+
k
+1)!
( )
x
2
n
+2
k
=
k
=
n
(1
)
k
k
!Γ(
n
+
k
+1)!
( )
x
2
n
+2
k
=
l
=0
(1
)
l
+
n
(
l
+
n
)!
l
!
( )
x
2
n
+2
l
= (1 (
x
).)
n
J
n
Γ(
l
) = ± 1/Γ(
l
) = 0
1/Γ(
l
)
l
=
n
+
k
[
(
x
)
]
d
dx
x
ν
J
ν
=
{ }
2
ν
d
dx
k
=0
(1
)
k
k
!Γ(
ν
+
k
+1)
( )
x
2
2
k
=
2
ν
k
=1
(1
)
k
(
k
1)!Γ(
ν
+
k
+1)
( )
x
2
2
k
1
=
2
ν
l
=0
(1
)
l
(
l
)!Γ(
ν
+
l
+2)
( )
x
2
2
l
+1
=
2
ν
l
=0
(1
)
l
(
l
)!Γ(
ν
+ 1 +
l
+1)
( )
x
2
2
l
+1
=
x
ν
l
=0
(1
)
l
(
l
)!Γ(
ν
+ 1 +
l
+1)
( )
x
2
2
l
+
ν
+1
= (
x
).
x
ν
J
ν
+1
pf2

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Niels Walet 5/14/2020 (^) 10.5.1 CC-BY-NC-SA https://math.libretexts.org/link?

10.5: Properties of Bessel functions

Bessel functions have many interesting properties:

Let me prove a few of these. First notice from the definition that is even or odd if is even or odd,

Substituting in the definition of the Bessel function gives if , since in that case we have the sum of positive powers of , which are all equally zero.

Let’s look at :

Here we have used the fact that since , [this can also be proven by defining a recurrence relation for ]. Furthermore we changed summation variables to.

The next one:

J 0 (0)

( x ) Jn ( x )

[ ( x )]

d dx x

ν (^)

[ ( x )]

d dx

^

[ ( x )]

d dx

Jν x Jν +1( x )

xν^ +1( x ) dx

^ −1( x ) dx

= 0 (if ν > 0), = (−1 ) n^ Jn ( x ),

= − xν^ +1( x ),

= ^ −1( x ),

= [ ( x ) − ( x )] ,

−1 + = 2 ν Jν ( x ) − x Jν −1( x ),

= − xν^ ( x ) + C ,

= ^ ( x ) + C.

Jn ( x ) n

Jn ( x ) = ∑. k =

∞ (^) (−1) k k !( n + k )!

x 2

n +2 k (10.5.1)

x = 0 0 ν > 0 0 Jn

Jn ( x )= ∑ k =

∞ (^) (−1) k k !Γ(− n + k + 1)!

(^ x ) 2

n +2 k

k = n

∞ (^) (−1) k k !Γ(− n + k + 1)!

x 2

n +2 k

l =

∞ (^) (−1) l + n ( l + n )! l!

x 2

n +2 l

= (−1 ) n^ Jn ( x ). Γ(− l ) = ±∞ 1/Γ(− l ) = 0 1/Γ( l ) l = − n + k

d [ ( x )] dx

xν^ = 2 − ν^ d { } dx

k =

∞ (^) (−1) k k !Γ( ν + k + 1)

(^ x ) 2

2 k

= 2 − ν^ ∑ k =

∞ (^) (−1) k ( k − 1)!Γ( ν + k + 1)

x 2

2 k

= − 2 − ν^ ∑ l =

∞ (^) (−1) l ( l )!Γ( ν + l + 2)

x 2

2 l +

= − 2 − ν^ ∑ l =

∞ (^) (−1) l ( l )!Γ( ν + 1 + l + 1)

x 2

2 l +

= − xν^ ∑ l =

∞ (^) (−1) l ( l )!Γ( ν + 1 + l + 1)

x 2

2 l + ν +

= − xν^ +1( x ).

Niels Walet 5/14/2020 (^) 10.5.2 CC-BY-NC-SA https://math.libretexts.org/link?

Similarly

The next relation can be obtained by evaluating the derivatives in the two equations above, and solving for :

Multiply the first equation by and the second one by and add:

After rearrangement of terms this leads to the desired expression.

Eliminating between the equations gives (same multiplication, take difference instead)

Integrating the differential relations leads to the integral relations.

Bessel function are an inexhaustible subject – there are always more useful properties than one knows. In mathematical physics one often uses specialist books.

[ ( x )]

d dx x

ν (^) = (^) −1( x ).

( x ) xν^ J (^) ν ′ ( x ) − ν xν −1^ ( x ) ^ ( x ) + ν xν −1^ ( x )

= − xν^ +1( x ), = ^ −1( x ). ^ xν

−2 ν ( x ) = − ( x ) + ( x ).

x

Jν Jν +1

2 J ν ′ ( x )= +1 ( x ) + −1( x ).