Radial Delta Function, and the 3-D Fourier-Bessel - Gauge-institute ...

Gauge Institute Journal 

H. Vic Dannon 

Radial Delta Function and the 

3-D Fourier-Bessel Transform 

H. Vic Dannon 

vic0@comcast.net 

October, 2010 

Abstract In [Dan6], we have shown that 

δ(,, xyz) = δ()()() 

xδ 

yδ 

z 

is the 3-dimensional Fourier Transform of the function 1. Here, 

we show that its Radial form is the 3-dimensional Fourier-Bessel 

Transform of the function 1, 

ν=∞ 

ν= 

0 

( πνr 

) 

sin 2 

δ( x) δ( y) δ( z) = 4∫ νdν, 

r 

2 2 

r = x + y + z 

2 . 

Thus, 

δ( x) δ( y) δ( z) 

is a Radially Symmetric Delta Function that we 

will denote δ r . 

Similarly, 

Radial () 

ν=∞ 

2 1 

δ( x −ξ) δ( y −η) δ( z − ζ) = 4 sin( 2πνr) sin( 2πνσ) 

rσ 

∫ dν 

ν= 

0 

is Radially symmetric, and defines the Radial Delta δ Radial ( r − σ) 

. 

The formulas above are exclusively Hyper-real. They cannot be 

derived by means of the Calculus of Limits, and are unknown in 

the Calculus of Limits. 

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H. Vic Dannon 

The Radial Delta Function 

δ Radial (r − σ) 

is a Discontinuous, 

Hyper-real function, that spikes at r = σ , and vanishes for σ ≠ r . 

The Fourier-Bessel Integral Theorem for Radially symmetric 

function f () r guarantees that the 3-dimesional Fourier-Bessel 

Transform and its Inverse are well defined operations, so that 

inversion yields the original function that generated the 

Transform. 

It is believed to hold in the Calculus of Limits under given 

conditions. In fact, it does not hold in the Calculus of Limits 

because the integration of the Fourier Integral requires the 

ν=∞ 

∫ 

( ) ( ) 

integration of sin 2πνσ sin 2πνr dν 

that diverges at σ = r . 

ν= 

0 

Only in Infinitesimal Calculus, we can integrate over 

singularities, and the Fourier Integral Theorem holds. 

⎛ 

⎞ ⎟ 

f () r = 4 ⎜4 f() 

d ⎟ 

⎝ 

⎠ 

( πνσ ) ( πνr 

) 

ν=∞ σ=∞ 

sin 2 sin 2 

⎜ 

2 2 

σ σ σ 

∫ ∫ ν d 

νσ 

νr 

ν= 0⎜ 

σ= 

0 

⎟ 

σ=∞ ⎛ ν=∞ 

⎞ 2 1 

2 

= f ( σ) 4 sin( 2πνσ) sin( 2πνr) 

d d 

rσ 

σ= 0 

⎜ 

⎝ ν= 

0 

⎠⎟ 

∫ ∫ ν σ σ 

σ=∞ ⎛ ν=∞ 

⎞ 2 


d d 

σ= 0 

⎜ 

⎝ ν= 

0 

⎠⎟ 

∫ ∫ ν σ. 

Keywords: Infinitesimal, Infinite-Hyper-real, Hyper-real, 

ν 

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H. Vic Dannon 

Cardinal, Infinity. Non-Archimedean, Non-Standard Analysis, 

Calculus, Limit, Continuity, Derivative, Integral, 

2000 Mathematics Subject Classification 26E35; 26E30; 

26E15; 26E20; 26A06; 26A12; 03E10; 03E55; 03E17; 03H15; 

46S20; 97I40; 97I30. 

Contents 

Introduction 

1. Hyper-real Line 

2. Hyper-real Integral 

3. Delta Function 

4. The Fourier Transform 

δ( r −r 

) δ( θ −θ ) δ( 

φ−φ 

0 

) 

5. 

0 0 

6. Radial Delta δ Radial 

() r 

7. Radial Delta δ ( − σ) 

Radial 

r 

8. δRadial( 

r − r0 

) and δ( r − r0 

) 

9. 3-D Fourier-Bessel Transform 

10. Fourier-Bessel Integral Theorem holds only in Infinitesimal 

Calculus 

References 

3


H. Vic Dannon 

Introduction 

It is believed that a radially symmetric f () r , satisfies the 3- 

dimensional Fourier-Bessel Integral Theorem 

⎛ 

⎞ ⎟ 

f () r = 4 ⎜4 f() 

d ⎟ 

⎝ 

⎠ 


) 

ν=∞ σ=∞ 

sin 2 sin 2 

⎜ 

2 2 

σ σ σ 

∫ ∫ ν d 

νσ 

νr 

ν= 0⎜ 

σ= 

0 

⎟ 

σ=∞ ⎛ ν=∞ 

⎞ 2 1 

2 


d d 

rσ 

σ= 0 

⎜ 

⎝ ν= 

0 

⎠⎟ 

∫ ∫ ν σ σ 

However, in the Calculus of Limits, at σ = r , 

ν=∞ 

ν= 

0 

( ) ( ) 

∫ sin 2πνσ sin 2πνr dν 

=∞. 

ν 

Avoiding the singularity at 

σ = r 

does not recover the Fourier- 

Bessel Integral Theorem, because without the singularity the 

integral equals zero. 

Thus, the Fourier Integral Theorem cannot be written in the 

Calculus of Limits. 

In Infinitesimal Calculus [Dan4], the singularity can be integrated 

over, and defines the Spherical Delta Function δ Radial ( r − σ) 

. 

Then, for any Hyper-real function f () r , the Fourier-Bessel 

Integral Theorem is the sifting property of δ Radial ( r − σ) 

. 

In the calculus of limits, 

δ Radial (r − σ) 

cannot be defined. 

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H. Vic Dannon 

1. 

Hyper-real Line 

Each real number α can be represented by a Cauchy sequence of 

rational numbers, ( r , r , r ,...) so that r → α . 

1 2 3 

The constant sequence ( ααα , , ,...) is a constant Hyper-real. 

In [Dan2] we established that, 

1. Any totally ordered set of positive, monotonically decreasing 

n 

to zero sequences 

infinitesimal Hyper-reals. 

( ι1, ι2, ι3,...) 

constitutes a family of 

2. The infinitesimals are smaller than any real number, yet 

strictly greater than zero. 

1 1 1 

3. Their reciprocals ( , , ,... 

ι 1 

ι 2 

ι 3 

) are the infinite Hyper-reals. 

4. The infinite Hyper-reals are greater than any real number, 

yet strictly smaller than infinity. 

5. The infinite Hyper-reals with negative signs are smaller 

than any real number, yet strictly greater than −∞. 

6. The sum of a real number with an infinitesimal is a 

non-constant Hyper-real. 

7. The Hyper-reals are the totality of constant Hyper-reals, a 

family of infinitesimals, a family of infinitesimals with 

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H. Vic Dannon 

negative sign, a family of infinite Hyper-reals, a family of 

infinite Hyper-reals with negative sign, and non-constant 

Hyper-reals. 

8. The Hyper-reals are totally ordered, and aligned along a 

line: the Hyper-real Line. 

9. That line includes the real numbers separated by the nonconstant 

Hyper-reals. Each real number is the center of an 

interval of Hyper-reals, that includes no other real number. 

10. In particular, zero is separated from any positive real 

by the infinitesimals, and from any negative real by the 

infinitesimals with negative signs, −dx . 

11. Zero is not an infinitesimal, because zero is not strictly 

greater than zero. 

12. We do not add infinity to the Hyper-real line. 

13. The infinitesimals, the infinitesimals with negative 

signs, the infinite Hyper-reals, and the infinite Hyper-reals 

with negative signs are semi-groups with 

respect to addition. Neither set includes zero. 

14. The Hyper-real line is embedded in , and is not 

∞ 

homeomorphic to the real line. There is no bi-continuous 

one-one mapping from the Hyper-real onto the real line. 

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H. Vic Dannon 

15. In particular, there are no points on the real line that 

can be assigned uniquely to the infinitesimal Hyper-reals, or 

to the infinite Hyper-reals, or to the non-constant Hyperreals. 

