VISTAS OF SPECIAL FUNCTIONS

By ROBERT MARTINEZ

CONTENTS 1 THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS 2 THE THEORY OF THE GAMMA AND RELATED FUNCTIONS 2 1 GAMMA FUNCTION 2 2 THE EULER DIGAMMA FUNCTION 3 THE THEORY OF THE HURWITZ-LERCH ZETA-FUNCTIONS 3 1 INTRODUCTION 3 2 INTEGRAL REPRESENTATIONS 3 3 A FORMULA OF RAMANUJAN 3 4 SOME DE?NITE INTEGRALS 3 5 THE FUNCTIONAL EQUATION 4 THE THEORY OF BERNOULLI POLYNOMILAS VIA ZETA-FUNCTIONS 5 THE THEORY OF THE GAMMA AND RELATED FUNCTIONS VIA ZETA-FUNCTIONS 5 1 DERIVATIVES OF THE HURWITZ ZETA-FUNCTION 5 2 ASYMPTOTIC FORMULAS FOR THE HURWITZ AND RELATED ZETA- FUNCTIONS IN THE SECOND VARIABLE 5 3 AN APPLICATION OF THE EULER DIGAMMA FUNCTION 5 4 THE ?RST CIRCLE 6 THE THEORY OF BESSEL FUNCTIONS AND THE EPSTEIN ZETA-FUNCTIONS 6 1 INTRODUCTION AND THE THEORY OF BESSEL FUNCTIONS 6 2 THE THEORY OF EPSTEIN ZETA-FUNCTIONS 6 3 LATTICE ZETA-FUNCTIONS 6 4 BESSEL SERIES EXPANSIONS FOR EPSTEIN ZETA-FUNCTIONS 7 FOURIER SERIES AND FOURIER TRANSFORMS 7 1 FOURIER SERIES 7 2 INTEGRAL TRANSFORMS 7 3 FOURIER TRANSFORM 7 4 MELLIN TRANSFORM 8 AROUND DIRICHLET’S L-FUNCTIONS 8 1 THE THEORY OF PERIODIC DIRICHLET SERIES 8 2 THE DIRICHLET CLASS NUMBER FORMULA 8 3 PROOF OF THE THEOREMS CHAPTER THE BERNOULLI POLYNOMIAL BN(X) OF DEGREE N (N = 0, 1, 2, ) CAN BE DE?NED BY EITHER OF THE FOLLOWING DE?NING CONDITIONS (D?) (APPELL SEQUENCE 1832) B ? N(X) = N BN?1(X) (1 1) WITH INITIAL VALUE B0(X) = 1 AND WITH NORMALIZATION Z BN(X) DX = (N ?N) IF WE KNOW THE DI?ERENTIATION FORMULA (D?), IT IS IMMEDIATE TO CALCULATE THE K-TH DERIVATIVE N K BK(0) = 0, N ?2 (1 3) WE DENOTE THE VALUE BN(0) BY BN AND REFER TO IT AS THE N-TH BERNOULLI NUMBER THROUGHOUT IN WHAT FOLLOWS WE UNDERSTAND THEY ARE DE?NED BY (1 3) ONCE AND FOR ALL ANOTHER DE?NITION BY (1 6) LEADS TO THE SAME RECURRENCE AS (1 3) AND THESE TWO DE?NITIONS ARE CONSISTENT (CF REMARK 1 1) THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS N K BN?K XK (1 7) (K) KUBERT IDENTITY (RAABE 1851) KR (1 10) THEN Z(N, R) = R + 1 (BR+1(N + 1) ?BR+1) , (R ?N) (1 11) AND Z(N, R) = R + K BK (N + 1)R+1?K, (R ?N) (1 12) THIS WAS KNOWN TO JACOB BERNOULLI IN HIS ARS CONJECTANDI, WE NOTE THAT THE FUNCTION Z EZ ?1 IS ANALYTIC IN |Z| < 2? (INCLUDING THE ORIGIN CF THEOREM A 8), SO THAT IT HAS THE TAYLOR EXPAN- SION (1 6), WHICH IMPLIES THE RECURRENCE (LEFT-HAND SIDE OF (1 3))= 0, N ?2 = 1, N = 1, IN CONFORMITY WITH (1 3) ON THE OTHER HAND, Z EZ ?1 EXZ HAS ITS THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS N K BK, 2, B2 = 2, B2(X) = B0X2 + 2B1X + B2 = X2 ?X + 2X2 + 2X4 + 6X, B6(X) = X6 ?3X5 + 42, B6 = 42 WE SHALL PROVE THE EQUIVALENCE OF SOME OF THE CONDITIONS (D?) ?(H), SOME BEING LEFT UNPROVED (D?)?