AMath350.F12.A9.Sol

# Exists 0 2 lim thus the z z 0 0 t x 1 i x z zz z z

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Unformatted text preview: ! lim = . Thus the Z Z 0 0 t x) (1 i )x ! Z ZZ Z Z 01 s 0 1 xi! i! )e i 1. (c) Z imit(x)ie x idx dx exist, sox e iZ x dxi = dx diverges. i xx dx x dxlim e e x dx di- = (a) L f edoes not = x convergese xe x dx xe i xlim and e(1 Hence both converge. xi = 0 0 s ixs s 1 i! x Z 2. (a) verges x)g (ytheZiiFourierx Transform (ofee i(x) = e. idoes not exist. te i t e i s t 1 f ( and s e ts dxi = g (y ) ) f x) f x dxx x t ee xe Z i x dx1= lim xe= Z 1 =. lim the Fourier 2Transform of f (x) exists. 2 dx + = lim t lim Thus exists the integral on thei! + cont 1 02 t t 0 0 2 Since thex2 Fourier Transform tof f (x) i! right2 ! s Z 1 | dx = 2 lim i! x dx = 2 lim ! e x = 1 lim e x = 1. i! 1Z (b) |xe xe t verges, does not exist, so on Z xe left dx tdiverges. Hence thus the integral0 the i x converges 2for any value tofe yi, x dx we and di0 Z imit L x 2 ˆ ˆ Thusˆxe gx ))=absolutely (Clearlyfuy )! , y )x it is y )f ! ). Further, 2. (a) havefu(andythe ZFourier ). integrable. (Since = Z. (also (continuous, by Theorem (x! , ( is g (x dx(= Transformˆof e (x) = g does not exist. ) y e i y )f ! g y0 ) (x f i dxx verges 2 5.31 x, y = Notes y ) i x (x exists of xe x dx = F{uy (the )Fourier f (x)Z t(the Fourier Transform f (x)t i xexists. g (y )f cong e Since of the}CourseTransforme of f dx)= g (y ) the integral on theˆright2ˆ(! ). 0 Z 2 2 2. Let the Fourierx2TransformZ of the function= (xlim Z 0 1 e xF {f (x)} =im(! ).x i =x 1. Z Z 0 (b) verges, thusidx =integral on Zthedx fconverges and any= 1 i x off ye (1 )we |xe | the 2 lim xxe i xx left 2 i) xexist (1 i )x value l ,e and for Z x ix ee 0 (c) dx = dx(x)e lim dx + g (ˆ ) 0dx e lim . e2 y = tdx = (b) (f (x)e+ g (y ))e tˆ dx = f ( x) f =s t s have (u(! ,x2)is absolutely integrable. (Since =show)that). Further, Trans-! s ˆ y )y =(Z (y )f ). Use the deﬁnition ) it sis (y f (continuous, by Theorem g )g (y (! ). Clearly uy ! , y to Z also ! the Fourier 1 i ˆ g a) Let uﬁrst integral s the right converges since the Fourier Transform of f (x) Thus xe The x,1 = sf x on i x F{u of u(x, y )e y ) Course Notes 1 )e i exists for (y ) (x dx Transform f x e i...
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