16. No neighbourhood of a Hyper-real is homeomorphic to 

an 

n 

ball. Therefore, the Hyper-real line is not a manifold. 

17. The Hyper-real line is totally ordered like a line, but it 

is not spanned by one element, and it is not one-dimensional. 

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H. Vic Dannon 

2. 

Hyper-real Integral 

In [Dan3], we defined the Hyper-real integral of a Hyper-real 

Function. 

Let f () x be a Hyper-real function on the interval [ ab] , . 

The interval may not be bounded. 

f () x may take infinite Hyper-real values, and need not be 

bounded. 

At each 

a 

≤ 

x 

≤b, 

there is a rectangle with base 

dx dx 

[ x − , x + 2 

], height f () x , and area 

2 

f ( xdx. ) 

We form the Integration Sum of all the areas for the x ’s that 

start at x = a, and end at x = b, 

∑ f ( xdx ) . 

x∈[ a, b] 

If for any infinitesimal dx , the Integration Sum has the same 

Hyper-real value, then f () x is integrable over the interval [ ab] , . 

Then, we call the Integration Sum the integral of f () x from x = a, 

to x 

= b, and denote it by 

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H. Vic Dannon 

x= 

b 

∫ f ( xdx ) . 

x= 

a 

If the Hyper-real is infinite, then it is the integral over [, ab] , 

If the Hyper-real is finite, 

x= 

b 

∫ fxdx ( ) = real part of the hyper-real . 

x= 

a 

2.1 The countability of the Integration Sum 

In [Dan1], we established the equality of all positive infinities: 

We proved that the number of the Natural Numbers, 

Card , equals the number of Real Numbers, 

2 Card 

Card = , and 

we have 

2 Card 

2 

Card 

Card = ( Card) = .... = 2 = 2 = ... ≡ ∞. 

In particular, we demonstrated that the real numbers may be 

well-ordered. 

Consequently, there are countably many real numbers in the 

interval [ ab] , , and the Integration Sum has countably many terms. 

While we do not sequence the real numbers in the interval, the 

summation takes place over countably many f ( xdx. ) 

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H. Vic Dannon 

The Lower Integral is the Integration Sum where f ( x ) is replaced 

by its lowest value on each interval 

2.2 

∑ 

x∈[ a, b] 

⎛ 

⎜⎝ 

dx dx 

2 2 

[ x − , x + ] 

⎞ 

inf f ( t) 

dx 

⎠⎟ 

x− ≤t≤ x+ 

dx dx 

2 2 

The Upper Integral is the Integration Sum where f ( x ) is replaced 

by its largest value on each interval 

2.3 

∑ 

x∈[ a, b] 

⎛ 

⎜⎝ 

dx dx 

2 2 

[ x − , x + ] 

⎞ sup f ( t) 

dx 

⎠⎟ 

x− ≤t≤ x+ 

dx dx 

2 2 

If the integral is a finite Hyper-real, we have 

2.4 A Hyper-real function has a finite integral if and only if its 

upper integral and its lower integral are finite, and differ by an 

infinitesimal. 

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H. Vic Dannon 

3. 

Delta Function 

In [Dan5], we defined the Delta Function, and established its 

properties 

1. The Delta Function is a Hyper-real function defined from the 

Hyper-real line into the set of two Hyper-reals 

⎧ 

⎪ 1 ⎫ 

⎨0, ⎪ 

⎬ 

⎪⎩ 

dx 

⎭⎪ . The 

Hyper-real 0 is the sequence 

0, 0, 0,... . The infinite Hyperreal 

1 

dx 

depends on our choice of dx . 

2. We will usually choose the family of infinitesimals that is 

spanned by the sequences 

1 

n , 1 

2 

n 

, 

1 

n 

3 

,… It is a 

semigroup with respect to vector addition, and includes all 

the scalar multiples of the generating sequences that are 

non-zero. That is, the family includes infinitesimals with 

negative sign. Therefore, 

1 

dx 

will mean the sequence n . 

Alternatively, we may choose the family spanned by the 

sequences 

1 

2 n , 

1 

3 n , 

1 

4 n ,… Then, 1 

dx 

will mean the 

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H. Vic Dannon 

sequence 2 n 

. Once we determined the basic infinitesimal 

dx , we will use it in the Infinite Riemann Sum that defines 

an Integral in Infinitesimal Calculus. 

3. The Delta Function is strictly smaller than ∞ 

4. We define, 

1 

χ δ ( x) ≡ dx ( ) 

, 

dx x 

dx 

⎡ ⎤ , 

⎢− 

⎣ 2 2 ⎥⎦ 

where 

χ ⎡ 

⎢− 

⎣ 

dx, 

dx 

2 2 

⎧ dx dx 

1, x ∈ ⎡− 

, ⎤ 

( x) 

= ⎪ ⎢ 2 2 ⎥ 

⎨ ⎣ ⎦ . 

⎪⎪ 0, otherwise 

⎩ 

⎤ 

⎥⎦ 

5. Hence, 

for x < 0 , δ ( x) = 0 

at 

for 

dx 

x =− , δ( x) 

jumps from 0 to 

2 

dx dx 

⎢ ⎣ 

, 

2 2 ⎥ ⎦ , 1 

( x) 

x ∈ ⎡− 

⎤ 

δ = . 

dx 

1 

dx , 

at x = 0 , 

δ (0) = 

1 

dx 

at 

dx 

x = , δ( x) 

drops from 

2 

for x > 0 , δ ( x) = 0. 

xδ ( x) = 0 

1 

dx to 0. 

6. If dx = 

1 

, ( x) = 1 1( x),2 1 1( x),3 1 1( x )... 

n 

[ − , ] [ − , ] [ − , ] 

δ χ χ χ 

2 2 4 4 6 6 

7. If dx = 

2 

, 

n 

1 2 3 

δ ( x) = , , ,... 

2 2 2 

2 cosh x 2 cosh 2x 2 cosh 3x 

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H. Vic Dannon 

8. If dx = 

1 

, 

n 

− x − 2x − 3x 

[0, ∞) [0, ∞) [0, ∞) 

δ( x) = e χ ,2 e χ , 3 e χ ,... 

x =∞ 

∫ 

9. δ( xdx ) = 1. 

x =−∞ 

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H. Vic Dannon 

4. 

The Fourier Transform 

In [Dan6], we defined the Fourier Transform and established its 

properties 

1. F { x } 

δ ( ) = 1 

2. δ( x) 

= the inverse Fourier Transform of the unit function 1 

ω=∞ 

1 

= 

2π 

∫ 

ν=∞ 

ω=−∞ 

e 

2πix 

iωx 

dω 

= ∫ e dν 

, ω = 2πν 

ν=−∞ 

3. 

1 

2π 

ω=∞ 

∫ 

ω=−∞ 

e 

iωx 

dω 

x = 0 

= 

1 = an infinite Hyper-real 

dx 

ω=∞ 

∫ 

ω=−∞ 

e 

iωx 

dω 

x ≠0 

= 

0 

4. Fourier Integral Theorem 

k =∞ ⎛ ξ=∞ 

⎞ 

1 

−ikξ 

f ( x) = f( ξ) 

e ξ 

2π 

k =−∞ ⎜ 

⎝ξ=−∞ 

⎠⎟ 

ikx 

∫ ∫ d e dk 

does not hold in the Calculus of Limits, under any 

conditions. 

