Internat. J. Math. & Math. Sci.
Vol. I0 No. 2 (1987) 287-294
287
AN OPERATIONAL PROCEDURE FOR HANKEL TYPE INTEGRALS
C. NASIM
Department of Mathematics and Statistics
The University of Calgary
Calgary, Alberta T2N 1N4 Canada
(Received March 28, 1986)
ABSTRACT.
In this paper, an operational procedure is established to evaluate Hankel
d
type integrals.
is constructed, which defines the
First, an operator L(8), 8 e -x
Then making use of some basic properties of this operator, an elementary
procedure is developed for evaluating integrals for a special class of analytic
functions. A fe example are given to illustrate the technique.
integral.
KEY WORDS AND PHRASES.
operator
Differential
of
infinite
order,
Imkel
transform,
Mellin transform, 8essel functions, Hypergeometric functions.
SUBJECT CLASSIFICATION: 44A15
1.
INTRODUCTION.
We consider the singular integral of the type
o
where
Ju
f(xt)Ju(2t)dt,
(1.1)
is the usual Bessel function of the first kind of order u, u
1
5’
and f(x) is
suitable function.
This integral can be viewed as defining the Hankel transform of
d
the function f. In this note, our main aim is to construct an operator L(8), 8
-x
a
,
so that L(8)[f(x)] defines the integral (I.I), [cf.
I, t9.5;
2].
We
then
establish
of the operator L(8), and with these help of the properties, we shall obtain
an operational procedure to evaluate the integral (1.1).
2. TI OPERATOR.
It is an easy matter to see that the differential operator 8 where 8 m -x d and
properties
n,
n a positive integer, is such that
sn[xa]
for some constant a.
Then
pn(8),
(-a)nx
a polynomial of n
th
pn(e)[xa] pn(-a)xa.
Consequent ly,
degree in 8, gives
288
C. NASIM
p(8)[xa]
lira
pn(p)[xa]
lira
Pn (-a)x
p(-a)x
where
p(s)
lim
Pn(S),
of
limit
a
polynomial.
Thus
the
operator p(O), is a
differential operator of infinite order and has the property
that when applied to a
function, simply replaces it with a multiplier. With this understanding, we may
write, symbolically,
N
-0
-0 In
(-In n) k
e
n
lim Z
lim
(O ), say.
k!
power
ok
PN
Then,
n
-o
x-S
lim
N-
pn(0)[x-S]
lim
N-
pn(S)x-S
(2.I)
-s
n
for some
If further
n
pn(O)
E (-l+2k+O), and s
k=l
pn(o)[x-s]
o + it, then
(v-l+2k+0)[x-s]
k=l
(2.2)
(p-l+2k+s)x
k=l
Next we write
a- e
a- 0
1
(a-O)[x-s
a- O
or
1
a- 8
for some constants a and s.
1
’a-s’x-
a-
(_s)x-s,
Ix-s]
(2.3)
1
_---,
a
This defines an operator of the type
in
the
sense
that when applied to a power function, simply reproduces it with a multiplier. This
property parallels that of the operator which is a polomial or limit of a olFnomial
in O. By a repeated application of an operator of the tpe (2.3), we have
n
n
-s
1
1
ix-S]
(2.4)
(p_l’+2..)
(vl+2ks)X
k=l
k=l
for some s, except where s
o
1 + 2k, k
,n.
1,2
Combining the results (2.1), (2.2) and (2.4), we construct the operator
k*(l-O)
n
n
I n-8
k=l
.-l+2kO.
_1+2k_8)
(2.5)
(l-s) x-s
(2.6)
such that
k*n
(l-O) x-s
k*n
where
k*n
n
(l-s)
I -s
n
k=l
u-s+2k+s.
(.._-zrmrz__)
s
o + it,
<
<
(R).
289
OPERATIONAL PROCEDURE FOR HANKEL TYPE INTEGRALS
-
k*
Next, we shall establish properties of the function
I.
[k*n
(l-o-i)[
n
This is quite obvious.
2.
PHOOF.
lira
k*n
(l-s).
as [r[
0(I)
(l-s) exists and is uniform on every compact set of the s-axis.
From above,
k*n
n
1 e -s In n
(l-s)
k=l
1
=-e
t-l+2k+s
"-l2k-s"
s(l+ "’’+IInn n)
n
k--1
28
(I + u-l+2k-s
e-S/k
Consider the product
28
J (I+ u_l+2k_s
k=l
n
e_S/k
[I + (I +
/
k=l
_l+2k_s) e-S/k
28
I]
n
ak(s)].
