piessens, r. “the hankel transform.” the transforms and ... · two-dimensional fourier...

Piessens, R. “The Hankel Transform.”The Transforms and Applications Handbook: Second Edition.Ed. Alexander D. PoularikasBoca Raton: CRC Press LLC, 2000

9The Hankel Transform

9.1 Introductory Definitions and Properties 9.2 Definition of the Hankel Transform9.3 Connection with the Fourier Transform9.4 Properties and Examples9.5 Applications

The Electrified Disc • Heat Conduction • The Laplace Equation in the Halfspace z > 0, with a Circularly Symmetric Dirichlet Condition at z = 0 • An Electrostatic Problem

9.6 The Finite Hankel Transform9.7 Related Transforms 9.8 Need of Numerical Integration Methods9.9 Computation of Bessel Functions Integrals over a

Finite Interval Integration between the Zeros of Jv(x) • Modified Clenshaw–Curtis Quadrature

9.10 Computation of Bessel Function Integrals over an Infinite IntervalIntegration between the Zeros of Jv(x) and Convergence Acceleration • Transformation into a Double Integral • Truncation of the Infinite Interval

9.11 Tables of Hankel Transforms

ABSTRACT Hankel transforms are integral transformations whose kernels are Bessel functions. Theyare sometimes referred to as Bessel transforms. When we are dealing with problems that show circularsymmetry, Hankel transforms may be very useful. Laplace’s partial differential equation in cylindricalcoordinates can be transformed into an ordinary differential equation by using the Hankel transform.Because the Hankel transform is the two-dimensional Fourier transform of a circularly symmetricfunction, it plays an important role in optical data processing.

9.1 Introductory Definitions and Properties

Bessel functions are solutions of the differential equation

(9.1)

where p is a parameter.Equation (9.1) can be solved using series expansions. The Bessel function Jp(x) of the first kind and

of order p is defined by

x y xy x p y2 2 2 0′′ + ′ + − =( )

Robert PiessensKatholieke Universieit Leuven

© 2000 by CRC Press LLC

. (9.2)

The Bessel function Yp(x) of the second kind and of order p is another solution that satisfies

.

Properties of Bessel function have been studies extensively (see References 7, 22, and 26).Elementary properties of the Bessel functions are

1. Asymptotic forms.

(9.3)

2. Zeros. Jp(x) and Yp(x) have an infinite number of real zeros, all of which are simple, with thepossible exception of x = 0. For nonnegative p the sth positive zero of Jp(x) is denoted by jp,s . Thedistance between two consecutive zeros tends to π: ( jp,s+1 – jp,s) = π.

3. Integral representations.

. (9.4)

If p is a positive integer or zero, then

.

(9.5)

4. Recurrence relations.

(9.6)

(9.7)

(9.8)

. (9.9)

J x x

x

k p kp

p

k

k

( )! ( )

=

−

+ +=

∞

∑1

2

1

4

1

2

0Γ

W xJ x Y x

J x Y x xp p

p p

( ) det( ) ( )

( ) ( )=

′ ′

= 2

π

J xx

x p xp( ) ~ cos , .2 1

2

1

4ππ π− −

→ ∞

lims→∞

J x

x

px dp

p

p( )( / )

cos( cos )sin/

=

+ ∫1

2

1 21 2

2

0πθ θ θ

π

Γ

J x x p d

je p d

p

njx

( ) cos( sin )

cos( )cos

= −

=

∫

∫−

1

0

0

πθ θ θ

πθ θ

π

θπ

J xp

xJ x J xp p p− +− + =1 1

20( ) ( ) ( )

J x J x J xp p p− +− = ′1 1 2( ) ( ) ( )

′ = −−J x J xp

xJ xp p p( ) ( ) ( )1

′ = − ++J x J xp

xJ xp p p( ) ( ) ( )1


5. Hankel’s repeated integral. Let ƒ(r) be an arbitrary function of the real variable r, subject to thecondition that

is absolutely convergent. Then for p ≥ –1/2

(9.10)

provided that ƒ(r) satisfies certain Dirichlet condtions.

For a proof, see Reference 26. The reader should also refer to Chapter 1, Section 5.6 for moreinformation regarding Bessel functions.

9.2 Definition of the Hankel Transform

Let ƒ(r) be a function defined for r ≥ 0. The vth order Hankel transform of ƒ(r) is defined as

. (9.11)

If v > –1/2, Hankel’s repeated integral immediately gives the inversion formula

. (9.12)

The most important special cases of the Hankel transform correspond to v = 0 and v = 1. Sufficient butnot necessary conditions for the validity of (9.11) and (9.12) are

1. f (r) = O(r –k), r → ∞ where k > 3/2.2. f ′(r) is piecewise continuous over each bounded subinterval of [0, ∞).3. f (r) is defined as [ f (r+) + f (r–)]/2.

These conditions can be relaxed.

9.3 Connection with the Fourier Transform

We consider the two-dimensional Fourier transform of a function ϕ(x,y), which shows a circular sym-metry. This means that ϕ(r cos θ, r sin θ) � ƒ(r,θ) is independent of θ.

The Fourier transform of ϕ is

. (9.13)

We introduce the polar coordinates

and

.

f r r dr( )0

∞

∫

s ds f r J sr J su r dr f u f up p( ) ( ) ( ) [ ( ) ( )]= + + −∞∞

∫∫ 1

200

F s f r r f r J sr drν ν ν( ) { ( )} ( ) ( )≡ ≡∞

∫�0

f r F s sF s J sr ds( ) { ( )} ( ) ( )= ≡−∞

∫� ν ν ν ν1

0

Φ( , ) ( , ) ( , )ζ ηπ

ζ η= −

−∞

∞

−∞

∞

∫∫1

2f x y e dx dyj x y

x r y r= =cos , sinθ θ

ζ ϕ η ϕ= =s scos , sin


We have then

This result shows that F(s,ϕ) is independent of ϕ, so that we can write F(s) instead of F(s, ϕ). Thus, thetwo-dimensional Fourier transform of a circularly symmetric function is, in fact, a Hankel transform oforder zero.