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H. Vic Dannon 

5. Fourier Integral Theorem in Infinitesimal Calculus 

If f ( x ) is Hyper-real function, 

Then, 

the Hyper-real Fourier Integral Theorem holds. 

x =∞ 

−iαx 

∫ f ( xe ) dx converges to F( α) 

x =−∞ 

 

1 

2π 

α=∞ 

−iαx 

∫ F( α) 

e dα converges to f ( x ) 

α=−∞ 

6. 3-Dimesional Fourier Transform 

F 

{ } 

z=∞ y=∞ x=∞ 

−iω x−iω y−iω 

z 

x y z 

f (,, xyz) fxyze (,, ) dxdydz 

= ∫ ∫ ∫ 

z=−∞ y=−∞ x=−∞ 

z=∞ y=∞ x=∞ 

− 2 πi( ν x+ ν y+ 

ν z) 

x y z 

= ∫ ∫ ∫ f (,, x y z) e 

dxdydz , 

z=−∞ y=−∞ x=−∞ 

ω 

ω 

ω 

x 

y 

z 

= 2πν 

= 2πν 

= 2πν 

x 

y 

z 

8. 3-Dimesional Inverse Fourier Transform 

−1 

ωz=∞ ωy 

=∞ ωx=∞ 

1 

i( ωxx+ ωyy+ 

ωzz) 

x y z 3 

x y z x y z 

(2 π) 

ω =−∞ ω =−∞ ω =−∞ 

{ } = ∫ ∫ ∫ 

F F( ω , ω , ω ) F( ω , ω , ω ) e 

dω dω dω 

z y x 

ν =∞ ν =∞ ν =∞ 

z 

∫ ∫ ∫ 

2 πi( νxx+ νyy+ 

νzz) 

= F(2 πν ,2 πν ,2 πν ) e dν dν dν 

, 

ν =−∞ ν =−∞ ν =−∞ 

z y x 

y 

x 

x y z x y 

z 

ω 

ω 

ω 

x 

y 

z 

= 2πν 

= 2πν 

= 2πν 

x 

y 

z 

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H. Vic Dannon 

9. 3-Dimesional Fourier Integral Theorem 

ωz=∞ ωy 

=∞ ωx=∞ ⎛ ζ=∞ η=∞ ξ=∞ 

⎞ 

1 

−iωξ x 

−iωη y 

−iωζ 

z 

fxyz (,, ) = f(, ξηζ , ) e ddd ξ ηζ 

3 ∫ ∫ ∫ × 

(2 π) 

∫ ∫ ∫ 

ω =−∞ ω =−∞ ω =−∞ 

⎜⎝ζ=−∞ η=−∞ ξ=−∞ 

⎠⎟ 

z y x 

x 

i( ω x+ iω y+ 

iω 

z) 

x y z 

× e d d d 

y 

ω ω ω 

x y z 

ζ=∞ η=∞ ξ=∞ ⎛ ω 

1 x =∞ 

⎞ ⎛ ωy 

=∞ 

⎞ 

iω 

( ) 1 

i 

y( y ) 

( , , ) 

x 

x−ξ 

ω −η 

= ∫ ∫ ∫ f ξηζ e dω 2 

x 

dξ e dω 2 

y 

η× 

π 

∫ π 

∫ ζ=−∞ η=−∞ ξ=−∞ ⎜ d 

⎝ ω =−∞ ⎠⎟ ⎜⎝ ω =−∞ 

⎠⎟ 

⎛ ωz 

=∞ 

⎞ 1 

iωz 

( z−ζ) 

e dω 

×⎜ 

⎜ 

z 

ζ 

2π 

∫ 

d 

⎜ 

⎝ ω =−∞ 

⎠⎟ 

ζ=∞ η=∞ ξ=∞ ⎛ νx 

=∞ 

⎞ ⎛ νy 

=∞ 

⎞ 

2 πν i ( ) 

2 i 

y( y ) 

( , , ) 

x 

x−ξ 

πν −η 

= ∫ ∫ ∫ f ξηζ e dν x 

dξ e dν y 

dη× 

∫ ∫ 

ζ=−∞ η=−∞ ξ=−∞ ⎝⎜ ν =−∞ ⎠⎟ ⎜⎝ν 

=−∞ 

⎠⎟ 

x 

y 

z 

10. 3-Dimesional Delta Function 

⎛ νz 

=∞ 

⎞ 2 πν i 

z 

( z−ζ) 

e dν 

×⎜ 

⎜ 

z 

dζ 

∫ , 

⎜⎝ν 

=−∞ 

⎠⎟ 

z 

ω 

ω 

ω 

x 

y 

z 

= 2πν 

= 2πν 

= 2πν 

x 

y 

z 

⎛ ωx 

=∞ ⎞⎛ 

ωy 

=∞ ⎞⎛ 

ωz 

=∞ 

1 i 1 i 1 

yy 

( xyz , , 

xx i 

zz 

) 

x 

y 

2 e ω 

d 2 e ω 

d 2 

e ω 

δ = ω ω 

π π π 

d ω 

⎝⎜ 

ω =−∞ ⎠⎟ 

⎜⎝ 

ω =−∞ ⎠⎟ 

⎝⎜ 

ω =−∞ 

∫ ∫ ∫ z 

x y z 

⎞ 

⎠⎟ 

⎛ ν 

ν =∞ 

x=∞ ⎞⎛ 

y 

⎞⎛ 

νz=∞ 

πν 

πν 

2 i πν 

e dν 

2 

= 

xx i 

yy 

e dν 

2 i e 

zz 

d 

∫ 

⎜ 

∫ ∫ ν 

⎝ν =−∞ ⎠⎟⎜ 

⎝ν =−∞ ⎠⎟ 

⎝⎜ν 

=−∞ 

x y z 

x y z 

⎞ 

, 

⎠⎟ 

ω 

ω 

ω 

x 

y 

z 

= 2πν 

= 2πν 

= 2πν 

x 

y 

z 

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H. Vic Dannon 

5. 

δ( r −r 

) δ( θ −θ ) δ( 

φ −φ 

0 

) 

0 0 

5.1 

1 

δ ( r − r0 ) = χ[ dr 

0 , dr 

2 0 

] ( ) 

r − r + 

dr r , r ≥ 0 

2 

5.2 

1 

δθ ( − θ 

0 ) = χ[ dθ 

0 , dθ 

2 0 

] ( θ) 

, 0 ≤ θ ≤ π 

θ − θ + 

dθ 2 

5.3 

1 

δφ ( − φ 

0 ) = χ[ dφ 

( ) 

0 , dφ 

φ, 

0 ≤ φ ≤ 2π 

dφ φ − φ 

2 0+ 

] 2 

The product 

δ( r −r 

) δ( θ −θ ) δ( 

φ− 

0 0 

0 ) 

φ 0 ) 

defines a Delta Function 

that sifts along a line through its singularity at ( r , θ , φ ). 

0 0 0 

5.4 

0 0 0 0 0 

δ( r −r , θ −θ , φ−φ ) ≡ δ( r −r 

) δ( θ −θ ) δ( 

φ− 

φ 

= 

1 χ 

1 1 

[ dr 

0 , dr 

2 0 ] () r χ[ 

2 0 , 2 0 ] () () 

r r 

dθ dθ θ χ 

2 [ dφ 0 , dφ 

φ 

− + θ − θ + 

dr dθ 

dφ φ − φ 

2 0+ 

] 2 

Transforming between Spherical and Cartesian Coordinates 

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H. Vic Dannon 

x 

y 

z 

= r sin θcosφ 

= r sin θsin 

φ , y = r sin θ sin φ , 

= r cos θ 

x 

z 

= r 

sin θ 

0 0 0 

0 0 0 

= r 

cosθ 

0 0 0 

cosφ 

0 

0 

r 

0 

> 0 

1 

δ( − ) δ( − ) δ( − ) = δ( − ) δ( θ −θ ) δ( 

φ−φ 

0 

) 

5.5 x x0 y y0 z z0 r r 

2 

0 0 

r0 sin θ0 

Proof: 

δ( r −r ) δ( θ −θ ) δ( φ− φ ) drdθdφ = δ( x −x ) δ( y −y ) δ( z −z ) dxdydz 

0 0 0 0 0 0 

∂(,, xyz) 

δ( r −r0) δ( θ −θ0) δ( φ − φ0) = δ( x −x0) δ( y −y0) δ( z −z0) 

∂(, r θφ , ) 

 

Both sides vanish unless 

r = r +infinitesimal ≈ r , 

0 0 

and 

θ = θ +infinitesimal ≈ θ 

0 0 . 

