Show that the Fourier cosine transform (Chapter 7, Section 12) of J 0 ( x ) is 2 π 1 1 − α 2 , 0 ≤ α < 1 , 0 , α > 1. Hence show that ∫ 0 ∞ J 0 ( x ) d x = 1. Hints : Show that the integral in Problem 20 gives J 0 ( x ) = ( 2 / π ) ∫ 0 π / 2 cos ( x sin θ ) d θ . (Replace θ by π − θ in the π / 2 to π integral.) Let sin θ = α to find J 0 as a cosine transform; write the inverse transform. Now let α = 0 .
Show that the Fourier cosine transform (Chapter 7, Section 12) of J 0 ( x ) is 2 π 1 1 − α 2 , 0 ≤ α < 1 , 0 , α > 1. Hence show that ∫ 0 ∞ J 0 ( x ) d x = 1. Hints : Show that the integral in Problem 20 gives J 0 ( x ) = ( 2 / π ) ∫ 0 π / 2 cos ( x sin θ ) d θ . (Replace θ by π − θ in the π / 2 to π integral.) Let sin θ = α to find J 0 as a cosine transform; write the inverse transform. Now let α = 0 .
Show that the Fourier cosine transform (Chapter 7, Section 12) of
J
0
(
x
)
is
2
π
1
1
−
α
2
,
0
≤
α
<
1
,
0
,
α
>
1.
Hence show that
∫
0
∞
J
0
(
x
)
d
x
=
1.
Hints: Show that the integral in Problem 20 gives
J
0
(
x
)
=
(
2
/
π
)
∫
0
π
/
2
cos
(
x
sin
θ
)
d
θ
.
(Replace
θ
by
π
−
θ
in the
π
/
2
to
π
integral.) Let
sin
θ
=
α
to find
J
0
as a cosine transform; write the inverse transform. Now let
α
=
0
.
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
A Problem Solving Approach to Mathematics for Elementary School Teachers (12th Edition)
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