Boersma, J.

Chapman, R.J.

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 Solution to problem 94-12 : A nonharmonic trigonometricseriesCitation for published version (APA):Boersma, J., & Chapman, R. J. (1995). Solution to problem 94-12 : A nonharmonic trigonometric series. SIAMReview, 37(3), 443-448.Document status and date:Published: 01/01/1995Document Version:Publisher’s PDF, also known as Version of Record (includes final page, issue and volume numbers)Please check the document version of this publication:• A submitted manuscript is the version of the article upon submission and before peer-review. There can beimportant differences between the submitted version and the official published version of record. Peopleinterested in the research are advised to contact the author for the final version of the publication, or visit theDOI to the publisher's website.• The final author version and the galley proof are versions of the publication after peer review.• The final published version features the final layout of the paper including the volume, issue and pagenumbers.Link to publicationGeneral rightsCopyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright ownersand it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights.            • Users may download and print one copy of any publication from the public portal for the purpose of private study or research.            • You may not further distribute the material or use it for any profit-making activity or commercial gain            • You may freely distribute the URL identifying the publication in the public portal.If the publication is distributed under the terms of Article 25fa of the Dutch Copyright Act, indicated by the “Taverne” license above, pleasefollow below link for the End User Agreement:www.tue.nl/taverneTake down policyIf you believe that this document breaches copyright please contact us at:openaccess@tue.nlproviding details and we will investigate your claim.Download date: 16. Nov. 2023PROBLEMS AND SOLUTIONS 443 Furthermore, by using [2, p. 230, (1)] a computable representation for the G-function in (1) (and hence also for F,112(s)) may be given in terms of essentially a linear combina- tion of four hypergeometric functions OF3[b2s2 /16] provided that no two of the parameters {1, 1 1+v-P- 1v-P) differ by an integer. This result for Fv-112(s) is given (correctly) in [3, p. 353, (9)] so that the proposed integral is in fact already tabulated in terms of known computable functions. REFERENCES [1] I. S. GRADSHTEYN AND I. M. RYZHIK, Table of Integrals, Series, and Prodcicts, Academic Press, San Diego, 1994. [2] Y L. LUKE, The Special Functions and Their Approximations, Vol. I, Academic Press, San Diego, 1969. [3] A. P. PRUDNIKOV, Yu. A. BRYCHKOV, AND 0. I. MARICHEV, Integrals and Series, Vol. 2, Gordon and Breach, New York, 1986. Also solved by R. G. BUSCHMAN (Langlois, Oregon), M. L. GLASSER (Clarkson Uni- versity), CARL C. GROSJEAN (University of Ghent, Ghent, Belgium), NORBERT ORTNER AND PETER WAGNER (University of Innsbruck, Innsbruck, Austria), and JET WIMP (Drexel Uni- versity), and the preparers. A Nonharmonic Trigonometric Series Problem 94-12, by M. L. GLASSER (Clarkson University). The following function has been encountered in studying the dielectric function for a "quantum dot": E sin(a ,I) f (a) = T n=1 Find lima+1o f(a) and limaeo f'(a). Solution by J. BOERSMA (Eindhoven University of Technology, Eindhoven, The Netherlands). The two limits are determined by a Mellin transform ethod with the results lim f (a) = 7r, limf '(a) = ( (), aI~O aj~O2 where ? stands for the Riemann zeta function. We start with the Mellin transform roc M[sinx; s] sinx x'-1 dx = sin(rs/2)r(s), valid for -1 < Re s < 1, and absolutely convergent for -1 < Re s < 0. By Mellin's inversion formula [1, p. 46, Thm. 281 we have 1 rc+ioo sinx = 27r I sin(7rs/2)r(s)xsds, (x > 0), in which -1 < cl < 0, so that the integration path is in the strip of absolute convergence. In fact, the path may be shifted to Re s = C2 with -1 < C2 < 1/2. To show this, we use the asymptotic expansion of the r-function to establish the estimate sin(7rs/2)f(s) = O(Itia-112), (t -z*o ), 444 PROBLEMS AND SOLUTIONS where s = a + it. Then it follows