Therefore, we can replace r with , and θ with θ . 

r0 

0 

r 

2 

sin θ 

2 

0 0 0 0 0 0 0 φ 0 ) 

δ( x −x ) δ( y −y ) δ( z − z ) r sin θ = δ( r −r 

) δ( θ −θ ) δ( 

φ− 

. 

x = r sin θcosφ 

5.6 y = r sin θsin 

φ , 

z = r cos θ 

r 

0 

= 0 ⇒ 

1 

δ()()() x δ y δ z = δ() 

r 

2 

4πr 

Proof: 

δ()()() r δ θ δ φdrdθdφ = δ()()() 

x δ y δ z dxdydz 

= 

4 

1 

dr dr 

2 2 

χ 2 [ − , ] 

πr dr 

() r 

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H. Vic Dannon 

∂(,, xyz) 

δ()()() r δ θ δ φ = δ()()() 

x δ y δ z 

∂(, r θφ , ) 

 

r 

2 sin θ 

Since r = , φ may take any value in [0,2 π] 

, θ may take any 

0 

0 

value in [0, π] 

, and we integrate over them. Then, 

2 

φ= 2π φ= 

2π 

∫ 

r sin θdθ dφδ( x)()() δ y δ z = δ()() r δ θ dθ δ( φ) 

dφ 

2 

φ= 0 φ= 

0 

 

2π 

1 

θ= π θ= 

π 

2 πr δ( x)()() δ y δ z sin θdθ = δ() r δ() 

θ dθ. 

2 

∫ 

 

θ= 0 θ= 

0 

θ= 

π 

− cos θ = 2 

θ= 

0 

∫ 

4 πr δ( x)()() δ y δ z = δ() 

r . 

 

1 

∫ 

 

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H. Vic Dannon 

6. 

Radial Delta δ Radial () r 

6.1 

ν=∞ 

sin(2 πνr) 

δ( x) δ( y) δ( z) = 4∫ 

νdν 

r 

ν= 

0 

Proof: By 4.10, 

⎛ νx=∞ ⎞⎛ 

νy 

=∞ ⎞⎛ 

νz=∞ 

2 i 2 i 

yy 

2 

( ) ( ) ( ) πνxx d i 

zz 

x 

πν 

d y 

πν 

δ δ δ = ν ν d ν 

⎜ 

⎝νx=−∞ ⎠⎟⎜ 

⎝ 

νy=−∞ ⎠⎟ 

⎝⎜νz=−∞ 

Substitute 


νx 

= νsin 

γcos 

β 

y = r sin θsin 

φ νy 

= νsin 

γsin 

β 

z = r cos θ νz 

= νcos 

γ 

Then, 

ν x + ν y + ν z 

x y z 

= 

∫ ∫ ∫ z 

{ } 

= νr 

⎡sin θsin γ cosφcos β + sin φsin β + cos θcos 

γ⎤ 

⎣ 

⎦ 

= νr 

⎡sin θsin γcos( β − φ) + cosθcos 

γ⎤ 

⎣ 

⎦ 

Integrating with respect to ν , γ , and β , 

ν=∞ ⎛γ= π⎡β= 

2π 

⎤ ⎞ 2πν i r sinθsinγcos( β−φ) 2πν i r cosθcosγ 

2 

= e dβ 

e 

sin γdγ ∫ ν dν 

∫ ∫ 

ν= 0⎜ 

⎝ ⎢ 

⎥ 

γ= 0⎣ 

β= 

0 

⎦ 

⎠⎟ 

Denoting 

α = β − φ, 

⎞ 

⎟ 

⎠ 

20


H. Vic Dannon 

β= 2π α= 2π−φ 

2πν i r sin θsin γcos( β−φ) 2πν i r sin θsin γcos 

α 

∫ e dβ 

= ∫ e 

d α 

β= 0 

α=−φ 

2 i r sin sin cos 

Since e πν θ γ α is periodic in α with period 2 π , 

α= 

2π 

α= 

0 

α= 

2π 

= ∫ 

α= 

0 

e 

2πν i r sin θsin γcos 

α 

dα 

⎛ 

2 

2πν i rsinθsinγcos α (2πν i rsinθsinγcos α) 

⎞ 

= 1 . 

∫ ⎜ 

+ + + .. dα 

. 

⎜⎝ 

1 2! 

⎠⎟ 

The integrals of the odd powers vanish, and we have 

2 4 

(2πνr sin θ sin γ) (2πνr sin θ sin γ) (2πνr 

sin θ sin γ) 

= 2π − 2π + 2π − 2π 

+ ... 

2 2 2 2 2 2 

2 2 ⋅4 2 ⋅4 ⋅6 

6 

= 

2 πJ 

(2πνr 

sinθsinγ). 

0 

Therefore, 

ν=∞ ⎛γ= 

π 

⎞ 2πν i r cosθcosγ 

2 

δ( x) δ( y) δ( z) = 2 πJ0(2πνrsin θsin γ) e sin γdγ ∫ ν dν 

∫ 

ν= 0 

⎜ 

⎝γ= 

0 

⎠⎟ 

Denote 

Then, 

u = cos γ , 

du =−sin 

γdγ 

γ = 0 ⇒ u = 1 , 

γ = π ⇒ u = −1 

and the γ summation becomes 

21


H. Vic Dannon 

u= 

1 

∫ 

u=−1 

0 

2 [2πν 

i r cos θ] 

u 

2 π J ([2πνr sin θ] 1 − u ) e du = 

u= 

1 

∫ 

= 8 π J ([2πνr sin θ] 1 −u )cos([2πνr cos θ] u) 

du. 

u= 

0 

0 

2 

Denote 

Then, 

2 

u = 1 − v . 

v 

du =− dv 

2 

1 − v 

u = 0 ⇒ v = 1 , 

u = 1 ⇒ v = 0 

and the u summation becomes 

v= 

1 

∫ 

= 8 π J ([2πνr sin θ] v)cos([2πνrcos θ] 1 −v 

) 

v= 

0 

0 

2 

v 

1 − v 

2 

dv. 

Denoting, 

b 

c 

= 

= 

2πνr 

cosθ 

, 

2πνr 

sinθ 

v= 

1 

2 v 

∫ J0 

cv b v dv. 

2 

v= 

0 

1 − v 

= 8 π ( )cos( 1 − ) 

By [Prudnikov, Vol.2, p.201, 2.12.21 #5], the γ summation equals 

π 

= 8π 

+ + 

−1 

2 2 

( ) ( ) 

4 2 2 

1 

2 b c J b c 

2 

22


H. Vic Dannon 

Since 1 ( ) 

π 1 

= 8π 

J1 

2 ν 

2 2πνr 

2 

1 

= 4π 

J1 

2 ν 

νr 

2 

( π r) 

( π r) 

2 J x = sin x , the γ summation equals 

2 πx 

= 1 2 

4π 

sin 

νr 

π2πνr 

2 ν 

( π r ) 

Therefore, 

= 

1 

4 sin 2 r 

νr 

( πν ) 

ν=∞ 

ν= 

0 

( πνr 

) 

sin 2 

δ( x) δ( y) δ( z) = 4∫ νdν. 

 

r 

Thus, 

δ( x) δ( y) δ( z) 

is a Radially Symmetric Delta Function that we 

will denote δ r . 


6.2 The Radial Delta r Definition 


δ 


ν=∞ 

sin ( 2πνr 

) 

δ 

() r ≡ 4∫ νdν 

r 

ν= 

0 

23


H. Vic Dannon 

7. 