This result can be generalized: the N-dimensional Fourier transform of a circularly symmetric functionof N variables is related to the Hankel transform of order N/2 – 1. If ƒ(r,θ) depends on θ, we can expandit into a Fourier series

(9.14)

and, similarly

(9.15)

where

(9.16)

and

. (9.17)

Substituting (9.15) into (9.17) and using (9.14), we obtain

φ ϕ ϕ ϕπ

θ

πα

πθ ϕ

πα

( cos , sin ) ( , ) ( )

( )

( ) ( ) .

cos( )

cos

s s F s r dr e f r d

rf r dr e d

rf r J rs dr

j r s

j r s

≡ =

=

=

∫∫

∫ ∫

∫

∞− −

∞−

∞

1

2

1

2

0

2

0

0 0

2

00

f r f r enjn

n

( , ) ( )θ θ==−∞

∞

∑

F s r dr e f r d F s ej r sn

jn

n

( , ) ( , ) ( )cos( )ϕπ

θ θθ ϕ ϕ= =∞ ∞

− −

=−∞

∞

∫ ∫ ∑1

2 0 0

f r f r e dnjn( ) ( , )= ∫ −1

2 0

2

πθ θ

πθ

F s F s e dnjn( ) ( , )= ∫ −1

2 0

2

πϕ ϕ

πϕ

F s e d d f r e r dr

e d r dr e d f r e

r dr

njn j s r

jn j s rm

jm

m

( )( )

( , )

( )( )

( )

cos( )

cos( )

=

= ×

=

∫ ∫ ∫

∫ ∫ ∫ ∑

∫ ∫

−∞

−

−∞

−

=−∞

∞

∞

1

2

1

2

1

2

20

2

0

2

0

20

2

0 0

2

0 0

2

πϕ θ θ

πϕ θ

π

πϕ

πθ ϕ

πϕ

πθ ϕ θ

π

ee e f r d

rf r J sr dr

f r

jn j s rn

n n

n n

−

∞

=

=

∫

α α αcos ( )

( ) ( )

{ ( )}.

0

�


In a similar way, we can derive

. (9.18)

9.4 Properties and Examples

Hankel transforms do not have as many elementary properties as do the Laplace or the Fourier transforms.For example, because there is no simple addition formula for Bessel functions, the Hankel transform

does not satisfy any simple convolution relation.

1. Derivatives. Let

.

Then

. (9.19)

Proof

.

In general, the expression between the brackets is zero, and

.

Hence, we have (9.19).2. The Hankel transform of the Bessel differential operator. The Bessel differential operator

is derived from the Laplacian operator

after separation of variables in cylindrical coordinates (r, θ, z).

Let ƒ(r) be an arbitrary function with the property that ƒ(r) = 0. Then

. (9.20)

f r F sn n n( ) { ( )}=�

F s f xν ν( ) { ( )}=�

G s f x s F s F sν ν ν ννν

νν

( ) { ( )} ( ) ( )= ′ = + − −

− +�

1

2

1

21 1

G s xf x J sx dx

xf x J sx f xd

dxxJ sx dx

ν ν

ν ν

( ) ( ) ( )

[ ( ) ( )] ( ) [ ( )]

= ′

= −

∞

∞∞

∫

∫0

00

d

dxxJ sx

sxJ sx J sx[ ( )] [( ) ( ) ( ) ( )]ν ν νν

ν ν= + − −− +21 11 1

∆νν ν≡ + −

= −

d

dr r

d

dr r r

d

drr

d

dr r

2

2

2 21 1

∇ = ∂∂

+ ∂∂

+ ∂∂

+ ∂∂

22

2 2

2

2

2

2

1 1

r r r r zθ

limr→∞

� �ν ν ν{ ( )} { ( )}∆ f r s f r= − 2


This result shows that the Hankel transform may be a useful tool in solving problems with cylindricalsymmetry and involving the Laplacian operator.

Proof Integrating by parts, we have

This property is the principal one for applications of the Hankel transforms to solving differentialequations. See References 2, 3, 24, 25, and 28.

3. Similarity.

. (9.21)

4. Division by r.

. (9.22)

5.

. (9.23)

6.

. (9.24)

7. Parseval’s theorem. Let

and

.

Then

�

�

ν ν ν

ν ν ν

ν

ν

ν

ν

{ ( )} ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

{ ( )}.

∆ f rd

drr

df

dr rf r J sr dr

s J srs

xJ sr

rJ sr f r r dr

s rf r J rs dr

s f r

= −

= ′′ + ′ −

= −

= −

∞

∞

∞

∫

∫

∫

0

2

0

22

2

2

0

2

� ν ν{ ( )}f ara

Fs

a=

12

� ν ν νν{ ( )} [ ( ) ( )]r f r

sF s F s−

− += +11 12

� νν ν

νrd

drr f r sF s− −

−

= −1 1

1[ ( )] ( )

� νν ν

νrd

drr f r sF s− − +

+

=1 1

1[ ( )] ( )

F s f rν ν( ) { ( )}=�

G s g rν ν( ) { ( )}=�


(9.25)

Example 9.1

From the Fourier pair (see Chapter 2) and the Fourier transform

relationship , we obtain the Hankel transform

.

Example 9.2

From the relationship = [aJ1(as)]/s (see Chapter 1, Section 5.6),we conclude that

where pa(r) = 1 for |r| < a and zero otherwise.

Example 9.3

From the identity s > 0 (see Chapter 1, Section 5.6), we obtain

.

Example 9.4

Since (see Chapter 1, Section 2.4), we obtain

and because of symmetry

.

Convolution Identity

Let ƒ1(r) and ƒ2(r) have Hankel transforms F1(s) and F2(s), respectively. From Section 2.4.1 above, we have

0 0 0

0 0

0

∞ ∞ ∞

∞ ∞

∞

∫ ∫ ∫

∫ ∫

∫

=

=

=

F s G s s ds F s s ds r g r J sr dr

r g r dr sF s J sx ds

r g r f r dr

ν ν ν ν

ν ν

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) .

�{ } ( / )( ) ( )/e a ea x y a− + − +=2 2 2 2 4π ζ η

�{ ( )} ( ) ( )f x y F F s2 20

2 202 2+ = + ≡π ζ η π

�{ } ,/ea

e aar s a− −= >2 21

204

∫ = ∫0 0 0 11a a

r J sr dr s d dr rJ sr( ) ( / )( / )[ ( )]

� 01{ ( )}( )

p raJ as

sa =

∫ =∞0 0 1J sr dr s( ) / ,

� 0

1 1

r s

=

∫ − =∞0 0 0r r a J sr dr aJ asδ( ) ( ) ( )

�0 0 0{ ( )} ( ),δ r a aJ as a− = >

�0 0 0{ ( )} ( ),aJ ar s a a= − >δ


.

Hence, we have

.

Therefore, to find the inverse Hankel transform of 2πF1(s)F2(s), we convolve with

and in the answer we replace by r. We can also write the above relationship

in the form

.

