By Jurii V. Linnik, Iosif V. Ostrovskii

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**Example text**

F. 's of every order p, 1 < p <; 00 ,andprescribed type o, 0 <; o <; oo. f. f. 4). f. ip(t; F) is of order p and type o. 'Spectively, If p = oo, then we take { O, x~O, 1- exp { - xx In+ x}, x > 0, II. F. 'S 40 . {0, x-

F. 'l't,,(t), a ;;;;ii*· considered in Example 2 of §3. Since cp°' (t) = exp {ex [2eit ·:_ e2 it"+ 3e8it + 3e it 4 7) }, the function 'l'a(t) indeed has no zeros. f. f. f. ;;;; b, a< b), possesses an analytic continuation beyond the strip to a region in which it has zeros. · EXAMPLE 1. For w = a + ib, a, b > 0, set 00 cp p= 2 J· e-ax (1-cos bx) dx (t) = (1- it/w) (1-it/W) (1- it/a)2 I ' 1 ·-i-X2 0 J ' I 00 N (x) = -2 e-au (1-cos bu) u-1 du, O

We claim that b 1 ;;;r. b. Indeed, if b 1 < b, it is easy to see that there exists a disk ltl < b 2 , b 1 < b 2 < b, in which 1{'2 (t) may be analytically continued. , we conclude that 1{'2 (t) can be analytically continued into the strip IIm ti < b2 • But then 1{'2 (t) can be analytically continued into the half-plane Im t < b2 , contradicting the choice of the number b 1 Thus, the function 1{'2 (t) may be continued analytically into Im t < b. Since 1{'1 (t) is analytic in Im t > 0 and continuous in Im t ;;;r.