Radial Delta 

δ Radial ( r − σ) 

By 6.2, we have, 

7.1 

δ 


( r −σ) ≡ 4 

ν=∞ 

∫ 

ν= 

0 

( πν r − σ ) 

sin 2 [ ] 

r − σ 

νdν 

Alternatively, we may derive similarly to 6.1, 

ν=∞ 

2 1 

7.2 δ( x −ξ) δ( y −η) δ( z − ζ) = 4 sin(2 πνr)sin(2 πνσ) 

ν 

rσ 

∫ d 


δ( x −ξ) δ( y −η) δ( z − ζ) 

= 

ν= 

0 

ν=∞ 

2 1 

= 

2 ∫ 

r 

ν= 

0 

4 sin(2 πνr)sin(2 πνσ) 

dν 

⎛ νx 

=∞ ⎞⎛ 

νy 

=∞ 

⎞⎛ 

νz 

=∞ 

e 2 πν i ( ) 2 i 

y 

( y ) 

x 

x−ξ d 2 i 

z 

( z ) 

x 

e πν −η 

d y 

e πν −ζ 

= ν 

ν 

∫ ∫ ∫ d 

⎜⎝ ν =−∞ ⎟⎠⎜ 

⎝ν =−∞ ⎠⎟ 

⎜⎝ν 

=−∞ 

x y z 

ν z 

⎞ 

⎠⎟ 

Substitute 


y = r sin θsin 

φ 

z = r cos θ 

Then, 

ξ = σsin 

λcos 

μ 

η = σsin 

λsin 

μ 

ζ = σcosλ 

ν = νsin 

γcos 

β 

x 

ν = νsin 

γsin 

β 

y 

ν = νcos 

γ 

z 

24


H. Vic Dannon 

ν x + ν y + ν z = 

x y z 

νξ 

x 

+ νη 

y 

+ νζ 

z 

= νr 


γ⎤ 

⎣ 

⎦ 

= νr 


γ⎤ 

⎣ 

⎦ 

= 

{ } 

{ } 

= νσ ⎡sinλ sin γ cos μ cos β + sin μ sin β + cosλ cos γ ⎤ 

⎣ 

⎦ 

= νσ ⎡sin λ sin γ cos( β − μ) + cosλ cos γ ⎤ 

⎣ 

⎦ 


ν=∞ ⎛γ= π⎛β= 

2π 

⎞ 2πν i r sin θsin γcos( β−φ) −2πνσ i sin λsin γcos( β−μ) 

= e 

e 

dβ 

∫ × 

∫ ∫ 

ν= 0 ⎜ ⎝γ= 0⎜⎝ β= 

0 

⎠⎟ 

2πν i r cosθcos γ −2πνσ i cosλcos γ 

2 

× e e sin γdγ ν 

) 

dν. 

Denoting 

A = β − φ, B = β −μ, 

β= 

2π 

∫ 

β= 

0 

e 

2πν i r sin θsin γcos( β−φ) −2πνσ i sinλsin γcos( β−μ) 

e 

dβ 

= 

A= 2π− φ B= 2π−μ 

2πν i r sin θsin γcosA −2πνσ i sinλsin γcosB 

= ∫ ∫ 

e dA e 

A=− φ 

B=−μ 

dB 

2πν i rsinθsinγcosA 

−2πνσ i sinλsinγcosB 

e 

Since e , and are periodic with 

period 2 π , 

25


H. Vic Dannon 

A= 2π 

B= 

2π 

2πν i r sin θsin γcosA −2πνσ i sin λsin γcosB 

= ∫ e dA ∫ e 

dB 

A= 0 B= 

0 

Integrating, 

A= 

2π 

∫ 

A= 

0 

e 

2πν i r sinθsinγcosA 

dA 

= 

A= 

2π 

⎛ 

2 

2πν i rsinθsinγcos A (2πν i rsinθsinγcos A) 

⎞ 

= 1 ... 

∫ 

⎜ 

+ + + 

dA 

⎜⎝ 

1 2! 

⎠⎟ 

A= 

0 


2 4 



= 2π − 2π + 2π − 2π 

+ ... 

2 2 2 2 2 2 

2 2 ⋅4 2 ⋅4 ⋅6 

6 

= 

2 πJ 

(2πνrsinθsinγ). 

0 

Similarly, 

B= 

2π 

∫ 

B= 

0 

e 

−2πνσ i sinλsinγcosB 

dB 

= 

B= 

2π 

⎛ 

2 

2πνσ i sinλsin γcos B (2πν i rsinλsin γcos B) 

⎞ 

= 1 . 

∫ ⎜ 

+ + + .. 

dB 

⎜⎝ 

1 2! 

⎠⎟ 

B= 

0 


2 4 

(2πνσsin λsin γ) (2πνσsin λsin γ) (2πνσsin λsin γ) 

= 2π − 2π + 2π − 2π 

+ ... 

2 2 2 2 2 2 

2 2 ⋅4 2 ⋅4 ⋅6 

6 

= 

2 πJ 

(2πνσsinλsinγ). 

0 

26


H. Vic Dannon 

Therefore, the 

γ 

summation is 

γ= 

π 

∫ 

γ= 

0 

2 πJ (2πνr sinθsin γ) 

e 

0 

2πν i r cosθcosγ 

0 

× 

−2 i cos cos 

πνσ λ γ 

× 2 πJ (2πνσsinλsin γ) e sin γdγ 

= 

Denoting, 

b 

c 

1 

1 

= 2πνrcosθ 

b2 

= 2πνσ 

cosλ 

, , 

= 2πνr 

sinθ 

c = 2πνσ 

sinλ 

2 

γ= 

π 

= ∫ 

γ= 

0 

b cos γ 

0 1 0 2 

−b 

cos γ 

2 πJ ( c sin γ) e 

1 

2 πJ ( c sin γ) e 

2 

sin γdγ 

Denote 

Then, 

u = cos γ , 

du =−sin 

γdγ 

γ = 0 ⇒ u = 1 , 

γ = π ⇒ u = −1 


u= 

1 

∫ 

u=−1 

2 bu 

2 

0 1 0 2 

= 2 π ( 1 − ) 2 ( 1 − ) 

−bu 

J c u e 

1 

πJ c u e 

2 

du 

u= 1 u= 

1 

∫ 

2 bu 

π 

0 1 ∫ 0 2 

2 

u=− 1 u=−1 

= 2 ( 1 − ) 2 ( 1 − ) 

−bu 

J c u e 

1 

du πJ c u e 

2 

du 

The first integration is 

27


H. Vic Dannon 

u= 1 u= 

1 

∫ 

2 bu 

2 

0 1 ∫ 0 1 1 

u=− 1 u= 

0 

2 π ( 1 − ) 

1 

= 8 π ( 1 − )cos( ) 

J c u e du J c u bu du 

Denote 

Then, 

2 

u = 1 − v . 

v 

du =− dv 

2 

1 − v 

u = 0 ⇒ v = 1 , 

u = 1 ⇒ v = 0 


v= 

1 

2 v 

∫ J0 c1v b1 

v dv. 

2 

v= 

0 

1 − v 

= 8 π ( )cos( 1 − ) 

By [Prudnikov, Vol.2, p.201, 2.12.21 #5], the v summation equals 

b1 

= 2πνrcosθ 

Recalling , 

c = 2πνrsinθ 

1 

Since 1 ( ) 

π 

= 8π 

+ + 

−1 

2 2 

( ) ( ) 

4 2 2 

1 1 1 1 1 

2 b c J b c 

2 

π 1 

= 8π 

J1 

2 ν 

2 2πνr 

2 

1 

= 4π 

J1 

2 ν 

νr 

2 

( π r) 

( π r) 

2 J x = sin x , the v summation equals 

2 πx 

28


H. Vic Dannon 

= 1 2 

4π 

sin 

νr 

π2πνr 

2 ν 

( π r ) 

That is, 

1 

= 4 sin( 2πνr 

). 

νr 

u= 

1 

( π ) 

2 bu 

1 

∫ 2 π 

0( 1 

1 − ) = 4 sin 2 ν 

u=−1 

1 

J c u e du r 

νr 

Similarly, 

u= 

1 

∫ 

u=−1 

2 

2 ( 1 ) 

2 

4 sin 2 

0 2 

bu 

πJ c − u e du = 

1 

νσ 

( πνσ 

) 

Therefore, the 

γ 

summation is 

and the ν summation is 

4 1 sin 2 4 1 sin 2 

νr 

νσ 

( πνr 

) ( πνσ ) 

ν=∞ ν=∞ 

1 1 2 2 1 

4 sin( 2πνr) 4 sin( 2πνσ ) ν dν = 4 sin( 2πνσ ) sin( 2πν ) ν 

νr 

νσ rσ 

∫ ∫ r d 

ν= 0 ν= 

0 

Therefore, 

ν=∞ 

2 1 

δ( x −ξ) δ( y −η) δ( z − ζ) = 4 sin(2 πνr)sin(2 πνσ) 

dν 

rσ 

∫ . 

ν= 

0 

Thus, δ( x −ξ) δ( y −η) δ( 

z −ζ) 

is a Radially Symmetric Delta 

function that we will denote 

δ Radial ( r − σ) 

29


H. Vic Dannon 

7.3 The Radial Delta δ − ) Definition 

Radial (r 

ν=∞ 

ν= 

0 

σ ( ) ( r) 

2 1 

δRadial( r −σ) ≡ 4 sin 2πνσ sin 2πν 

rσ 

∫ dν 

spikes at r = σ , and vanishes for r ≠ σ . 