Example 9.5

If ƒ1(r) = ƒ2(r) = [J1(ar)]/r then from the convolution identity above, we obtain

where

.

Hence,

Example 9.6

From the definition, the Hankel transform of rvh(a – r), a > 0 is given by

since h(a – r) is the unit step function with value equal to 1 for r ≤ a and 0 for r > a. But ∫ t vJv–1(t) dt =t vJv(t) + C (see Chapter 1, Section 5.6, Table 5.6.1) and hence,

.

� ∫ ∫−∞∞

+ − + −

=f x y f x x y y dx dy F s F s1 1

212

2 12

12

1 12

1 24( ) ( ( ) ( ) ) ( ) ( )π

� �0 1 2 2 1 2 1 2

1

22{ ( ) ( )} { ( ) ( )} ( ) ( )( )f r f r f r f r F s F s★★ ★★= =

ππ

f x y12 2( )+

f x y22 2( ),+ x y2 2+

� 0 1 2 1 22{ ( ) ( )} ( ) ( )πf r f r F s F s= ★★

� 012

2 22

1πJ ar

r ap s p sa a

( )( ) ( )

= ★★

p s p ss

a

s

a

s

aaa a( ) ( ) cos ★★ = − −

−22

14

12

2

2

� 012

2

12

2 2

2

2 22

14

1 2

0

πJ ar

r

s

a

s

a

s

ap s

p ss a

a

a

( )cos ( )

( ) | |

= − −

=≤

−

otherwise .

� νν ν

ν νν

ν{ ( )} ( ) ( )r h a r r J sr drs

x J x dxa as

− = =∫ ∫++

+

0

1

20

11

� νν

ν

ν ν

ν

ν ν{ ( )}( )

( ) ( ), , r h a ras

sJ as

a

sJ as a− = = > > −

+

+ +

+

+

1

2 1

1

1 01

2


Example 9.7

The Hankel transform of r v–1e–ar, a > 0 is given by

where we set t = rs and L is the Laplace transform operator (see also Chapter 5). But

and, hence,

From Chapter 1, Section 2.5, the duplication formula of the gamma function gives the relationship

and, therefore, the Laplace transform relation becomes

.

The last series can be summed by using properties of the binomial series

where the relation

was used. The Laplace transform now becomes

� νν ν

ν νν

ν

νν

ν

{ } ( ) ( )

( );

r e r e J sr drs

t J t e dt

st J t p

a

s

ar aras

t− −∞

−+

∞ −

+

= =

= =

∫ ∫1

01

0

1

1

1L

t J tt

n n

n n

n

n

νν

ν

νν( )

( )

! ( )= −

+ +

+

+=

∞

∑ 1

1 2

2 2

2

0Γ

�{ ( ); }( )

! ( ){ ; }

( ) ( )

! ( ).

t J t pn n

t p

n

n n p

n

n

n

n

n

n nn

νν ν

ν

ν ν

ν

νν

= −+ +

= − + ++ +

++

=

∞

+ + +=

∞

∑

∑

1

1 2

1 2 2 1

1 2

2

2 2

0

2 2 2 10

Γ

ΓΓ

L

ΓΓ

Γ( )

( )

2 2 1

1

12

1

22 2n

nnn+ +

+ += + +

+νν π

νν

�{ ( ); }

( )

!t J t p

p

n

n p

nn

n

νν

ν

νπ

ν=

− + +

+

=

∞

∑21

1

2 12 1 2

0

Γ

( )( ) ( )

! ( ), | |1

11

00

+ =−

= − + <−

=

∞

=

∞

∑∑xb

nx

n b

n bx xb n

nn

nn

ΓΓ

−

= − + + − = − +b

n

b b b n

n

n b

n b

n n( ) ( ) ( )

!

( ) ( )

! ( )

1 1 1 1L ΓΓ


and, hence,

.

If we set v = 0 and a = 0 in the above equation, we obtain the results of Example 9.3. If we set v = 0, weobtain

.

Example 9.8

The Hankel transform �0{e–ar} is given by

since multiplication by r corresponds to differentiation in the Laplace transform domain.

Example 9.9

From Chapter 1, Section 5.6, we have the following identity:

,

where rn(x) = Jn(x)/x n.Using the Hankel transform property of the Bessel operator, we obtain the relationship

or

�{ ( ); }

( )

, Re( )t J t p

p a

s

pνν

ν

ν

ν

ν

ν

π

ν

π

=+

+=

+

+

>+ +

21

2

1

21

2

1

12

1

2 21

2

Γ Γ

� νν

ν

ν

ν

ν ν

ν

ν

π

ν

πν{ }

( )

,r es

a

s

s

a s

ar− −+ + +

=+

+

=+

+> −1

12

1

2 2 21

2

12

1

2

1

21

2 1

2

Γ Γ

� 01

2 2

10{ } ,r e

a saar− − =

+>

� 0 02 2

1

2

2 2 3 20

{ } { ( ); } ( )

[ ],

/

e r J sr r ad

das a

a

s aa

ar− −= → = − +

=+

>

�

d r x

dx x

dr x

dxr x nr xn n

n n

2

2 1

12

( ) ( )( ) ( )+ + = +

( ) ( ) ( )1 221− = +s R s nR sn n

R ss

nR s

s

nR sn n

n

n+ = − = = −1

2 2

1

1

2

1

2( ) ( )

( )

!( ) .L


But from Example 9.2, and, hence,

where p1(s) is a pulse of width 2 centered at s = 0.

Example 9.10

If the impulse response of a linear space invariant system is h(r) and the input to the system is ƒ(r), thenits output is g (r) = ƒ (r)��h(r) and, hence,

.

Since �0{J0(ar)} = [δ(s – a)]/a (see Example 9.4) and ϕ(s)δ(s – a) = ϕ(a)δ(s – a), we conclude that ifthe input is ƒ(r) = J0(ar), then

.

Therefore, the output is

.

9.5 Applications

9.5.1 The Electrified Disc

Let υ be the electric potential due to a flat circular electrified disc, with radius R = 1, the center of thedisc being at the origin of the three-dimensional space and its axis along the z-axis.

In polar coordinates, the potential satisfies Laplace’s equation

(9.26)

The boundary conditions are

(9.27)

. (9.28)

In (9.27), υ0 is the potential of the disc. Condition (9.28) arises from the symmetry about the plane z = 0.Let

� 01

1

J r

rp s

( )( )

=

� 0

2 1

1 1

1

2 1

J r

r

s

np sn

n

n

n

( ) ( )

( )!( )

= −

−

−

−

G s F s H s( ) ( ) ( )= 2π

G sa

s a H sH a

as a( ) ( ) ( )

( )( )= − = −2 2πδ π δ

g r H a J ar( ) ( ) ( )= 2 0π

∇ ≡ ∂∂

+ ∂∂

+ ∂∂

=22

2

2

2

10υ υ υ υ

r r r z.

