Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 2 docx

5
10
15
20 25
30
10000
20000
30000
40000
Figure 33.2: Plot of the integrand for Γ(10)
We see that the ”important” part of the integrand is the hump centered around x = 9. If we find where the
integrand of Γ(x) has its maximum
d
dx

e
−t
t
x−1

= 0
−
e
−t
t
x−1
+ (x − 1)
e
−t
t

x−2
= 0
(x − 1) − t = 0
t = x − 1,
we see that the maximum varies with x. This could complicate our analysis. To take care of this problem we introduce
1614
the change of variables t = xs.
Γ(x) =

∞
0
e
−xs
(xs)
x−1
x ds
= x
x

∞
0
e
−xs
s
x
s
−1
ds
= x
x


∞
0
e
−x(s−log s)
s
−1
ds
The integrands, (
e
−x(s−log s)
s
−1
), for Γ(5) and Γ(20) are plotted in Figure 33.3.
1
2
3
4
0.001
0.002
0.003
0.004
0.005
0.006
0.007
1
2 3
4
5·10
-10

1·10
-9
1.5·10
-9
2·10
-9
Figure 33.3: Plot of the integrand for Γ(5) and Γ(20).
We see that the important part of the integrand is the hump that seems to be centered about s = 1. Also note
that the the hump becomes narrower with increasing x. This makes sense as the
e
−x(s−log s)
term is the most rapidly
varying term. Instead of integrating from zero to infinity, we could get a good approximation to the integral by just
integrating over some small n eighborhood centered at s = 1. Since s − log s has a minimum at s = 1,
e
−x(s−log s)
has a maximum there. Because the important part of the integrand is the small area around s = 1, it makes sense to
1615
approximate s − log s with its Taylor series about that point.
s − log s = 1 +
1
2
(s − 1)
2
+ O

(s − 1)
3

Since the hump becomes increasingly narrow with increasi ng x, we will approximate the 1/s term in the i ntegrand with

its value at s = 1. Substituting these approximations into the integral, we obtain
Γ(x) ∼ x
x

1+
1−
e
−x(1+(s−1)
2
/2)
ds
= x
x
e
−x

1+
1−
e
−x(s−1)
2
/2
ds
As x → ∞ both of the integrals

1−
−∞
e
−x(s−1)
2

/2
ds and

∞
1+
e
−x(s−1)
2
/2
ds
are exponentially small. Thus instead of integrating from 1 −  to 1 +  we can integrate from −∞ to ∞.
Γ(x) ∼ x
x
e
−x

∞
−∞
e
−x(s−1)
2
/2
ds
= x
x
e
−x

∞
−∞

e
−xs
2
/2
ds
= x
x
e
−x

2π
x
Γ(x) ∼
√
2πx
x−1/2
e
−x
as x → ∞.
This is known as Stirling’s approximation to the Gamma function. In the table below, we see that the approximation
is pretty good eve n for relatively small argument.
1616
n Γ(n)
√
2πx
x−1/2
e
−x
relative error
5 24 23.6038 0.0165

15 8.71783 · 10
10
8.66954 · 10
10
0.0055
25 6.20448 · 10
23
6.18384 · 10
23
0.0033
35 2.95233 · 10
38
2.94531 · 10
38
0.0024
45 2.65827 · 10
54
2.65335 · 10
54
0.0019
In deriving Stirling’s approximation to the Gamma function we did a lot of hand waving. However, all of the steps can
be justified and better approximations can be obtained by using Laplace’s method for finding the asymptotic behavior
of integrals.
1617
33.6 Exercises
Exercise 33.1
Given that

∞
−∞

e
−x
2
dx =
√
π,
deduce the value of Γ(1/2). Now find the value of Γ(n + 1/2).
Exercise 33.2
Evaluate

∞
0
e
−x
3
dx in terms of the gamma function.
Exercise 33.3
Show that

∞
0
e
−x
sin(log x) dx =
Γ(ı) + Γ(−ı)
2
.
1618
33.7 Hints
Hint 33.1

Use the change of variables, ξ = x
2
in the integral. To find the value of Γ(n + 1/2) use the difference relation.
Hint 33.2
Make the change of variable ξ = x
3
.
Hint 33.3
1619
33.8 Solutions
Solution 33.1

∞
−∞
e
−x
2
dx =
√
π

∞
0
e
−x
2
dx =
√
π
2

Make the change of variables ξ = x
2
.