From 7.1, and 7.3, 

7.4 

ν 

( πν r − σ ) ( πνσ ) ( πνr) 

ν=∞ =∞ 

sin 2 [ ] sin 2 sin 2 

νdν 

= 4 

r − σ 

rσ 

∫ ∫ 

ν= 0 ν= 

0 

dν 

Common Tables do not have 7.4. 

30


H. Vic Dannon 

8. 

δ ( r − r ) and 

Radial 0 

δ( r − r ) 

0 

δ 

1 

( r − r ) = δ( r − r ) 

8.1 

Radial 0 2 0 

r0 


Proof: Denote y = rsin 

θsin 

φ , y = r sin θ sin φ , 

z = rcos 

θ 

Then, 

x 

z 

= r 

sin θ 

0 0 0 

0 0 0 

= r 

cosθ 

0 0 0 

cosφ 

δ( x −x ) δ( y −y ) δ( z − z ) dxdydz = δ( r −r )( dr)(sin θd θ) 

dφ 

0 0 0 0 

∂(,, xyz) 

δ( x −x0) δ( y −y0) δ( z − z0) = δ( 

r −r 

∂(, r −cos θφ , ) 

0 

0 

0 ) 

Since 

∂(,, xyz) 

∂(, r −cos θφ , ) 

= 

r 

2 

, 

δ( x −x0) δ( y −y0) δ( z − z0) = δ( 

r −r 

2 0 

r 

) 

δ ( r−r 

) 

δ 

Radial 0 

1 

( r − r ) = δ( r − r ). 

 

Radial 0 2 0 

r0 

1 

8.2 Sifting by δ Radial ( r σ) 

− 31


H. Vic Dannon 

σ=∞ 

2 

∫ δRadial( r − σ) 

σ dσ 

= 1 

σ= 

0 

σ=∞ σ=∞ 

∫ 

Proof: By 8.1, δ ( r − σ) σ 2 dσ = δ( r −σ) dσ 

= 1. 

∫ 


σ= 0 σ= 

0 

ν=∞ 

2 

8.3 δ( r − σ) = 4 ∫ sin(2 πνr)sin(2 πνσ) 

dν 

ν= 

0 


= 

4 

ν=∞ 

∫ 

ν= 

0 

( r − ) 

sin 2 πν[ σ] 

ν d ν 

r − σ 

2 

δ( r − σ) = σ δ ( r −σ) 


ν=∞ 

2 2 

= σ 4 ∫ 

ν= 

0 

sin(2 πνr)sin(2 πνσ) 

dν 

rσ 

For r ≠ σ , the integral vanishes. For r = σ , we have 

ν=∞ 

2 2 

= σ 4 ∫ 

By 7.3, 

= 4 

ν= 

0 

ν=∞ 

2 

ν= 

0 

sin(2 πνr)sin(2 πνσ) 

dν 

2 

σ 

= 4 ∫ sin(2 πνr)sin(2 πνσ) 

dν 

ν=∞ 

ν= 

0 

Common Tables do not have 8.3. 

( r − ) 

sin 2 πν[ σ] 

∫ ν d ν . 

r − σ 

32


H. Vic Dannon 

9. 

3-D Fourier-Bessel Transform 

The 3-dimensional Fourier-Bessel Transform is the 3-dimensional 

Fourier Transform applied to a Radially symmetric function f () r . 

Then, integration of the Exponential Function with respect to the 

azimuth angle, yields a Bessel Function. 

9.1 The 3-dimensional Fourier-Bessel Transform 

F 

Bess 

sin 2 

() = 4 ∫ f() 

r 

ν 

{ f r } 

r =∞ 

r = 0 

( πνr 

) 

rdr 

Proof: 

The 3-dimensional Fourier Transform of a radially 

symmetric function f (,, xyz) = fr () is 

F 

Bess 

{ } 

z=∞ y=∞ x=∞ 

−2πν i x−2πν i y−2πν 

i z 

x y z 

f () r f () r e dxdydz 

= ∫ ∫ ∫ 

z=−∞ y=−∞ x=−∞ 

Substitute 

Then, 

x 

y 

z 


= r sin θsin 

φ 

= r cosθ 

ν = νsin 

γcos 

β 

x 

ν = νsin 

γsin 

β 

y 

ν = νcos 

γ 

z 

33


H. Vic Dannon 

ν x + ν y + ν z = 

x y z 

{ } 

= νr 


γ⎤ 

⎣ 

⎦ 

= νr 


γ⎤ 

⎣ 

⎦ 

Integrating with respect to r , θ , and φ , 

F 

Bess 

r =∞ ⎛θ= π⎡φ= 

2π 

⎤ ⎞ −2πν i r sinθsinγcos( φ−β) −2πν i r cosθcosγ 

2 

f () r = f() r ∫ e dφ 

e sinθd 

∫ ∫ 

θ 

r dr 

⎢ 

⎥ 

r = 0 

⎜⎝θ= 0 φ= 

0 

⎠⎟ 

⎣ 

⎦ 

{ } 

Denoting 

α 

= φ− β, 

φ= 2π α= 2π−β 

−2πν i r sinθsinγcos( φ−β) −2πν i r sinθsinγcosα 

∫ ∫ 

e dφ 

= e 

φ= 0 

α=−β 

dα 

−2πν i r sinθsinγcosα 

Since e 

is periodic in α with period 2 π , 

α= 

2π 

α= 

0 

α= 

2π 

= ∫ 

α= 

0 

e 

−2πν i r sin θsin γcos 

α 

dα 

⎛ 

2 

2πν i rinθsinγcos α (2πν i rsinθsinγcos α) 

⎞ 

= 1 ... 

∫ ⎜ 

− + − 

dα 

. 

⎜⎝ 

1 2! 

⎠⎟ 


2 4 



= 2π − 2π + 2π − 2π 

+ ... 

2 2 2 2 2 2 

2 2 ⋅4 2 ⋅4 ⋅6 

6 

= 

2 πJ 


0 

Therefore, 

34


H. Vic Dannon 

F 

Bess 

⎛ 

r =∞ θ= 

π 

⎜ 

= ∫ 

∫ 0 

r = 0 

⎜ 

θ= 

0 

⎟ 

⎞ ⎟ 

r dr 

⎠ 

2 i r cos cos 2 

{ f () r } f()2 r ⎜ 

− πν θ γ 

π J (2πνrsinθsin γ) e sinθdθ⎟ 

⎝ 

Denoting, 

b 

c 

= 

= 

2πνr 

cosγ 

, 

2πνrsinγ 

the 

θ 

summation becomes 

θ= 

π 

−b 

cos θ 

2 π∫ J0( csin θ) e sinθdθ. 

θ= 

0 

Denote 

Then, 

u = cos θ . 

du =−sin 

θdθ 

θ = 0 ⇒ u = 1 , 

θ = π ⇒ u = −1 

and the θ summation becomes 

u= 1 u= 

1 

∫ 

2 bu 

0 

− = π∫ 

0 

− 

2 

u=− 1 u= 

0 

2 π J ( c 1 u ) e du 8 J ( c 1 u )cos( bu) 

du 

Denote 

Then, 

u 

2 

= 1 − v . 

35


H. Vic Dannon 

v 

du =− dv 

2 

1 − v 

u = 0 ⇒ v = 1 , 

u = 1 ⇒ v = 0 


v= 

1 

2 v 

= 8 π ∫ J0( cv)cos( b 1 −v ) dv 

2 

1 − v 

v= 

0 

By [Prudnikov, Vol.2, p.201, 2.12.21 #5], this equals 

b = 2πνr 

cosγ 

Recalling , 

c = 2πνrsinγ 

Since 1 ( ) 

π 

= 8π 

+ + 

−1 

2 2 

( ) ( ) 

4 2 2 

1 

2 b c J b c 

2 

π 1 

= 8π 

J1 

2 ν 

2 2πνr 

2 

1 

= 4π 

J1 

2 ν 

νr 

2 

( π r) 

( π r) 

2 J x = sin x , the v summation equals 

2 πx 

= 1 2 

4π 

sin 

νr 

π2πνr 

2 ν 

( π r ) 

Therefore, 

= 

1 

4 sin 2 r 

νr 

( πν ) 

36


H. Vic Dannon 

F 

Bess 

r =∞ 

{ f () r } f()4 r sin( 2πνr) 2 

= ∫ 

r = 0 

r =∞ 

r = 0 

1 

νr 

( πνr 

) 

r dr 

sin 2 

= 4 ∫ f ( r) 

rdr . 