υ υ( , ) ,r r0 0 10= ≤ <

∂∂

= >υz

r r( , ) ,0 0 1

V s z r z( , ) { ( , )}=� 0 υ


so that

.

The solution of this differential equation is

where A and B are functions that we have to determine using the boundary conditions.Because the potential vanishes as z tends to infinity, we have B(s) � 0. By inverting the Hankel

transform, we have

. (9.29)

The boundary conditions are now

(9.30)

. (9.31)

Using entries (9.8) and (9.9) of Table 9.1 (see Section 9.11), we see that A(s) = sin s/s2 so that

. (9.32)

In Figure 9.1, the graphical representation of υ(r,z) for υ0 = 1 is depicted on the domain 0 ≤ r ≤ 2, 0 ≤z ≤ 1. The evaluation of υ(r,z) requires numerical integration.

Equations (9.30) and (9.31) are special cases of the more general pair of equations

FIGURE 9.1 Electrical potential due to an electrified disc.

� 02 2

2

20{ } ( , ) ( , )∇ = − + ∂

∂=υ s V s z

V

zs z

V s z A s e B s esz sz( , ) ( ) ( )= +−

υ( , ) ( ) ( )r z sA s e J sr dssz=∞

−∫0 0

υ υ( , ) ( ) ( ) ,r sA s J rs ds r0 0 10

0 0= = ≤ <∞

∫∂∂

= = >∞

∫υz

r s A s J rs ds r( , ) ( ) ( ) ,0 0 10

20

υυ

( , )sin

( )r zs

se J sr dssz=

∞−∫2 0

00π


(9.33)

(9.34)

where a(x) is given and ƒ(x) is to be determined.The solution of (9.29) can be expressed as a repeated integral:12

(9.35)

. (9.36)

If a(x) = x β, and α < 1, 2α + β > –3/2, α + v > –1, v > –1, then

(9.37)

With β = v and α < 1, α + v > – 1, v > – 1 further simplification is possible:

. (9.38)

9.5.2 Heat Conduction

Heat is supplied at a constant rate Q per unit area and per unit time through a circular disc of radius ain the plane z = 0, to the semi-infinite space z > 0. The thermal conductivity of the space is K. The planez = 0 outside the disc is insulated. The mathematical model of this problem is very similar to that ofSection 5.1. The temperature is denoted by υ(r,z). We have again the Laplace Equation (9.26) in polarcoordinates, but the boundary conditions are now

(9.39)

The Hankel transform of the differential equation is again

. (9.40)

We can now transform also the boundary condition, using formula (3) in Table 9.1:

0

2 0 1∞

∫ = ≤ <f t t J xt dt a x x( ) ( ) ( ),αν

0

0 1∞

∫ = >f t J xt dt x( ) ( ) ,ν

f xx

s J xsd

dsa t t s t dt ds

s

( )( )

( ) ( ) ( ) ,=+

− − < <− −

− −+

+∫ ∫2

11 0

1

0

1

0

1 2 2α α

ν αν α

ν α

αα

Γ

f xx

s J xs a t t s t dt dss

( )( )

( )( ) ( ) ( ) ,= − < <

−− − +

++ −∫ ∫2

0 11

0

11

0

1 2 2 1α

ν αν α

ν α

αα

Γ

f x

x

t J t dtx

( ) ( ) .

( )

=+ +

+ + +

− + +

+ ++∫

Γ

Γ

12

2 12

2 1

0

1

β ν

α β ν

α β

α

α βν α

f xx

J x( )( )

( ) ( )( )= +

+ + + +ΓΓνν αα ν α

1

2 11

− ∂∂

= < =

= > =

Kr z

zQ r a z

r a z

υ( , ), ,

, , .

0

0 0

∂∂

− =2

2

2 0V

zs z s V s z( , ) ( , )


. (9.41)

The solution of (9.39) must remain finite as z tends to infinity. We have

.

Using condition (9.41) we can determine

.

Consequently, the temperature is given by

. (9.42)

9.5.3 The Laplace Equation in the Halfspace z > 0, with a Circularly Symmetric Dirichlet Condition at z = 0

We try to find the solution υ(r,z) of the boundary value problem

(9.43)

Taking the Hankel transform of order 0 yields

and

.

The solution is

so that

. (9.44)

For the special case

− ∂∂

=KV

zs Qa J as s( , ) ( )/0 1

V s z A s e sz( , ) ( )= −

A s Qa J as Ks( ) ( )/( )= 12

υ( , ) ( ) ( )r zQa

Ke J as J rs s dssz=

∞− −∫0 1 0

1

∂∂

+ ∂∂

+ ∂∂

= > < < ∞

=

2

2

2

2

10 0 0

0

υ υ υ

υr r r z

z r

r f r

, ,

( , ) ( ).

∂∂

− =2

2

2 0V

zs z s V s z( , ) ( , )

V s rf r J sr dr( , ) ( ) ( )00

0=∞

∫

V s z e rf r J sr drsz( , ) ( ) ( )= −∞

∫0 0

υ( , ) ( ) ( ) ( )r z s e J sr ds p f p J sp dpsz=∞

−∞

∫ ∫00

00

f r h a r( ) ( )= −


where h(r) is the unit step function, we have the solution

. (9.45)

9.5.4 An Electrostatic Problem

The electrostatic potential Q(r,z) generated in the space between two grounded horizontal plates at z =± � by a point charge q at r = 0, z = 0 shows a singular behavior at the origin. It is given by

(9.46)

where ϕ(r,z) satisfies Laplace’s Equation (9.26). The boundary conditions are

. (9.47)

Taking the Hankel transform of order 0, we obtain

(9.48)

(9.49)

(see formula (18) in Table 9.1).The solution is

where A(s) and B(s) must satisfy

Hence,

and

.

Hence,

υ( , ) ( ) ( )r z a e J sr J as dssz=∞

−∫0 0 1

υ( , ) ( , ) ( ) /r z r z q r z= + + −ϕ 2 2 1 2

ϕ( , ) ( ) /r q r± + + =−l l2 2 1 2 0

∂∂

− =2

2

2 0Φ Φ

zs z s s z( , ) ( , )

Φ( , )sqe

s

s

± = −−

ll

A s e B s esz sz( ) ( )− +

A s e B s eq e

s

A s e B s eq e

s

s ss

s ss

( ) ( )

( ) ( )

.