∞
0
e
−ξ
1
2
ξ
−1/2
dξ =
√
π
2
Γ(1/2) =
√
π
Recall the difference relation for the Gamma function Γ(z + 1) = zΓ(z).
Γ(n + 1/2) = (n − 1/2)Γ(n − 1/2)
=
2n − 1
2
Γ(n − 1/2)
=
(2n − 3)(2n − 1)
2
2
Γ(n − 3/2)

=
(1)(3)(5) ···(2n − 1)
2
n
Γ(1/2)
Γ(n + 1/2) =
(1)(3)(5) ···(2n − 1)
2
n
√
π
1620
Solution 33.2
We make the change of variable ξ = x
3
, x = ξ
1/3
, dx =
1
3
ξ
−2/3
dξ.

∞
0
e
−x
3
dx =


∞
0
e
−ξ
1
3
ξ
−2/3
dξ
=
1
3
Γ

1
3

Solution 33.3

∞
0
e
−x
sin(log x) dx =

∞
0
e
−x

1
ı2

e
ı log x
−
e
−ı log x

dx
=
1
ı2

∞
0
e
−x

x
ı
− x
−ı

dx
=
1
ı2
(Γ(1 + ı) − Γ(1 − ı))
=

1
ı2
(ıΓ(ı) − (−ı)Γ(−ı))
=
Γ(ı) + Γ(−ı)
2
1621
Chapter 34
Bessel Functions
Ideas are angels. Implementations are a bitch.
34.1 Bessel’s Equation
A commonly encountered differential equation in applied mathematics is Bessel’s equation
y

+
1
z
y

+

1 −
ν
2
z
2

y = 0.
For our purposes, we will consi der ν ∈ R
0+

. This equation arises when solving certain partial differential equations
with the method of s eparation of variables in cylindrical coordinates. For this reason, the solutions of this equation are
sometimes called cylindrical functions.
This equation cannot be solved directly. However, we can find series representations of the solutions. There is
a regular singular point at z = 0, so the Frobenius method is applicable there. The point at infinity is an irregular
singularity, so we will look for asymptotic series about that point. Additionally, we will use Laplace’s method to find
definite integral representations of the solutions.
1622
Note that Bessel’s equation depends only on ν
2
and not ν alone. Thus if we find a solution, (which of course
depends on this parameter), y
ν
(z) we know that y
−ν
(z) is also a solution. For this reason, we will consider ν ∈ R
0+
.
Whether or not y
ν
(z) and y
−ν
(z) are linearly independent, (distinct solutions), remains to be seen.
Example 34.1.1 Consider the differential equation
y

+
1
z
y


+
ν
2
z
2
y = 0
One solution is y
ν
(z) = z
ν
. Since the equation depends only on ν
2
, another solution is y
−ν
(z) = z
−ν
. For ν = 0, these
two solutions are linearly independent.
Now consider the differential equation
y