ν 

9.2 3-D Inverse Fourier-Bessel Transform 

F 

−1 

Bess 

ν=∞ 

ν= 

0 

( πνr 

) 

sin 2 

F(2 πν) = 4 ∫ F(2 πν) 

νdν 

r 

Proof: The 3-dimensional Inverse Fourier-Bessel Transform of a 

radially symmetric function 

F( ν , ν , ν ) = F(2 πν) 

x y z 

is 

F 

νz=∞ νy 

=∞ νx=∞ 

−1 

2 πi[ νxx+ νyy+ 

νzz] 

Bess { F(2 πν) } = ∫ ∫ ∫ F(2 πν) 

e 

dνxdνydνz 

ν =−∞ ν =−∞ ν =−∞ 

Substitute 

z y x 

Then, 

x 

y 

z 


= r sin θsin 

φ 

= r cosθ 

ν = νsin 

γcos 

β 

x 

ν = νsin 

γsin 

β 

y 

ν = νcos 

γ 

z 

ν x + ν y + ν z 

x y z 

= 

{ } 

= νr 


γ⎤ 

⎣ 

⎦ 

= νr 


γ⎤ 

⎣ 

⎦ 


37


H. Vic Dannon 

ν=∞ ⎛γ= π⎡β= 

2π 

⎤ ⎞ 

1 2 i r sin sin cos( ) 2 i r cos cos 2 

Bess { F(2 )} 

− 

πν θ γ β−φ πν θ γ 

F πν = F(2 πν) ∫ e dβ e sin γdγ ν ν 

∫ ∫ 

d 

ν= 0 

⎜ ⎢ 

⎥ 

⎝γ= 0 β= 

0 

⎠⎟ 

⎣ 

⎦ 

Denoting 

α 

= β −φ, 

β= 2π α= 2π−φ 

2πν i r sin θsin γcos( β−φ) 2πν i r sin θsin γcos 

α 

∫ ∫ 

e dβ 

= e 

β= 0 

α=−φ 

dα 

2 i r sin sin cos 

Since e πν θ γ α is periodic in α with period 2 π , 

α= 

2π 

α= 

0 

α= 

2π 

= ∫ 

α= 

0 

e 

2πν i r sin θsin γcos 

α 

dα 

⎛ 

2 

2πν i rsinθsinγcos α (2πν i rsinθsinγcos α) 

⎞ 

= 1 . 

∫ ⎜ 

+ + + .. dα 

. 

⎜⎝ 

1 2! 

⎠⎟ 


2 4 



= 2π − 2π + 2π − 2π 

+ ... 

2 2 2 2 2 2 

2 2 ⋅4 2 ⋅4 ⋅6 

6 

= 

2 πJ 


0 

Therefore, 

ν=∞ ⎛γ= 

π 

⎞ 

−1 2 i r cos cos 

(2 ) (2 ) 2 2 Bess F F J 0(2 r sin sin ) e πν θ γ 

F πν = πν π πν θ γ sin γ d γ ∫ 

ν dν 

∫ 

ν= 0 

⎜⎝γ= 

0 

⎠⎟ 

Denoting, 

b 

c 

= 

= 

2πνr 

cosθ 

, 

2πνr 

sinθ 

the 

γ 

summation is 

38


H. Vic Dannon 

Denote 

Then, 

γ= 

π 

b cos γ 

∫ 2 πJ0( csin γ) e sin γdγ. 

γ= 

0 

u = cos γ 

du =−sin 

γdγ 

γ = 0 ⇒ u = 1 , 

γ = π ⇒ u = −1 


u= 1 u= 

1 

∫ 

2 bu 

0 

− = π∫ 

0 

− 

2 

u=− 1 u= 

0 

2 π J ( c 1 u ) e du 8 J ( c 1 u )cos( bu) 

du 

Denote 

Then, 

2 

u = 1 − v . 

v 

du =− dv 

2 

1 − v 

u = 0 ⇒ v = 1 , 

u = 1 ⇒ v = 0 


v= 

1 

2 v 

= 8 π ∫ J0( cv)cos( b 1 −v ) dv 

2 

1 − v 

v= 

0 

By [Prudnikov, Vol.2, p.201, 2.12.21 #5], this equals 

π 

= 8π 

+ + 

−1 

2 2 

( ) ( ) 

4 2 2 

1 

2 b c J b c 

2 

39


H. Vic Dannon 

Since 1 ( ) 

π 1 

= 8π 

J1 

2 ν 

2 2πνr 

2 

1 

= 4π 

J1 

2 ν 

νr 

2 

2 J x = sin x , 

2 πx 

( π r) 

( π r) 

= 1 2 

4π 

sin 

νr 

π2πνr 

2 ν 

( π r ) 

Therefore, 

= 

1 

4 sin 2 r 

νr 

( πν ) 

ν=∞ 

ν= 

0 

( πνr 

) 

sin 2 

−1 

FBessF(2 πν) = 4 ∫ F(2 πν) 

νd ν . 

r 

40


H. Vic Dannon 

10. 

Fourier-Bessel Integral Theorem 

Holds only in Infinitesimal Calculus 

The 3-dimensional Fourier-Bessel Integral Theorem guarantees 

that the 3-dimensional Fourier-Bessel Transform and its Inverse 

are well defined operations, so that inversion yields the original 

function that generated the Transform. 

That is, 

ν=∞ ⎛ σ=∞ 

sin 2 

⎞ sin 2 

f () r = 4 4 f() 

σ 

σdσ 

ν 

r 

ν= 0⎜ 

⎝ σ= 

0 

⎠⎟ 


) 

∫ ∫ d 

But in the Calculus of Limits, this integral is singular, and the 

Fourier-Bessel Integral Theorem does not hold in the Calculus of 

Limits under any conditions. 

ν ν 

10.1 The 3-D Fourier-Bessel Integral Theorem Fails 

in the Calculus of Limits 

Proof: By the 3-dimensional Fourier-Bessel Integral Theorem 

41


H. Vic Dannon 

⎛ 

⎞ ⎟ 

f () r = 4 ⎜4 f() 

d ⎟ 

⎝ 

⎠ 


) 

ν=∞ σ=∞ 

sin 2 sin 2 

⎜ 

2 2 

σ σ σ 

∫ ∫ ν d 

νσ 

νr 

ν= 0 ⎜ σ= 

0 

⎟ 

σ=∞ ⎛ ν=∞ 

⎞ 2 1 

2 


d d 

rσ 

σ= 0 ⎜ 

⎝ ν= 

0 

⎠⎟ 

However, at σ = r , 

∫ ∫ ν σ σ 

ν=∞ ν=∞ 

∫ 

2 

( ) ( r) d = ( ) 

sin 2πνσ sin 2πν ν sin 2πνσ dν 

ν= 0 ν= 

0 

∫ 

ν=∞ 

1 

= − 

2 

∫ 

ν= 

0 

{ 1 cos( 4πνσ 

)} 

ν =∞ 

⎡ sin(4 πνσ) 

⎤ 

= ν − = ∞. 

⎣ 

⎢ 4πσ 

⎦ 

⎥ 

ν = 0 

dν 

ν 

Hence, the 3-dimensinal Fourier-Bessel Integral diverges, and the 

3-dimensional Fourier-Bessel Integral Theorem does not hold in 

the Calculus of Limits. 

Avoiding the singularity at σ = ρ does not recover the Theorem, 

because without the singularity the integral equals zero. 