+ −−

−−

+ = −

+ = −

l ll

l ll

A s B sq e

s s

s

( ) ( )

cosh( )= = −

− l

l2

Φ( , ) cosh( )

cosh( )s z

q e

s

sz

s

s

= −− l

l


. (9.50)

9.6 The Finite Hankel Transform

We consider the integral transformation

. (9.51)

A property of this transformation is that

where �v is the Bessel differential operator.If α is equal to the sth positive zero jv,s of Jv(x), we have

.

If α is equal to the sth positive root βv,s of

where h is a nonnegative constant, we have

.

The transformation (9.51) with α = j v,s , s = 1, 2, … is the finite Hankel transform. It maps the functionƒ(r) into the vector (Fv(jv,1), Fv(jv, 2), Fv(j v,3) …). The inversion formula can be obtained from the well-known theory of Fourier-Bessel series

. (9.52)

The transformation (9.51) with α = βv ,s , s = 1, 2, … is the modified finite Hankel transform. The inversionformula is

. (9.53)

If h = 0, βv,s is the sth positive zero of denoted by .Formulas for the computation of jv,s and are given by Olver.15 Values of jv,s and are tabulated

in Reference 1. A Fortran program for the computation of jv,s and is given in Reference 18.

ϕ( , )cosh( )

cosh( )( )r z

q

r zq e

sz

sJ sr dss=

+−

∞−∫2 2 0

0l

l

F H f rf r J r drν ν να α α( ) { , } ( ) ( )= =∫01

H f F J f J fν ν ν ν να α α α α α( , ) ( ) [ ( ) ( ) ( ) ( )]∆ = − + ′ − ′2 1 1

H f j j H f j j J j fs s s s sν ν ν ν ν ν ν ν ν( , ) ( , ) ( ) ( ), , , , ,∆ = − + +2

1 1

hJ x x J xν ν( ) ( )+ ′ = 0

H f H f J hf fs s s sν ν ν ν ν ν ν νβ β β β( , ) ( , ) ( )[ ( ) ( )], , , ,∆ = − + + ′2 1 1

f rF j

J jJ j rs

s

s

s

( )( )

( )( ),

,

,=+=

∞

∑21

21

ν ν

ν ννν

f rF

h

J r

Js s

s

s

ss

( )( ) ( )

( ), ,

,

,

,

=+ −=

∞

∑22

2 2 2 21

β ββ ν

ββ

ν ν ν

ν

ν ν

ν ν

′J xν( ), ′j sν ,

′j sν , ′j sν ,

′j sν ,


ApplicationWe calculate the temperature υ(r,t) at time t of a long solid cylinder of unit radius. The initial temperatureis unity and radiation takes place at the surface into the surrounding medium maintained at zerotemperature.

The mathematical model of this problem is the diffusion equation in polar coordinates

(9.54)

The initial condition is

. (9.55)

The radiation at the surface of the cylinder is described by the mixed boundary condition

(9.56)

where h is a positive constant.Transformation of (9.54) by the modified finite Hankel transform yields

(9.57)

where

so that

. (9.58)

The solution of (9.57), with the initial condition (9.58), is

.

Using the inversion formula, we obtain

. (9.59)

∂∂

+ ∂∂

= ∂∂

≤ < >2

2

10 1 0

υ υ υr r r t

r t, ,

υ( , ) ,r r0 1 0 1= ≤ ≤

∂∂

= −υ υr

t h t( , ) ( , )1 1

dV

dtt V ts s s( , ) ( , ), , ,β β β0 0

20= −

V t r r t J r dr( , ) ( , ) ( )α α= ∫01

0υ

V rJ r drJ

( , ) ( )( )

α ααα

00

1

01= =∫

V tJ

ess

s

ts( , )( )

,,

,

,βββ

β0

1 0

0

02

= −

υ( , )( ) ( )

( ), , ,

,

,

,

r t eJ

h

J r

Jst s s

s

s

sj

=+

−

=

∞

∑2 02

0 1 0

202

0 0

02

01

β β ββ

ββ


9.7 Related Transforms

For some applications, Hankel transforms with a more general kernel may be useful. We give one example.We consider the cylinder function

. (9.60)

Using this function as a kernel, we can construct the following transform pair:

(9.61)

(9.62)

The inversion formula follows immediately from Weber’s integral theorem (see Watson26):

. (9.63)

For this reason, we will refer to (9.61) and (9.62) as the Weber transform. This transform has the followingimportant property:

If

(9.64)

then

. (9.65)

We may expect that this transform is useful for solving Laplace’s equation in cylindrical coordinates, witha boundary condition at r = 1.

Example

We want to compute the steady-state temperature u(r,z) in a horizontal infinite homogeneous slab ofthickness 2�, through which there is a vertical circular hole of radius 1. The horizontal faces are held attemperature zero and the circular surface in the hole is at temperature T0.

The mathematical model is

(9.66)

Taking the Weber transform of order zero, we have

.

The solution of this ordinary differential equation, satisfying the boundary condition, is

Z s r J sr Y s Y sr J sν ν ν ν ν( , ) ( ) ( ) ( ) ( )= −

F s rf r Z s r drν ν( ) ( ) ( , )=∞

∫1f r sF s

Z s r

J s Y sds( ) ( )

( , )

( ) ( )=

+

∞

∫0 2 2νν

ν ν

1 0

2 21

2

∞ ∞

∫ ∫ = + + + −u du f s Z r u Z s u s ds J r Y r f r f r ( ) ( , ) ( , ) [ ( ) ( )][ ( ) ( )]ν ν ν ν

f x g xx

g xx

g x( ) ( ) ( ) ( )= ′′ + ′ −1 2

2

ν

F s s G s gν ν π( ) ( ) ( )= − −2 2

1

∂∂

+ ∂∂

+ ∂∂

=

= − =

=

2

2

2

2

0

10

0

1

u

r r

u

r

u

z

u r u r

u z T

( , ) ( , )

( , ) .

l l

∂∂

− =2

02

20 0

2U

zs z s U s z T( , ) ( , )

π


.

Consequently, we have

(9.67)

9.8 Need of Numerical Integration Methods

When using the Hankel transform for solving partial differential equations, the solution is found as anintegral of the form

(9.68)

where a is a positive real number or infinite. In most cases, analytical integration of (9.68) is impossible,and numerical integration is necessary. But integrals of type (9.68) are difficult to evaluate numerically if

1. The product ap is large.2. a is infinite.3. ƒ (x) shows a singular or oscillatory behavior.

In cases 1 and 2, the difficulties arise from the oscillatory behavior of Jv(x) and they grow when theoscillations become stronger.