[i +
k:l
Now,
lim
k
k2ak (s)
lim
k
k2..l
lira k
k-#=
lira
k--,=
1
Or,
ak(s)
(l-u)o(1),
f(
2
28
_I+2_s) e-S/k
+
., k-
+ 1
28
-l+2k-) (I
rl_g) + s 2 (,-
[(1 +
2
I k(-l+2k-s)2
2
k2[s(u_1+2k_s
I]
+ O(
2k
k
)]
s(l-u)
k
as
--,
all finite S, hence the infinite product
for
convergence uniformly on every compact set of the s-axis.
in n)
n
---, eTs as n
e
I]
s(1+" "+-I
Also
m,
T being the Ruler’s constant; hence the result.
$
l n --S
In fact,
lim
lim
(l-s)
kn
(2.7)
using Ruler’s product for F-functions, [3, p.
II].
Note that the function
(2.8)
,l[Ju(2x),s],
the ellin transform of
Ju(2x),
Now, we define the operator
s
o + it,
<
T
<
(R),
and
0
<
o
< +I,
[4,
p. 326].
C. NASIM
290
k*(1-0)
r(
k*(l-s)x -s, where k*(l-s) is
where
lira k*(1-8)[x -s]
n
K*(1-O)[x -s]
k*(1-e)[x -s]
r-functions,
lira n
k:(1-p) =1/2 nn-
k*(1-,)
lira
-
defined above.
by
using Euler’s product for
n
---1-p_),
--1
as in (2.5) above.
Now, using the results (2.6) and (2.7) we have
k*(l-e)[x-s] lira (l-S)[x-s]
k*
lim k*
n
(1-s)x
-s
k*(l-s)x-s,
as desired.
3.
THE INTEGRAL.
THEOREM I.
_>
J(2x),
k(x)
Let
f(x)
and 0
be
<
o
such
<
+ I.
k*(l-S) If(x)
k*(1-@)
where
PROOF.
f*(s)
that
and
Then
[o f(xt)Ju(2t)dt,
(3.1)
is the operator defined by (2.8) above.
f*(s) defines the Mellin transform of f(x), we may write
I
f*(s)x-Sds
k*(1-e)[f(x)] k*(1-8)
Since
o+is
o-is
f,(s)x_Sds
k:(1-e) +i(R)
o-is
-s
1
lim-f io+is f, (s) k:(l-S Ix ds
:lira
n-
f*(s)k (1-s)x-d,
lim
due
to
Lemma 3.
To jtify bringing the operator
k(1-e)
inside the integral si, we
have simply to show that the resulting integral
+i(R). f,(s)k:(l_s)x_Sd
(3.2)
s
is uniformly convergent for all finite x.
This is so, since
---i’r
[+is f*(s)k:(l-s)x-Sds[ I-f*(o+ir)k:(l--ir)x id’l
o-is
0(n-)x
> O,
> O,
f*(s)
-
If*(o+ir) ldr <
s,
In
(o-is,o+is) and by using tahe results of La 1.
fact the integral (3.2) converges absolutely, as well. Together with the results of
Lemma 2 and 3, we can then apply Lebesgue’s limit theorem, to obtain
for x
o
since
I.
OPERATIONAL PROCEDURE FOR HANKEL TYPE INTEGRALS
k*(l-e)[f(x)]
I
lira
f+i(R) f, s)k:
1
l-s) x-Sds
k*(l-s)x-Sdsn
f*(s)[lim
+i* f*(s)k*(l-s) x-sds
I
O
f(xt)Ju(2t)dt,
II],
due to the Parseval theorem for Mellin transforms [5, chapt.
A[Ju(2t)
Thus
the
s].
291
and
since
k*(s)
Hence the theorem.
equation
(3.1)
k*(1-#)
defines
as an integral operator, having the property
that
k*(l+a)xa
k*(1-0)[xa]
for soe a, due to Lepta 3. In light of the operational calculus generated by the
operator
one can no evaluate the integrals of the Hankel type, using the
operational procedures. For instance if f(x) is analytic and expressed in a poer
k*(1-0),
series
Z C xn+
f(x)
Ixl <
r
n:o
for soe r and a, then
k*(1-O)[f(x)]
k*(1-e)
Z
n=o
Z C
C x
n
x’
k*(1-e)[xn+a]
n=o
n+a
Z Cn k*(l+n+a)x
n=o
Ix[
<
R,
for some R, where
k* (l+n+a)
1
The above analysis is justified provided
0(1).
C
n
Hence we nm have an operational procedure for evaluating the given integrals, in fact,
lim
o
where
f(x)
f(xt)Jp(2t)dt
Z Cn
and
n=o
4.
Z
n:o
Cnk*(l+n+a)xn+a
k*(s) l[Jp(t):s]
(3 3)
r(l)
SPECIAL CASgS AND EXAMPLES.
