FOURIER INTEGRALS  1   sin v  dv jI3 j @ j f x 0 j   v L 1 in Visual Studio .NET

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FOURIER INTEGRALS  1   sin v  dv jI3 j @ j f x 0 j   v L 1
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sin v dv both converge, we can choose L so large that jI2 j @ =3, jI3 j @ =3. v Also, we can choose so large that jI1 j @ =3. Then from (4) we have jIj <  for and L su ciently large, so that the required result follows. This result follows by reasoning exactly analogous to that in part (a). Since j f x j dx and
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14.12. Prove Fourier s integral formula where f x satis es the conditions stated on Page 364.
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We must prove that lim 1 L!1  L 1
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0 u 1
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f u cos x u du d
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 1  1   f u cos x u du @ j f u j du, which converges, it follows by the Weierstrass test Since    1 1 1 f u cos x u du converges absolutely and uniformly for all . Thus, we can reverse the that
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order of integration to obtain 1  L d
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0 u 1
f u cos x u du
L 1 1 f u du cos x u d  u 1 0 1 1 sin L u x du f u  u 1 u x 1 1 sin Lv dv f x v  u 1 v 0 1 sin Lv 1 1 sin Lv dv dv f x v f x v  1 v  0 v
where we have let u x v. f x 0 f x 0 Letting L ! 1, we see by Problem 14.11 that the given integral converges to as 2 required.
MISCELLANEOUS PROBLEMS 14.13. Solve @U @2 U 2 subject to the conditions U 0; t 0; U x; 0 @t @x bounded where x > 0; t > 0.
0<x<1 , U x; t is xA1
We proceed as in Problem 13.24, 2 13. A solution satisfying the partial di erential equation and the rst boundary condition is given by Be  t sin x. Unlike Problem 13.24, 13, the boundary conditions do not prescribe the speci c values for , so we must assume that all values of  are possible. By analogy with that problem we sum over all possible values of , which corresponds to an integration in this case, and are led to the possible solution 1 2 B  e  t sin x d 1 U x; t
where B  is undetermined.
By the second condition, we have & 1 1 0<x<1 B  sin x d f x 0 xA1 0
from which we have by Fourier s integral formula 2 1 2 1 2 1 cos  f x sin x dx sin x dx B   0  0 
FOURIER INTEGRALS
[CHAP. 14
so that, at least formally, the solution is given by   2 1 1 cos  2 t U x; t e sin x dx  0  See Problem 14.26.
14.14. Show that e x
Since e
p 1 x2 =2 is even, its Fourier transform is given by 2= e cos x dx. 0 p Letting x 2 u and using Problem 12.32, 12, the integral becomes p p 2  2 =2 2 1 u2 2 p e cos 2 u du p e =2 e  0  2
x =2
is its own Fourier transform.
which proves the required result.
14.15. Solve the integral equation y x g x
1 y u r x u du
where g x and r x are given.
Suppose that the Fourier transforms of y x ; g x ; and r x exist, and denote them by Y ; G ; and R , respectively. Then taking the Fourier transform of both sides of the given integral equation, we have by the convolution theorem Y G & p 2 Y R or Y 1 G p 2 R
Then
y x f 1
' G 1 1 G p p p e i x d 2 1 1 2 R 1 2 R
assuming this integral exists.
Supplementary Problems
THE FOURIER INTEGRAL AND FOURIER TRANSFORMS & 1=2 jxj @  14.16. (a) Find the Fourier transform of f x 0 jxj >  (b) Determine the limit of this transform as  ! 0 and discuss the result. Ans: 1 sin  ; a p 2  1 b p 2 jxj < 1 jxj > 1
& 1 x2 14.17. (a) Find the Fourier transform of f x 0  1 x cos x sin x x (b) Evaluate cos dx. 2 x3 0 r   2 cos sin 3 Ans: a 2 ; b  16 3
CHAP. 14] &
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