We give here a survey of numerical methods that are especially suited for the evaluation of I (a, p, v)when ap is large or a is infinite. We restrict ourselves to cases where ƒ(x) is smooth, or where ƒ(x) =x α g (x) where g (x) is smooth and α is a real number.

9.9 Computation of Bessel Function Integrals Over a Finite Interval

Integral (9.68) can be written as

. (9.69)

We assume that α + v > – 1. If (α+v) is not an integer, then there is an algebraic singularity of theintegrand at x = 0. If ap is large, then the integrand is strongly oscillatory.

If (α + v) is an integer and ap is small, classical numerical integration methods, such as Rombergintegration, Clenshaw-Curtis integration of Gauss-Legendre integration (see Davis and Rabinowitz 4) areapplicable. If (α + v) is not an integer and ap is small, the only difficulty is the algebraic singularity atx = 0, and Gauss-Jacobi quadrature or Iri-Moriguti-Takesawa (IMT) integration 4 can be used. If ap islarge, special methods should be applied that take into account the oscillatory behavior of the integrand.We describe two methods here.

9.9.1 Integration between the Zeros of Jv(x)

We denote the sth positive zero of Jv (x) by j v , s and we set jv, 0 = 0. Then

U s zT

s

sz

s002

21( , )

cosh

cosh= −

π l

u r zT

s

sz

s

Z s r

J s Y sds r( , )

cosh

cosh

( , )

( ) ( )( ) .= −

+

>∞

∫2 11 10

0

0

02

02π l

I a p J px f x dxa

( , , ) ( ) ( )ν ν= ∫0

I a p a x J apx g ax dx( , , ) ( ) ( )ν α αν= + ∫1 0

1


(9.70)

where

(9.71)

and where N is the largest natural number for which jv,N ≤ ap. This means that N is large when ap is large.Using a transformation attributed to Longman,10 the summation in Equation (9.70) can be written as

Assuming now that N and p are large and that high-order differences are small, the last bracket may beneglected and

(9.72)

The summations in (9.72) may be truncated as soon as the terms are small enough. For the evaluationof Ik , k = 1, 2, …, classical integration methods (e.g., Lobatto’s rule) can be used, but special Gaussquadrature formulas (see Piessens16) are more efficient. If the integral I1 has an algebraic singularity atx = 0, then the Gauss-Jacobi rules or the IMT rule are recommended.

9.9.2 Modified Clenshaw-Curtis Quadrature

The Clenshaw-Curtis quadrature method is a well-known and efficient method for the numerical eval-uation of an integral I with a smooth integrand. This method is based on a truncated Chebyshev seriesapproximation of the integrand. However, when the integrand shows a singular or strongly oscillatorybehavior, the classical Clenshaw-Curtis method is not efficient or even applicable, unless it is modifiedin an appropriate way, taking into account the type of difficulty of the integrand. We call this methodthen a modified Clenshaw-Curtis method (MCC method). The principle of the MCC method is thefollowing: the integration interval is mapped onto [–1, +1] and the integrand is written as the productof a smooth function g(x) and a weight function w(x) containing the singularities or the oscillating factorsof the integrand; that is,

. (9.73)

I a p I J px f x dxkk

j p

a

k

N

N

( , , ) ( ) ( ) ( ),

νν

ν= − ++

=∫∑ 1 1

1

I J px f x dxkj p

j p

k

k

=−

∫ν

ν

ν,

,

| ( )| ( )1

S I I I I I

I I I I

I

kk

p p p

k

N

NN N N

p pn p

p p p

= − = − + + + −

+ − + + + +

+ −

+ − − −

=

−− −

− −− +

−

∑( ) ( )

( )

( ) [

11

2

1

4

1

81 2

11

2

1

4

1

82

2 1

11 1

21

1 11

1

11

22

11

∆ ∆ ∆

∆ ∆ ∆

∆

L

L

11 2 311− + − + − − −

−∆ ∆ ∆p p N p pN pI I IL ( ) ].

S I I I

I I INN N N

� 1

2

1

4

1

8

11

2

1

4

1

8

1 12

1

11

22

− + −

+ − + + +

−− −

∆ ∆

∆ ∆

L

L( ) .

I w x g x dx=−

+

∫ 1

1

( ) ( )


The smooth function is then approximated by a truncated series of Chebyshev polynomials

. (9.74)

Here the symbol Σ′ indicates that the first term in the sum must be halved. For the computation of thecoefficients ck in Equation (9.74) several good algorithms, based on the Fast Fourier Transform, areavailable.

The integral in (9.73) can now be approximated by

(9.75)

where

are called modified moments.The integration interval may also be mapped onto [0, 1] instead of [–1, 1], but then the shifted

Chebyshev polynomials are to be used.We now consider the computation of the integral (9.69),

. (9.76)

If

(9.77)

then

(9.78)

where

. (9.79)

These modified moments satisfy the following homogeneous, linear, nine-term recurrence relation:

g x c T x xk k

k

N

( ) ( ),� ′ − ≤ ≤=∑

0

1 1

I c Mk k

k

N

� ′=∑

0

M w x T x dxk k=−

+

∫ 1

1

( ) ( )

T xk★ ( )

I x J x g x dx= ∫01

αν ω( ) ( )

g x c T xk k

k

N

( ) ( )� ′ ′=∑

0

I c Mk k

k

N

� ′=∑ ( , , )ω ν α

0

M x J x T x dxk k( , , ) ( ) ( )ω ν α ωαν= ∫0

1★


(9.80)

Because of the symmetry of the recurrence relation of the shifted Chebyshev polynomials, it is convenientto define

and consequently

.

To start the recurrence relation with k = 0, 1, 2, 3, … we need only M0, M1, M2, and M 3 . Using theexplicit expressions of the shifted Chebyshev polynomials, we obtain

(9.81)

where

. (9.82)

Because

(9.83)

we need only G(ω, v, α) and G(ω , v, α + 1).Luke12 has given the following formulas:

1. A Neumann series expansion that is suitable for small ω

ω α α ν ω

ν α α

ν α α ω

ν α

2

42 2

2

2

2 21

2 2 22

2 2

163 3 2

4

4 2 2 2 1

2 4 6 2 2 13

8

4

M k k M

k M

k M

k k

k

k

+ +

+

+ + + + + − −

+ − − + −

− − + − − − −

+ −

( )( )

[ ( ) ( )( )]

( ) ( ) ( )

[ ( )) ( )( )]

( )( ) .