+ ν
2
y = 0
One solution is y
ν
(z) = cos(νz). Therefore, another solution is y
−ν
(z) = cos(−νz) = cos(νz). However, these two

solutions are not linearly independent.
34.2 Frobeneius Series Solution about z = 0
We note that z = 0 is a regular singular point, (the only singular point of Bessel’s equation in the finite complex plane.)
We will use the Frobenius method at that point to analyze the solutions. We assume that ν ≥ 0.
The indicial equation is
α(α − 1) + α − ν
2
= 0
α = ±ν.
If ±ν do not differ by an integer, (that is if ν is not a half-integer), then there will be two series solutions of the
Frobenius form.
y
1
(z) = z
ν
∞

k=0
a
k
z
k
, y
2
(z) = z
−ν
∞

k=0
b

k
z
k
1623
If ν is a half-integer, the second s olution may or may not be in the Frobenius form. In any case, then will always be at
least one solution in the Frobenius form. We will determine that series solution. y(z) and it derivatives are
y =
∞

k=0
a
k
z
k+ν
, y

=
∞

k=0
(k + ν)a
k
z
k+ν−1
, y

=
∞

k=0

(k + ν)(k + ν −1)a
k
z
k+ν−2
.
We substitute the Frobenius series into the differential equation.
z
2
y

+ zy

+

z
2
− ν
2

y = 0
∞

k=0
(k + ν)(k + ν −1)a
k
z
k+ν
+
∞


k=0
(k + ν)a
k
z
k+ν
+
∞

k=0
a
k
z
k+ν+2
−
∞

k=0
ν
2
a
k
z
k+ν
= 0
∞

k=0

k
2

+ 2kν

a
k
z
k
+
∞

k=2
a
k−2
z
k
= 0
We equate powers of z to obtain equations that determine the coefficients. The coefficient of z
0
is the equation
0 · a
0
= 0. This corroborates that a
0
is arbitrary, (but non-zero). The coefficient of z
1
is the equation
(1 + 2ν)a
1
= 0
a
1

= 0
The coefficient of z
k
for k ≥ 2 gives us

k
2
+ 2kν

a
k
+ a
k−2
= 0.
a
k
= −
a
k−2
k
2
+ 2kν
= −
a
k−2
k(k + 2ν)
From the recurrence relation we see that all the odd coefficients are zero, a
2k+1
= 0. The even coefficients are
a

2k
= −
a
2k−2
4k(k + ν)
=
(−1)
k
a
0
2
2k
k!Γ(k + ν + 1)
1624
Thus we have the series solution
y(z) = a
0
∞

k=0
(−1)
k
2
2k
k!Γ(k + ν + 1)
z
2k
.
a
0

is arbitrary. We choose a
0
= 2
−ν
. We call this solution the Bessel function of the first kind and order ν and denote
it with J
ν
(z).
J
ν
(z) =
∞

k=0
(−1)
k
k!Γ(k + ν + 1)

z
2

2k+ν
Recall that the Gamma function i s non-zero and finite for all real arguments except non-positive integers. Γ(x)
has singularities at x = 0, −1, −2, . . Therefore, J
−ν
(z) is well-defined when ν is not a positive integer. Since
J
−ν
(z) ∼ z
−ν

at z = 0, J
−ν
(z) is clear linearly independent to J
ν
(z) for non-integer ν. In particular we note that there
are two solutions of the Frobenius form when ν is a half odd integer.
J
−ν
(z) =
∞

k=0
(−1)
k
k!Γ(k −ν + 1)

z
2

2k−ν
, for ν ∈ Z
+
Of course for ν = 0, J
ν
(z) and J
−ν
(z) are identical. Consider the case that ν = n is a positive integer. Since
Γ(x) → +∞ as x → 0, −1, −2, . . . we see the the coefficients in the series for J
−nu
(z) vanish for k = 0, . . . , n − 1.