Consider the csse when u
__1_ [
J o
1
5"
f(xt)
Then from (3.1), we have
sin(2t)_
Jt
dt
k*(1-e)[f(x)]
(4.1)
C. NASIM
292
where
k*(1-e)21 r()
r()
,
If for instance, we let f(x)
1
I1 <
x
]
then
3 1
o
r() x/a
-
or,
o
Putting
1
t"- sin(2t)dt
4
r()
r(/>
gives us the classical result,
O
sin(2t) dt
t
5"
I xn (3.1), we have
Similarly, by letting u
_I_
J
f(xt)
o
cos(2t)_
4t
dt
k*(l-e)[f(x)],
(4.2)
where
And if f(x)
x/, II <
1/2,
we have from above
f t-
=-
cos(2t)dt
o
Putting
0, give us the well-known result,
f c.OS(2t)_
o
dt
Jt
r(
=-
We no consider a few examples to illustrate the procedure given in formula (3.3).
-x
1. Let f(x)
Then from (3 3) we have
-
xe
(xt)e-xt
o
J (2t)dt
u
n=o
n’
r (i.I
The
series on the right-hand side converges absolutely for
luJ
W
Using the functional equation r(z)r(l-z)
into odd and even terms.
(4.3)
)x
2, where lul < l+# and
and splitting the series
Ixl <
sinz
We obtain from the right hand side of (4.3),
2n+l
=0
Hence,
1
1
nr(n)
293
OPERATIONAL PROCEDURE FOR HANKEL TYPE INTEGRALS
0
te-XtJu(2t) dt
(4.4)
2
+
sin
due to the result (3.3).
can easily be derived.
:<o>rf
Some special cases of this result such as when
The range of the result (4.4) can be extended
0
u and
to
x
> O,
by
analytic continuation.
2.
Let f(x)
.
x-AJ(n).
Then from (3.3), we have
(-l)nr(1/2++++n)x2n+-
(4.5)
(-I)msi(I-u+-A)
(-I)
"r()r()
Therefore the right hand side of (4.5), then gives
Z
II 1 1
1 1 1 1
r (=+.-^+)r
(++.-^+)
n
n.’r(l++n)
n=o
2
2FI(A,A; I+#;),
where
-
Ix[ < 2, Re(+u) + I > Re
As a special case if A
0
u-/a-l, then
^ >-I,
_
[6, p.48].
2-- i x2-u+ 1
(xt)-u++Ij#(xt)Ju (2t)dt
2
2Fl(l-u+’l+;l+;
r(u-)’
2 u-#-I
2--1x2+1
(I
r(u-)
Let f(x)
k
I J __(aix), M
i=l *i
__I
k
X
i=l
-xu’’
Or
"-x
f(x)
k
-#i
x
i=l
k
J/ai(aix)
Z
i=l n=O
Then
(xt)
o
u-M-1
J
i=l
k
i=l
/i (aixt)Ju(2t)dt
(R)
n=oZ
ai 2m+#i
(-1) n (-.-)
n:(’n+i+’l)"
a.x
k*(l-O)
>
)
0 and 0
<
Ixl <
Re u
< PM
2.
+
+
2n+ i
1
(-I) n ai
2n+u-I
x
n!r(n+#i+l
n
Z
i-1 n=o
k*(2n+ )x2n+u_l
ai
2n+#i
n:r(n+i+l)
x2n+-1]
294
C. NASIM
(_l)n(_)2n+,ir(u+n)a
k
1
n=O
Z
n:r(n+i+l)r(1-n)
i 1
1
x_l[( )ir(
k
il
r(i+l
k
-1
4.
x
2n+u-1
a
z
+
n=l
(-1)n(-)i
2n+
ir(+n)
n !r(n+i+l)F(l-n)
x2n+,
1
ai i
C-)
[6, p.54]
il P("i+I)
Finally we will derive a general result formally.
Let
x2 A Gm’n
p,q
f(x)
[ex21al
b
I
ap]
,bqJ
p + q
<
2(+n).
Then
cSx2S+2^ds
with the usual conditions on parameters, [6, p.91].
I.
WIDDER,
(1971).
D.V.
An
Introduction
to Transform Theory, Academic Press, New York,
2.
NASIM, C. An inversion formula for Hankel transforms, Pacific J. Vol. 57 (1), 1975
pp. 255-258.
3.
RAINVILLE, E.D.
4.
ERDELYI, A. et al.
(1954).
5.
Special Functions, Macillan, New York,
Tables of integral transforms Vol. I,
(1967).
McGraw-Hill,
New York,
An Introduction to the Theory of Integral Transforms, Second
TITCIIARSH, E.C.
Edition, Oxford University Press, (1948).