+ − −

+ − − − + − −

+ =

−

− −

2 2 2 1

3 3 24 16

0

1

2 22

2

2

4

k M

k k M M

k

k k

α

α α ν ω ω

T x T x kk k− = = …★ ★( ) ( ), , , , 1 2 3

M Mk k− =( , , ) ( , , )ω ν α ω ν α

M G

M G G

M G G G

M G G G

0

1

2

3

2 1

8 2 8 1

32 3 48 2 18

=

= + −

= + − + +

= + − + +

( , , )

( , , ) ( , , )

( , , ) ( , , ) ( , , )

( , , ) ( , , ) ( ,

ω ν α

ω ν α ω ν α

ω ν α ω ν α ω ν α

ω ν α ω ν α ω , ) ( , , )ν α ω ν α+ −1 G

G x J x dx( , , ) ( )ω ν α ωαν= ∫0

1

ω ω ν α ν α ω ν α

α ν ω ω ων ν

2 2 2

1

2 1

1

G G

J J

( , , ) [ ( ) ] ( , , )

( ) ( ) ( )

+ = − +

+ + + − −


(9.84)

2. An asymptotic expansion that is suitable for large ω

(9.85)

where

and

If α and v are integers, the following formulas are useful1:

For the evaluation of

G

k

J

k

k

k

( , , )( )

( )

( )ω ν αω α ν

ν ν α

ν αων=

+ +

+ + − +

+ +

+ +=

∞

∑2

1

2 11

2

3

2

2 1

0

G g g( , , ) ( cos sin )ω ν αω

ν α

ν α πωθ θ

α

α=

+ +

− +

− ++

2

1

2

1

2

21 3 1 2

Γ

Γ

θ ω νπ π= − +/ /2 4

g a

g a

ak

b

b

bk k

k k

kk

k

k

kk

k

k

kk kk k

k

1 22

0

2 2 12 1

0

0

1

1

1

1 2 1 2

2

1

12 1 1 2

1 2 1

~ ( ) ,

~ ( ) ,

( / ) ( / )

!

( )( / )

( / )( /

− →∞

− →∞

= − +

=

= + + − −− − + +

−

=

∞

+− −

=

∞

+

∑

∑

ω ω

ω ω

ν ν

αν ν 22)

.bk

0

1

20

1

0 2 1

0

1

0

1

2 10

2

1

2

1 2

∫ ∫ ∑

∫ ∑

= −

= − −

+=

−

+=

J x dx J x dx J

J x dxJ

J

k

k

k

k

ν

ν

ν

ν

ω ωω

ω

ω ωω ω

ω

( ) ( ) ( )

( )( )

( ) .

0

1

0∫ J x dx( )ω


Chebyshev series approximations are given by Luke.11 We now discuss the numerical aspect of therecurrence formula (9.80). The numerical stability of forward recursion depends on the asymptoticbehavior of Mk(ω, υ, α) and of eight linearly independent solutions yi,k , i = 1, 2, …, 8, k → ∞.

Using the asymptotic theory of Fourier integrals, we find

(9.86)

and

(9.87)

The asymptotically dominant solutions are y7,k and y8,k . The asymptotically minimal solutions are y5,k andy6,k. We may conclude that forward and backward recursion are asymptotically unstable. However, theinstability of forward recursion is less pronounced if k ≤ ω/2. Indeed, practical experiments demonstratethat Mk(ω, v, α) can be computed accurately using forward recursion for k ≤ ω/2. For k > ω/2 the lossof significant figures increases and forward recursion is no longer applicable. In that case, Oliver’salgorithm14 has to be used. This means that (9.80) has to be solved as a boundary value problem withsix initial values and two end values. The solution of this boundary value problem requires the solutionof a linear system of equations having a band structure.

An important advantage of the MCC method is that the function evaluations of g, needed for thecomputation of the coefficients ck of the Chebyshev series expansion, are independent of the value of ω.Consequently, the same function evaluations may be used for different values of ω, and have to becomputed only once.

Numerical examples can be found in References 19 and 20.

9.10 Computation of Bessel Function Integrals over an Infinite Interval

In this section we consider methods for the computation of

. (9.88)

y k

y k

y k

y k

nk

yk

e k

y yk

e k

k

k

k

k

k

k

k

k k

k

k

12

24

32 1 2

42 1 2

6

7 8

0

0

4

4

,

,

,( )

,( )

,

, ,

~

~

~

~ ,

,

~ ~

~ ~

−

−

− + −

− + +

−

−

≠

=

α ν

α ν

α

α

α

ν

ν

ω

ω

if

~ k if

y

2( +1)

5,k

l

M J k

k

k

k

( , , ) ~ ( )

( )( )

cos ( ) ( ) .

ω ν α ω

ων

π α α

ν

ν αν

α ν

−

+ −+

+[ ] +

−

− − − − − −

1

2

1 21

1 2 2

2

3 2 1 2 2 2

ΓΓ

I p J px f x dx( , ) ( ) ( )ν ν=∞

∫0

9.10.1 Integration between the Zeros of Jv (x) and Convergence Acceleration

We have

(9.89)

where

(9.90)

Using Euler’s transformation,4 the convergence of series (9.89) can be accelerated

. (9.91)

It is not always desirable to start the convergence acceleration with I1, but with some later term, say Im,so that

.

Other convergence accelerating methods, for example the �-algorithm,23 are also applicable (for anexample, see Reference 21).

9.10.2 Transformation into a Double Integral

Substituting the integral expression

(9.92)

into (9.88) and changing the order of integration, we obtain

(9.93)

where

. (9.94)

We assume that the integral in (9.94) is convergent. If we want to evaluate (9.93) using an N-point Gauss-Jacobi rule, then we have to compute the Fourier integral (9.94) for N values of t. Because F(t) shows apeaked or even a singular behavior especially when ƒ(x) is slowly decaying, a large enough N has to bechosen.

This method is closely related to Linz’s method,8 which is based on the Abel transformation of I (p,v).