J
−n
(z) =
∞

k=n
(−1)
k
k!Γ(k −n + 1)

z
2

2k−n
J
−n
(z) =
∞

k=0
(−1)
k+n
(k + n)!Γ(k + 1)

z
2

2k+n
J
−n

(z) = (−1)
n
∞

k=0
(−1)
k
k!(k + n)!

z
2

2k+n
J
−n
(z) = (−1)
n
J
n
(z)
Thus we see that J
−n
(z) and J
n
(z) are not linearly independent for integer n.
1625
34.2.1 Behavior at Infinity
With the change of variables z = 1/ζ, w(z) = u(ζ) Bessel’s equation becomes
ζ
4

u

+ 2ζ
3
u

+ ζ

−ζ
2

u

+

1 − ν
2
ζ
2

u = 0
u

+
1
ζ
u

+


1
ζ
4
−
ν
2
ζ
2

u = 0.
The point ζ = 0 and hence the point z = ∞ is an irregular singular point. We will find the leading order asymptotic
behavior of the solutions as z → +∞.
Controlling Factor. We starti with Bessel’s equation for real argument.
y

+
1
x
y

+

1 −
ν
2
x
2

y = 0
We make the substitution y =

e
s(x)
.
s

+ (s

)
2
+
1
x
s

+ 1 −
ν
2
x
2
= 0
We know that
ν
2
x
2
 1 as x → ∞; we will assume that s

 (s

)

2
as x → ∞.
(s

)
2
+
1
x
s

+ 1 ∼ 0 as x → ∞
To simplify the equation further, we will try the possible two-term balances.
1. (s

)
2
+
1
x
s

∼ 0 → s

∼ −
1
x
This balance is not consistent as it violates the assumption that 1 is smaller
than the other terms.
2. (s


)
2
+ 1 ∼ 0 → s

∼ ±ı This balance is consistent.
3.
1
x
s

+ 1 ∼ 0 → s

∼ −x This balance is inconsistent as (s

)
2
isn’t smaller than the other terms.
Thus the only dominant balance is s

∼ ±ı. This balance is consistent with our initial assumption that s

 (s

)
2
.
Thus s ∼ ±ıx and the controlling factor is
e
±ıx

.
1626
Leading Order Behavior. In order to find the leading order behavior, we substitute s = ±ıx+t(x) where t(x)  x
as x → ∞ into the differential equation for s. We first consider the case s = ıx + t(x). We assume that t

 1 and
t

 1/x.
t

+ (ı + t

)
2
+
1
x
(ı + t

) + 1 −
ν
2
x
2
= 0
t

+ ı2t


+ (t

)
2
+
ı
x
+
1
x
t

−
ν
2
x
2
= 0
We use our assumptions about the behavior of t

and t

.
ı2t

+
ı
x
∼ 0
t


∼ −
1
2x
t ∼ −
1
2
ln x as x → ∞.
This asymptotic behavior is consistent with our assumptions.
Substituting s = −ıx + t(x) will also yield t ∼ −
1
2
ln x. Thus the leading order behavior of the solutions is
y ∼ c
e
±ıx−
1
2
ln x+u(x)
= cx
−1/2
e
±ıx+u(x)
as x → ∞,
where u(x)  ln x as x → ∞.
By substituting t = −
1
2
ln x+u(x) into the differential equation for t, you could show that u(x) → const as x → ∞.
Thus the full leading order behavior of the solutions is

y ∼ cx
−1/2
e
±ıx+u(x)
as x → ∞
where u(x) → 0 as x → ∞. Writing this in terms of sines and cosines yields
y
1
∼ x
−1/2
cos(x + u
1
(x)), y
2
∼ x
−1/2
sin(x + u
2
(x)), as x → ∞,
1627
where u
1
, u
2
→ 0 as x → ∞.
Result 34.2.1 Bessel’s equ ation for real argument is
y

+
1

x
y

+

1 −
ν
2
x
2

y = 0.
If ν is not an integer then the solutions behave as linear combinations of
y
1
= x
ν
, and y
2
= x
−ν
at x = 0. If ν is an integer, then the solutions behave as linear combinations of
y
1
= x
ν
, and y
2
= x
−ν