that the contributions ofthe closing line segments [cl + iT, c, ? i T] vanish as T -x o, if cl < c2 < 1/2. The next step is to replace sin(a,/ih) by the integral representation presented above, and to interchange the order of summation and integration i the series for f (a). As a result it is found that 1 f C+ioC (1) f(a) = . sin(7rs/2)r(s)w(1 +s/2)a'5ds, 27ri J_-i__ where the factor ((1 + s/2) stems from E=l n'I-s12 = ((1 + s/2), valid for Re s > 0; accordingly, one should take 0 < c < 1/2. The integrand (1) is an analytic function of s, except for simple poles at s = 0 with residue 7r, and at s = -2k - 1 with residues (l)k'(l/2 -k)a2k+l/(2k + 1)!, where k = 0, 1, 2, .... By shifting the path of integration to Re s = -2N (with N = 1, 2, 3, ... )5 we are led to the representation N-I (_l)k~.Q - k)akI+ N() k=i (2k + 1)! where the remainder term RN (a) is given by 1 r-2N+ic0 (2) RN (a) = I sin(7rs/2)r(s)(1 + s/2) a-s ds. 27r i J-2Nioo To justify the shift of the integration path, we need the estimate [2, pp. 81-82] ((1 + s/2) = T(1 + 4 (a + it)) = O(1t|1(a)+e), (I i x) valid for every e > 0, in which it(u) is determined by gu(u) =0 if a > 0; gu(a) < -a/4 if -2 < a <0; gt(c) = -(1 +o)/2 if a <-2. It is now clear that a - 1 + gt(o) < 0 if a < 4, and the integrand in (1) tends to zero as Im s -- ?ox, for fixed Re s <- . Hence, the contributions ofthe closing line segments [-2N?iT,c?iT]vanishas T-- x. Changing the integration variable s into -s in (2), we use the functional equation of the zeta function and some standard properties of the r-function to obtain the representation 27 3/.2 2N+ioc cos(7rs/4) (s/2) I a s R(a--27r i JN-ioe COS(7rs/2) srm( + s)/2) 2 V /2, s In the latter integral we set s = 2N(1 + it). Next we establish the inequalities cos(N7r(1 + itO/2) < 2 exp[- N7r It 1/2], 1~(N(l + it))l < N) cos(Nzr(1 + it)) and, by use of the asymptotic expansion of the r-function, < (27rf /2KNNNe 1 + ite N exp[N7r ItI/2], r(I + N(I + it))N where KN = 1 + O (N1). The remainder term RN (a) can now be estimated by IRN(a)l < 2l/2KNO(N)NN ( ) N +tIN+1 = 1 / 2KN (N)N-N /_ 2 N r(I1/2)r(N/2) - 1/ 2KN ,(N-J N -N ~1V~\~I~' 2,/2 2zre/ r((N + 1)/2) PROBLEMS AND SOLUTIONS 445 From the latter estimate it is clear that RN (a) -+ 0 as N -+ oo. Consequently, we have for f (a) the representation bya convergent power series f (a) =7r+ (2k + 1)! valid for all a > 0. The limit results tated at the outset are now obvious. Alternatively, -( I-k) may be expressed in terms of 1 + k) by means of the functional equation of the zeta function, leading to f(a)=zr+27r1/2 sin((k + ') 7)~(k+) (a) 2k+1I f (a) = 7r  2nr 1/E sn(+ 2) 2 2( a ) (k + )k! The latter series is convergent for all complex a. Remark 1. For a quick derivation of f (0+) = 7r, write f (a) as a Riemann sum: f (a) = E 0=1 g(a2n)a2, where g(x) = sin(jx)/x. It can be shown that lima,0o f (a) is equal to the Riemann integral j g(x)dx =j i (T ) dx = 2 sin dt = 7r. x ~~~~~~~t Remark 2. Similar results can be derived for the cosine-series, namely, 00 cos(a,/n-) I C+iOO0 E c( ) - 2 i 1 oO cos(7rs/2)I(s)((1 +s/2)a-sds n=1 = -2loga-+y + E (2)! a2k = 2loga-y + -+ 7r 1/2 L 1~2 )(~~4 , (a > 0), where4 E mr(2m + Ee constant where y is Euler's constant. REFERENCES [ 1 ] E. C. TITCHMARSH, Introduction tothe Theory of Fourier Integrals, 3rd ed., Chelsea, New York, 1986. [2] , The Theory of the Riemann Zeta Function, Clarendon Press, Oxford, 1951. Solution by ROBIN JOHN CHAPMAN (University of Exeter, UK). We start by proving a more general result. Let (1) f,q(a) = L n), (0< < 1). n=1 I claim that lima,,o ffi(a) = zr/(2,6). I remark that this is also true when ,B = 1, for by the theory of Fourier series ? sin(na) zr-a n=h n 2 when O < a < 27r. 446 PROBLEMS AND SOLUTIONS Note that if ?(x) is a continuously differentiable function then integration by parts gives rn n (x)dx = ?(n) - (x - n + l)&'(x) dx. Applying this to ?