I p Ikk

k

( , ) ( )ν = − +

=

∞

∑ 1 1

1

I J px f x dxkj p

j p

k

k

=−

∫ν

ν

ν,

,

( ) ( ) .1

I p I I I I( , )ν = − + − +1

2

1

4

1

8

1

161 12

13

1∆ ∆ ∆ L

I p J px f x dx I I Ij p

mm m m

m

( , ) ( ) ( ) ( ),

νν

ν= + − − + −

−

∫ −

0

1 21

11

2

1

4

1

8∆ ∆ L

J xx

t xt dtν

νν

ν π( )

( / )

( / )( ) cos( )/=

+−∫ −2

2

1 21

0

12 1 2

Γ

I pp

t F t dt( , )( / )

( / )( ) ( )/ν

ν π

νν=

+−∫ −2 2

1 21

0

12 1 2

Γ

F t x f x pxt dx( ) ( )cos( )=∞

∫0 ν


9.10.3 Truncation of the Infinite Interval

If a is an arbitrary positive real number, we can write

(9.95)

where

. (9.96)

The first integral in the right side of (9.95) can be computed using the methods of Section 9.9.If a is sufficiently large and ƒ is strongly decaying, then we may neglect R(p,a).If

(9.97)

is an asymptotic series approximation which is sufficiently accurate in the interval [a, ∞), then

. (9.98)

Longman9 has tabulated the values of the integrals in (9.98) for some values of v and ap.Using Hankel’s asymptotic expansion,1 for x → ∞

(9.99)

where χ = px – (v/2+1/4)π, and where Pv(x) and Qv(x) can be expressed as a well-known asymptoticseries, R(p,a) can be written as the sum of two Fourier integrals.

Especially, if a = (8 + v/2 + 1/4)π/p, we have

(9.100)

For the computation of the Fourier integrals in (9.100), tailored methods are available.4

9.11 Tables of Hankel Transforms

Table 9.1 lists the Hankel transform of some particular functions for the important special case v = 0.Table 9.2 lists Hankel transforms of general order v. In these tables, h(x) is the unit step function, Iv andKv are modified Bessel functions, L0 and H0 are Struve functions, and Ker and Kei are Kelvin functionsas defined in Abramowitz and Stegun.1 Extensive tables are given by Erdélyi et al.,6 Ditkin and Prudnikov,5

Luke,11 Wheelon,27 Sneddon,24 and Oberhettinger.13

I p J px f x dx R p aa

( , ) ( ) ( ) ( , )ν ν= +∫0

R p a J px f x dxa

( , ) ( ) ( )=∞

∫ ν

f xc

x

c

x( ) ~ 1 2

2+ +L

R p a cJ px

xdxk

ak

k

( , )( )

�∞

∫∑ ν

J pxpx

P px Q pxν ν νπχ χ( ) ~ [ ( )cos ( )sin ]

2 −

R p ap u a

P p u a f u a pu du

p u aQ p u a f u a pu du

( , )( )

( ( )) ( )cos

( )( ( )) ( )sin .

=+

+ +

−+

+ +

∞

∞

∫

∫

0

0

2

2

π

π

ν

ν


TABLE 9.1 Hankel Transforms of Order 0.


TABLE 9.2 Hankel Transforms of General Order v.


References

1. M. Abramowitz and I. N. Stegun, Handbook of Mathematical Functions, Dover Publications, NewYork, 1965.

2. R. N. Bracewell, The Fourier Transform and Its Applications, McGraw-Hill, New York, 1978.3. B. Davies, Integral Transforms and Their Applications, Springer-Verlag, New York, 1978.4. P. J. Davis and P. Rabinowitz, Methods of Numerical Integration, Academic Press, New York,

1975.5. V. A. Ditkin and A. P. Prudnikov, Integral Transforms and Operational Calculus, Pergamon Press,

Oxford, 1965.6. A. Erdélyi, W. Magnus, F. Oberhettinger, and F. G. Tricomi, Tables of Integral Transforms, Vol. 2,

McGraw-Hill, New York, 1954.7. A. Gray and G. B. Mathews, A Treatise on Bessel Functions and Their Applications to Physics, Dover

Publications, New York, 1966.8. P. Linz, A method for computing Bessel function integrals, Math. Comp., 20, pp. 504-513, 1972.9. I. M. Longman, A short table of J0(t)t–n dt and J1(t)t–n dt, Math. Tables Aids Comp., 13, pp.

306-311, 1959.10. I. M. Longman, A method for the numerical evaluation of finite integrals of oscillatory functions,

Math. Comp., 14, pp. 53-59, 1960.11. Y. L. Luke, The Special Functions and their Approximations, Vol. 2, Academic Press, New York,

1969.12. Y. L. Luke, Integrals of Bessel Functions, McGraw-Hill, New York, 1962.13. F. Oberhettinger, Table of Bessel Transforms, Springer-Verlag, Berlin, 1971.14. J. Oliver, The numerical solution of linear recurrence relations, Num. Math., 11, pp. 349-360,

1968.15. F. W. J. Olver, Bessel Functions, Part III: Zeros and Associated Values, Cambridge University Press,

Cambridge, U.K., 1960.16. R. Piessens, Gaussian quadrature formulas for integrals involving Bessel functions, Microfiche

section of Math. Comp., 26, 1972.17. R. Piessens, Automatic computation of Bessel function integrals, Comp. Phys. Comm., 25, pp. 289-

295, 1982.18. R. Piessens, On the computation of zeros and turning points of Bessel functions, in E. A. Lipitakis

(Ed.), Advances on Computer Mathematics and Its Applications, World Scientific, Singapore, pp. 53-57, 1993.

19. R. Piessens and M. Branders, Modified Clenshaw-Curtis method for the computation of Besselfunction integrals, Bit 23, pp. 370-381, 1983.

20. R. Piessens and M. Branders, A survey of numerical methods for the computation of Bessel functionintegrals, Rend. Sem. Mat. Univers. Politecn. Torino, Fascicolo speciale, pp. 250-265, 1985.

21. R. Piessens, E. de Doncker-Kapenga, C. W. Ueberhuber, and D. K. Kahaner, Quadpack, A SubroutinePackage for Automatic Integration, Springer-Verlag, Berlin, 1983.

22. F. E. Relton, Applied Bessel Functions, Dover Publications, New York, 1965.23. D. Schanks, Non-linear transformations of divergent and slowly convergent sequences, J. Math.

Phys., 34, pp. 1-42, 1955.24. I. N. Sneddon, The Use of Integral Transforms, McGraw-Hill, New York, 1972.25. C. J. Tranter, Integral Transforms in Mathematical Physics, Methuens & Co., London, 1951.26. G. N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge University Press, Cambridge,

U.K., 1966.27. A. D. Wheelon, Table of Summable Series and Integrals Involving Bessel Functions, Holden-Day, San

Francisco, 1968.28. K. B. Wolf, Integral Transforms in Science and Engineering, Plenum Press, New York, 1979.

∫ ∞x

∫ ∞x


piessens, r. “the hankel transform.” the transforms and ... · two-dimensional fourier...

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