+ cx
ν
log x
at x = 0. The solutions are asymptotic to a linear combin ation of
y
1
= x
−1/2
sin(x + u
1
(x)), and y
2
= x
−1/2
cos(x + u
2
(x))
as x → +∞, where u
1
, u
2
→ 0 as x → ∞.
34.3 Bessel Functions of the First Kind
Consider the function exp(
1
2
z(t − 1/t)). We can expand this function in a Laurent series in powers of t,
e
1
2

z(t−1/t)
=
∞

n=−∞
J
n
(z)t
n
,
1628
where the coefficient functions J
n
(z) are
J
n
(z) =
1
ı2π

τ
−n−1
e
1
2
z(τ−1/τ)
dτ.
Here the path of integration is any positive closed path around the origin. exp(
1
2

z(t−1/t)) is the generating function
for Bessel function of the first kind.
34.3.1 The Bessel Function Satisfies Bessel’s Equation
We would like to expand J
n
(z) in powers of z. The first step in doing this is to make the substitution τ = 2t/z.
J
n
(z) =
1
ı2π


2t
z

−n−1
exp

1
2
z

2t
z
−
z
2t

2

z
dt
=
1
ı2π

z
2

n

t
−n−1
e
t−z
2
/4t
dt
We differentiate the expression for J
n
(z).
J

n
(z) =
1
ı2π
nz
n−1
2

n

t
−n−1
e
t−z
2
/4t
dt +
1
ı2π

z
2

n

t
−n−1

−2z
4t

e
t−z
2
/4t
dt
=
1

ı2π

z
2

n


n
z
−
z
2t

t
−n−1
e
t−z
2
/4t
dt
J

n
(z) =
1
ı2π

z
2


n


n
z

n
z
−
z
2t

+

−
n
z
2
−
1
2t

−
z
2t

n
z
−

z
2t


t
−n−1
e
t−z
2
/4t
dt
=
1
ı2π

z
2

n


n
2
z
2
−
nz
2zt
−
n

z
2
−
1
2t
−
nz
2zt
+
z
2
4t
2

t
−n−1
e
t−z
2
/4t
dt
=
1
ı2π

z
2

n



n(n − 1)
z
2
−
2n + 1
2t
+
z
2
4t
2

t
−n−1
e
t−z
2
/4t
dt
1629
We substitute J
n
(z) into Bessel’s equation.
J

n
+
1
z

J

n
+

1 −
n
2
z
2

J
n
=
1
ı2π

z
2

n


n(n − 1)
z
2
−
2n + 1
2t
+

z
2
4t
2

+

n
z
2
−
1
2t

+

1 −
n
2
z
2

t
−n−1
e
t−z
2
/4t
dt
=

1
ı2π

z
2

n


1 −
n + 1
t
+
z
2
4t
2

t
−n−1
e
t−z
2
/4t
dt
=
1
ı2π

z

2

n

d
dt

t
−n−1
e
t−z
2
/4t

dt
Since t
−n−1
e
t−z
2
/4t
is analytic in 0 < |t| < ∞ when n is an integer, the integral vanishes.
= 0.
Thus for integer n, J
n
(z) satisfies Bessel’s equation.
J
n
(z) is called the Bessel function of the first kind. The subscript is the order. Thus J
1

(z) is a Bessel function
of order 1. J
0
(x) and J
1
(x) are plotted in the first graph in Figure 34.1. J
5
(x) is plotted in the second graph in
Figure 34.1. Note that for non-negative, integer n, J
n
(z) behaves as z
n
at z = 0.
34.3.2 Series Expansion of the Bessel Function
We expand exp(−z
2
/4t) in the integral expression for J
n
.
J
n
(z) =
1
ı2π

z
2

n


t
−n−1
e
t−z
2
/4t
dt
=
1
ı2π

z
2

n

t
−n−1
e
t

∞

m=0

−z
2
4t

m

1
m!

dt
1630
2
4
6 8 10
12
14
-0.4
-0.2
0.2
0.4
0.6
0.8
1
5
10
15
20
-0.2
-0.1
0.1
0.2
0.3
Figure 34.1: Plots of J
0
(x), J
1