(A) INDEED, (A) IS NOTHING BUT THE TAYLOR EXPANSION OF BN(X + Y) IN Y, AND THE TAYLOR COE?CIENT IS B(K) N (X), WHICH IS N! (N ?K)!BN?K(X) BY (1 2), WHENCE THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS N ?L K BN?L?K XK, (1 15) BY THE CHANGE OF VARIABLE NOW THE INNER SUM IS BN?L(X) BY (1 7) (U)?(DE) IS PROVED IN THE FOLLOWING K S + S BK?1?S XS = KBK?1(X) PROOF OF Z BN(X) DX = 0 FOR N ?N Z N K BN?K Z XK DX = N + K + BN?K = K BN+1?K THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS BY (1 4); THIS IS IN TURN BN+1(0) ?BN+1 BY (1 16) AND IS 0 NOTE THAT R E2?I X M R RN MN E2?I R M K EKZ THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS (?1)N N + Z(N, R) (1 18) WE SHALL DEDUCE (1 18) FROM (1 12) WITH THE HELP OF (1 19) WHICH GIVES THE CLOSED FORM FOR THE STIRLING NUMBER OF THE SECOND KIND (CF (1 20)) S(N, K) = (?1)J K J (K ?J)N (1 19) = (?1)K?I K I IN = K! ?K0N, 1 ?K ?N, K BK(N + 1)R+1?K = (?1)N R + (N + 1)R?K = (?1)R+J R + J (R + 1 ?J)R?K ON CHANGING THE ORDER OF SUMMATION AND THEN WRITING R ?N = J DIVIDE THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS THE SUM OVER K INTO TWO: 0 ?K ?R ?1 AND K = R RHS = (?1)J R + K BK(?1)R(R + 1)!S(R ?K, R + 1) + (?1)RBR, = (?1)RBR = BR, WHERE WE USED THE FACT THAT S(N, R) = K2 = 3 (B3(N + 1) ?B3) = 2(N + 1) + 6(N + 1)(2N2 + 4N + 2 ?3N ?3 + 1) = (J ?1)! S(R + 1, J) N J (1 21) S(R + 1, J)N (N ?J + 2), THE ?RST TERM IS Z(N, R) BY HYPOTHESIS AND THE SECOND CAN BE WRITTEN AS S(R + 1, J)(N + 1)J, WHICH IS N + 1(N + 1)R+1 = (N + 1)R, ON APPLYING (1 20) ? THE SECOND PROOF WE ?RST PREPARE AUXILIARY RESULTS FIRST, N X J=K J K = N + XK+1, WE OBTAIN (1 22) BY COMPARING THE COE?CIENTS OF XK+1 SECONDLY, WE ALSO NEED THE TRIANGULAR RECURRENCE FORMULA FOR S(R, J): S(R + 1, J) = S(R, J ?1) + J S(R, J), 1 ?J ?R + 1 (1 23) THIS MAY BE PROVED BY WRITING XR+1 IN (1 20) IN TWO WAYS: THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS K J = N J + N J + WE TAKE UP THE SPECIAL CASE OF (1 21) Z(N, 2) = (J ?1)! N J S(3, J) VISTAS OF SPECIAL FUNCTIONS SINCE FROM (1 19) WE HAVE S(3, 1) = 1, S(3, 2) = 3, S(3, 3) = 1, IT FOLLOWS THAT Z(N, 2) = S(3, 1)N + N S(3, 2) + 2! N S(3, 3) = N + 2N(N ?1) + 3N(N ?1)(N ?2) = 6N(2N2 + 3N + 1) = 2 = X + B1 DENOTING THE RIGHT-HAND SIDE OF (1 9) BY BN(X), WE OBTAIN N D DX BN(X) = BN?1(X), WHENCE N! DN DXN BN(X) = B1(X), 0 < X < 1 INTEGRATING D DXB2(X) = B1(X), WE DEDUCE THAT 2 B2(X) = Z B1(X) DX = 2 X2 + B1X + C, WHERE C = 2 B2(0) = N2 = ?2 ?(2) = BY (5 66) REPEATING THIS PROCEDURE, EACH TIME USING (5 66), WE ARRIVE AT (U) NOW WE SHALL FOLLOW LEHMER [LEH2] TO DEDUCE SOME OF THE ABOVE DE?NING CONDITIONS FROM (K) FIRST WE STATE A LEMMA LEMMA 1 1 (LEHMER) FOR A GIVEN N ?N THERE IS A UNIQUE MONIC POLYNOMIAL OF DEGREE N SATISFYING (1 8) THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS RR X + K M = RR(X) IDENTIFYING THE COE?CIENTS OF XR ON BOTH SIDES, WE OBTAIN MN M R A0 = A0, WHICH CONTRADICTS THE FACT THAT R < N AND A0 ?