Furthermore, 

10.2 Calculus of Limits Conditions are insufficient 

for the 3-D Fourier-Bessel Integral Theorem 

42


H. Vic Dannon 

Proof: According to [Watson, p.458], The following conditions 

guarantee the 3-dimensional Fourier-Bessel Integral Theorem in 

the Calculus of Limits 

r =∞ 

1. convergence of f () r 

∫ 

r =−∞ 

rdr 

r =∞ ⎛ω= 

λ 

⎞ 2. Existence of lim f ( r) J1( ωσ) J1( ωr) 

ωdω 

r 

λ→∞ 2 2 

r = 0 ⎜ 

⎝ω 

= 0 

⎠⎟ 

∫ ∫ dr 

It is clear from 7.1 that Condition 2. never holds. No conditions 

on f ( x ) can resolve the singularity, at 

ω=∞ ν=∞ 

2 

1 1 

2 2 

ω= 0 ν= 

0 

σ = r , of 

∫ ∫ . 

r J ( ωσ) J ( ωr) ωdω = 4 sin(2 πνσ)sin(2 πνr) 

dν 

Therefore, the Calculus of Limits Conditions are insufficient for 

the 3-dimensional Fourier Integral Theorem. 

In Infinitesimal Calculus, the 3-dimensional Fourier-Bessel 

Integral Theorem holds for any radial Hyper-real function 

10.3 3-dimensional Fourier Integral Theorem 

If f (,, xyz) = fr () is Hyper-real function, 

Then, the Fourier-Bessel Integral Theorem holds. 

⎛ 

⎞ ⎟ 

f () r = 4 ⎜4 f() 

d ⎟ 

⎝ 

⎠ 


) 

ν=∞ σ=∞ 

sin 2 sin 2 

⎜ 

2 2 

σ σ σ 

∫ ∫ ν d 

νσ 

νr 

ν= 0 

⎜ 

σ= 

0 

⎟ 

ν 

43


H. Vic Dannon 

σ=∞ ⎛ ν=∞ 

⎞ 2 1 

2 


d d 

rσ 

σ= 0 

⎜ 

⎝ ν= 

0 

⎠⎟ 

∫ ∫ ν σ σ. 

Proof: 

f () r = f(,, x y z ) 

In Infinitesimal Calculus, 

By 7.2, 

ζ=∞ η=∞ ξ=∞ 

∫ ∫ ∫ d 

= f ( ξηζδ , , ) ( x −ξδ ) ( y −ηδ ) ( z − ζ) 

dξdηζ 

ζ=−∞ η=−∞ ξ=−∞ 

ν=∞ 

2 1 

δ( x −ξ, y −η, z − ζ) = 4 sin( 2πνσ) sin( 2πνr) 

d 

rσ 

∫ ν, 

ν= 

0 


ζ=∞ η=∞ ξ=∞ ⎛ ν=∞ 

⎞ 2 1 

f ( r) = f( ξηζ , , ) 4 sin( 2πνσ) sin( 2πνr) 

dν ∫ ∫ ∫ ξ ηζ 

rσ 

∫ d d d 

ζ=−∞ η=−∞ ξ=−∞ ⎜⎝ 

ν= 

0 

⎠⎟ 

Put 

f ( ξηζ , , ) = f ( σ) 

. 

Integrating with respect to σ , 


ddd ξ η ζ = σ dσ, 

⎛ 

⎞ f ( r) = f( σ) 4 sin( 2πνσ) sin( 2πνr) 

d 

ν 

rσ 

σ 

σ= 0 ⎜ 

⎝ ν= 

0 

⎠⎟ 

σ=∞ ν=∞ 

⎜ 

2 1 

2 

∫ ∫ d 

By changing the Summation order, 

2 

σ 

44


H. Vic Dannon 

⎛ 

⎞ ⎟ 

f () r = 4 ⎜4 f() 

d ⎟ 

⎝ 

⎠ 


) 

ν=∞ σ=∞ 

sin 2 sin 2 

⎜ 

2 2 

σ σ σ ∫ dν 

∫ ν . 

νσ 

νr 

ν= 0 ⎜ σ= 

0 

⎟ 

Then, the 3-dimensional Fourier-Bessel Transform of f () r , 

σ=∞ 

sin(2 πνσ) 

4 ∫ f ( σ) 

σdσ, 

ν 

σ= 

0 

converges to a Hyper-real function 

F(2 πν) 

be infinite Hyper-reals, like the Delta Function. 

And the Inverse Fourier-Bessel Transform of F(2 πν) 

ν=∞ 

sin(2 πνσ) 

4 ∫ F(2 πν) 

νdν 

σ 

ν= 

0 

converges to the Hyper-real function f ( ρ ). 

, some of its values may 

10.4 If f () r is Hyper-real function, 

Then, 

the Hyper-real integral 

σ=∞ 

sin(2 πνσ) 

4 ∫ f ( σ) 

σdσ converges to 

ν 

σ= 

0 

F(2 πν) 

45


H. Vic Dannon 

the Hyper-real integral 

ν=∞ 

sin(2 πνσ) 

4 ∫ F(2 πν) 

νdν 

converges 

σ 

ν= 

0 

to f () r 

46


H. Vic Dannon 

References 

[Bowman] Bowman, Frank, Introduction to Bessel Functions, Dover, 1958 

[Dan1] Dannon, H. Vic, “Well-Ordering of the Reals, Equality of all Infinities, 

and the Continuum Hypothesis” in Gauge Institute Journal Vol.6 No 2, May 

2010; 

[Dan2] Dannon, H. Vic, “Infinitesimals” in Gauge Institute Journal Vol.6 No 

4, November 2010; 

[Dan3] Dannon, H. Vic, “Infinitesimal Calculus” in Gauge Institute Journal 

Vol.7 No 1, February 2011; 

[Dan4] Dannon, H. Vic, “Riemann’s Zeta Function: the Riemann Hypothesis 

Origin, the Factorization Error, and the Count of the Primes”, in Gauge 

Institute Journal of Math and Physics, November 2009. 

[Dan5] Dannon, H. Vic, “The Delta Function” in Gauge Institute Journal 

Vol.7 No 2, May 2011; 

[Dan6] Dannon, H. Vic, “The Delta Function and Fourier Transform” in 

Gauge Institute Journal Vol.7 No 3, August 2011; 

[Dirac] Dirac, P. A. M. The Principles of Quantum Mechanics, Second Edition, 

Oxford Univ press, 1935. 

[Gray] Gray, Andrew, and Mathews, G.B., A Treatise on Bessel Functions and 

their applications to Physics, Dover 1966 

[Hen] Henle, James M., and Kleinberg Eugene M., Infinitesimal Calculus, 

MIT Press 1979. 

[Hosk] Hoskins, R. F., Standard and Nonstandard Analysis, Ellis Horwood, 

1990. 

47


H. Vic Dannon 

[Keis] Keisler, H. Jerome, Elementary calculus, An Infinitesimal Approach, 

Second Edition, Prindle, Weber, and Schmidt, 1986, pp. 905-912 

[Laug] Laugwitz, Detlef, “Curt Schmieden’s approach to infinitesimals-an eyeopener 

to the historiography of analysis” Technische Universitat Darmstadt, 

Preprint Nr. 2053, August 1999 

[Mikus] Mikusinski, J. and Sikorski, R., “The elementary theory of 

distributions”, Rosprawy Matematyczne XII, Warszawa 1957. 

[Rand] Randolph, John, “Basic Real and Abstract Analysis”, Academic Press, 

1968. 

[Riemann] Riemann, Bernhard, “On the Representation of a Function by a 

Trigonometric Series”. 

(1) In “Collected Papers, Bernhard Riemann”, translated from 

the 1892 edition by Roger Baker, Charles Christenson, and 

Henry Orde, Paper XII, Part 5, Conditions for the existence of a 

definite integral, pages 231-232, Part 6, Special Cases, pages 

232-234. Kendrick press, 2004 

(2) In “God Created the Integers” Edited by Stephen Hawking, 

Part 5, and Part 6, pages 836-840, Running Press, 2005. 

[Schwartz] Schwartz, Laurent, Mathematics for the Physical Sciences, 

Addison-Wesley, 1966. 

[Temp] Temple, George, 100 Years of Mathematics, Springer-Verlag, 1981. 

pp. 19-24. 

[Watson] Watson, G. N., A Treatise on the Theory of Bessel Functions, 2 nd 

Edition, Cambridge, 1958. 

48

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