(x) and J
5
(x).
For the path of integration, we are free to choose any contour that encloses the origin. Consider the circular path on
|t| = 1. Since the integral is uniformly convergent, we can interchange the order of integration and summation.
J
n
(z) =
1
ı2π

z
2

n
∞

m=0
(−1)
m
z
2m
2
2m
m!

t
−n−m−1
e
t

dt
1631
Let n be a non-negative integer.
1
ı2π

t
−n−m−1
e
t
dt = lim
z→0

1
(n + m)!
d
n+m
dz
n+m
(
e
z
)

=
1
(n + m)!
We have the series expansion
J
n

(z) =
∞

m=0
(−1)
m
m!(n + m)!

z
2

n+2m
for n ≥ 0.
Now consider J
−n
(z), (n positive).
J
−n
(z) =
1
ı2π

z
2

−n
∞

m=1
(−1)

m
z
2m
2
2m
m!

t
n−m−1
e
t
dt
For m ≥ n, the integrand has a pole of order m − n + 1 at the origin.
1
ı2π

t
n−m−1
e
t
dt =

1
(m−n)!
for m ≥ n
0 for m < n
The expression for J
−n
is then
J

−n
(z) =
∞

m=n
(−1)
m
m!(m − n)!

z
2

−n+2m
=
∞

m=0
(−1)
m+n
(m + n)!m!

z
2

n+2m
= (−1)
n
J
n
(z).

Thus we have that
J
−n
(z) = (−1)
n
J
n
(z) for integer n.
1632
34.3.3 Bessel Functions of Non-Integer Order
The generalization of the f actorial function is the Gamma function. For integer values of n, n! = Γ(n+1). The Gamma
function is defined for all complex-valued arguments. Thus one would guess that if the Bessel function of the first kind
were defined for non-integer order, it would have the definition,
J
ν
(z) =
∞

m=0
(−1)
m
m!Γ(ν + m + 1)

z
2

ν+2m
.
The Integrand for Non-Integer ν. Rec all the definition of the Bessel function
J

ν
(z) =
1
ı2π

z
2

ν

t
−ν−1
e
t−z
2
/4t
dt.
When ν is an integer, the integrand is single valued. Thus if you start at any point and follow any path around the
origin, the integrand will return to its original value. This property was the key to J
n
satisfying Bessel’s equation. If ν
is not an integer, then this property does not hold for arbitrary paths around the origin.
A New Contour. First, since the integrand is mu ltiple -value d, we need to define what branch of the function we
are talking about. We will take the principal value of the integrand and introduce a branch cut on the negative real
axis. Let C be a contour that starts at z = −∞ below the branch cut, circles the origin, and returns to the point
z = −∞ above the branch cut. This contour is shown in Figure 34.2.
Thus we define
J
ν
(z) =

1
ı2π

z
2

ν

C
t
−ν−1
e
t−z
2
/4t
dt.
Bessel’s Equation. Substituting J
ν
(z) into Bessel’s equation yields
J

ν
+
1
z
J

ν
+


1 −
ν
2
z
2

J
ν
=
1
ı2π

z
2

ν

C
d
dt

t
−ν−1
e
t−z
2
/4t

dt.
Since t

−ν−1
e
t−z
2
/4t
is analytic in 0 < |z| < ∞ and |arg(z)| < π, and it vanishes at z = −∞, the integral is zero.
Thus the Bessel function of the first kind satisfies Bessel’s equation for all complex orders.
1633
Figure 34.2: The Contour of Integration.
Series Expansion. Because of the
e
t
factor in the integrand, the integral defining J
ν
converges uniformly. Expanding
e
−z
2
/4t
in a Taylor series yields
J
ν
(z) =
1
ı2π

z
2

ν

∞

m=0
(−1)
m
z
2m
2
2m
m!