(x) = sin(ax 0)/x and summing ives N sin(anO) [N sin(axo) dx + (x cos(axO) _ sin(axO) ) (2) E x j{x Iap x2- x2 )dx, n=I n O X r where {x} = x - Lxj denotes the fractional part of x. Putting y = axo gives jN sin(axf) 1 jaN sin I ~~~dx =-Idy, JO x j O y which tends to zr/(21i) as N -+ oo. If E > 0 the convergence is uniform for a > E. The integrand of the second integral is O(x-l) as x -O 0 and O(xO2) as x -+ oc. Hence (3) ffi(a)~~7 = + cos(axfi) _sin(axO) (3) o(a)  2p + 1 x}I (ap 2 -# 2,) dx. If K > E > 0 the second integrand in (2) is bounded by (Kfi + 1)x -2 so the series (1) converges uniformly for a E [e, K]. The integrand in (3) tends pointwise to zero as a J, 0. If 0 < a < I it is bounded in absolute value by 2+r(x), where xI I ifx < 1, (x) -x2 if X > 1 is a positive integrable function. Hence by Lebesgue's theorem of dominated convergence 7r lim f (a) = a40 2 as claimed. We now return to the case ,B = 1/2. Using (3) we write f(a) = fl/2(a) = r + 'agi(a)- g2(a), where f0 co s(a~f gi(a) = ]{x} (3a2 dx and 92(a) = j{x}- ) dx. Differentiating the integral for go formally with respect to a gives the integral for gl. As the integrand in the integral for gI is bounded above by the L1 function x H-+ {x}/x3/2, then g'(a) = g, (a). Now consider gI. Integration by parts gives, for a suitable function 4, jn { xhPI(x) dx= 4 (n) - X {x}2 0'(x) dx. n- 12 _ 2 PROBLEMS AND SOLUTIONS 447 Applying this to ?(x) = x-3/2 cos(a Vx) and summing ives gi (a 1 = cos(a3'/i) 0; {x}2 a sin(a %,/i-) 3 cos(a),/ix) d 2E 1 n312 J 2 2x2 2x5/2 / The formal derivative of this with respect o a is _1 sin(a%/;i) f X {x}2 (a cos(a/X') _ 2 sin(a xi) h (a) = - = + J 4 3 12 2 /dx. We have already shown that he sum here is uniformly convergent when a lies in a set of the form [e, K]; also the integrand isbounded by a multiple of the L1 function x -+ X}2 (X-2 + x-3/2) when a < K. Hence gj(a) = h(a), and so f'(a) = - gI(a) + a2h(a). By Lebesgue's dominated convergence theorem, and the fact that lima,o f(a) exists, it follows that h(a) tends to a finite limit as a J, 0. Hence, again by Lebesgue's theorem, lim f'(a) = 1 d{x} It only remains to identify this final integral. If s = a + it is a complex number with real part a > 1, then E = 1+ E s= 1+ }; s -, dx E_ X1j ?sjN{ +1d and so aocxf{XI s {x}1 (4) ((s) = I + JX-s J '1 dx= -s f { dx. But the integral on the right-hand side of this is an analytic function for a > 0, and so by analytic ontinuation (4) is valid for all s =$ 1 with a > 0. Suppose that s is real and 0 < s < 1. Then {x} 'dx 1 A Xs+d 1 Xs 1-s Hence for 0 < s < 1 we have a(S) = -sj ' } dx and so ((1/2) =[- X dx as required. Also solved by CARL C. GROSJEAN (University of Ghent, Ghent, Belg,ium), W. B. JORDAN (Scotia, NY), KEE-WAI LAU (Hong Kong), G. LOHOFER (German Aerospace Research Establishment, Koln, Germany), 0. P. LOSSERS (Technical University, Eindhoven, The Nether- lands), G. F. NEWELL (University of California, Berkeley), PETER WAGNER (University of Inns- bruck, Innsbruck, Austria), JAMES A. WILSON (Iowa State University), and the proposer. 448 PROBLEMS AND SOLUTIONS PETER WAGNER also found the generalization obtained by ROBIN CHAPMAN. The proposer incorporated both the results of Boersma and Chapman in the following eneralization. Let fp(a) = sin(anP) (a > 0). n=1 n Then the Mellin transform is Fp(s) = L fxs-1sin(npx)dx = r(s) sin(7rs/2) (l + ps), (O < Res < l, p > 0). For p > 0, 1 C+io _ fp(a) = 2si f(asFS) sin(rs/2) (1 + ps)ds, (O < c < For 0 < p < 1 we can close the contour to the left (justified as in Boersma's solution) and sum the residues of the simple poles to get fp(a) = + J ((2+1)! -(1 -(2k+ l)p). For p = 1, this result reduces to fi (a) = (7r - a)/2, valid for 0 < a < 27r. The generalization to the sum with nY in the denominator can be obtained in the same way, but there are several cases that must be looked at separately. A Limit Problem from the Theory of Delay Equations Problem 94-13*, by YUNKANG Liu (Cambridge University, Cambridge, England). Suppose that a and b are complex constants, Re b > 0 and A E (0, 1). Prove or disprove (1) li Y(t)eb = - ( (l~~~~~~~~ )0 k=o b where ? tn n-I y(t) = 1 + E L H(b + axk) n= 1*k=O is the unique solution of the delay differential equation y'(t) = ay(AI) + by(t), I > 0, y(O) = 1. The existence of the limit was proved by Kato and McLeod [1] in the case of b > 0 and later extended by Kato [2] to the case of Re b > 0. The proof that (1) holds in the case of b > 0 is trivial. 

Solution to problem 94-12 : A nonharmonic trigonometric series

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