C
t
−ν−m−1
e
t
dt
Since
1
Γ(α)
=
1
ı2π

C
t
−α−1
e
t
dt,

we have the series expansion of the Bessel function
J
ν
(z) =
∞

m=0
(−1)
m
m!Γ(ν + m + 1)

z
2

ν+2m
.
1634
Linear Independence. We use Abel’s formula to compute the Wronskian of Bessel’s equation.
W (z) = exp

−

z
1
ζ
dζ

=
e
−log z

=
1
z
Thus to within a function of ν, the Wronskian of any two solutions is 1/z. For any given ν, there are two linearly
independent solutions. Note that Bessel’s equation is unchanged under the transformation ν → −ν. Thus both J
ν
and
J
−ν
satisfy Bessel’s equation. Now we must determine if they are linearly independent. We have already shown that
for integer values of ν they are not independent. (J
−n
= (−1)
n
J
n
.) Assume that ν is not an integer. We compute the
Wronskian of J
ν
and J
−ν
.
W [J
ν
, J
−ν
] =





J
ν
J
−ν
J

ν
J

−ν




= J
ν
J

−ν
− J
−ν
J

ν
We substitute in the expansion for J
ν
=

∞


m=0
(−1)
m
m!Γ(ν + m + 1)

z
2

ν+2m

∞

n=0
(−1)
n
(−ν + 2n)
n!Γ(−ν + n + 1)2

z
2

−ν+2n−1

−

∞

m=0
(−1)

m
m!Γ(−ν + m + 1)

z
2

−ν+2m

∞

n=0
(−1)
n
(ν + 2n)
n!Γ(ν + n + 1)2

z
2

ν+2n−1

Since the Wronskian is a function of ν times 1/z the coefficients of all of the powers of z except 1/z must vanish.
=
−ν
zΓ(ν + 1)Γ(−ν + 1)
−
ν
zΓ(−ν + 1)Γ(ν + 1)
= −
2

zΓ(ν)Γ(1 − ν)
1635
Using an identity for the Gamma function simplifies this expression.
= −
2
πz
sin(πν)
Since the Wronskian is nonzero for non-integer ν, J
ν
and J
−ν
are independent functions when ν is not an integer. In
this case, the general solution of Bessel’s equation is aJ
ν
+ bJ
−ν
.
34.3.4 Recursion Formulas
In showing that J
ν
satisfies Bessel’s equation for arbitrary complex ν, we obtained

C
d
dt

t
−ν
e
t−z

2
/4t

dt = 0.
Expanding the integral,

C

t
−ν
+
z
2
4
t
−ν−2
− νt
−ν−1

e
t−z
2
/4t
dt = 0.
1
ı2π

z
2


ν

C

t
−ν
+
z
2
4
t
−ν−2
− νt
−ν−1

e
t−z
2
/4t
dt = 0.
Since J
ν
(z) =
1
ı2π
(z/2)
ν

C
t

−ν−1
e
t−z
2
/4t
dt,


2
z

−1
J
ν−1
+

2
z

z
2
4
J
ν+1
− νJ
ν

= 0.
J
ν−1

+ J
ν+1
=
2ν
z
J
ν
1636
Differentiating the integral expression for J
ν
,
J

ν
(z) =
1
ı2π
νz
ν−1
2
ν

C
t
−ν−1
e
t−z
2
/4t
dt +

1
ı2π

z
2

ν

C
t
−ν−1

−
z
2t

e
t−z
2
/4t
dt
J

ν
(z) =
ν
z
1
ı2π


z
2

ν

C
t
−ν−1
e
t−z
2
/4t
dt −
1
ı2π

z
2

ν+1

C
t
−ν−2
e
t−z
2
/4t
dt
J


ν
=
ν
z
J
ν
− J
ν+1
From the two relations we have derived you can show that
J

ν
=
1
2
(J
ν−1
+ J
ν+1
) and J

ν
= J
ν−1
−
ν
z
J
ν

.
1637