= 0 THIS COMPLETES THE PROOF ? THEOREM 1 2 (LEHMER) FOR N, M ?N, THERE EXISTS A UNIQUE MONIC POLYNOMIAL OF DEGREE N SATISFYING THE FUNCTIONAL EQUATION F X + K M = M?NF(X) (1 24) PROOF SINCE FOR M = 1, (1 24) REDUCES TO A TRIVIAL EQUALITY, WE MAY ASSUME THAT M > 1 WE SUBSTITUTE A CANDIDATE POLYNOMIAL PN(X) = B0XN + B1XN?1 + + BN, B0 ?= WITH BK AS INDETERMINATES, INTO (1 24) IF F SATIS?ES (1 24), THEN SO DOES ANY MULTIPLE OF F, SO THAT WE MAY ASSUME B0 = 1 THE LEFT-HAND SIDE OF (1 24) BECOMES XN?RBRM?N BK N ?K R ?K MK?1Z(M ?1, R ?K) (1 25) NOW, FROM Z(N, 1) = N(N+1) , IT FOLLOWS THAT B1 = N R ?K + R K BK (1 26) THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS R ?L + R ?K ?L + = ( 1, L = R 0, L ?= R BY (U) AND (1 16) SUCCESSIVELY SINCE THE LAST MEMBER IS 0 EXCEPT WHEN L = R, WE INFER THAT S2 = BR = BR (1 28) ON THE OTHER HAND, S1 = Z R K BKXR?K DX = Z (BR(X) ?BR) DX = R + 1(BR+1(1) ?BR+1(0)) ?BR = ?BR HENCE LHS = ?M?N+R N R (?BR) ?M?N N R BR OR M?N(MR ?1)BR = ?(M?N ?M?N+R) N R BR, WHENCE BR = N R BR, N! GN(X)TN, |T| < 2? (1 29) WITH GN(X) A POLYNOMIAL OF DEGREE N, GN(X) = A(N) 0 XN + + A(N) N , SAY WE MAY DETERMINE A(N) AS FOLLOWS: REPLACING X BY N! YNGN N! A(N) 0 TN (SINCE YNGN Y ?A(N) 0 ): HENCE A(N) = 1 AND GN IS MONIC NOW, N!GN X + K M TN = F X + K M , T = E X+K M T = M T E X M T E T M ?1, THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS N! GN(X) MN TN HENCE, COMPARING THE COE?CIENTS OF N!TN, WE-CONCLUDE THAT M K BK ZM?K, B1 ?= M J M ?J K ?J ZM?K USING (1 14), WE GET M J M ?J K ?J = M M ?J M ?J M ?K = M M ?K K K ?J = M K K J M J AJZM?J + F(Z) = AM (1 34) THIS IS INDEED THE CASE BECAUSE THE COE?CIENT MATRIX IS LOWER TRIAN- GULAR, SO THAT ITS DETERMINANT IS THE PRODUCT OF ALL DIAGONAL COMPONENTS AS IS REMARKED IN [MI, P 284], THE R + 1-ST BERNOULLI POLY- NOMIAL CAN BE CHARACTERIZED AS THE UNIQUE POLYNOMIAL SATISFYING (1 11) FOR EVERY NATURAL NUMBER R HENCE (S) CAN BE ALSO USED AS THE DE?NITION AS (K) AND (DE) THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS BM?1(Z) (M ?1)! XM, WHENCE WE HAVE Z + K + Z + K Q + (M + 1)QMZM (1 40) Z + K Q = F(Z), (1 41) SAY, WHENCE F Z + Q = F(Z) I E F(Z) IS PERIODIC OF PERIOD X+K M = N BN?1(X) BY (1 1), WHENCE (1 8) ENSUES FINALLY, WE ADD THE FOLLOWING AS A MEMORIAL OF THE WORK OF L LAGRANGE, WHO WAS A CONTEMPORARY OF EULER, WHICH IS OFTEN USED IN NUMBER-THEORETIC SETTINGS THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS L LAGRANGE X ??L I AND GI(X) = Y K?=I GK(X), 1 ?I ?N (1 43) THEN THE POLYNOMIAL LI(X) = FI(X) X ??K I ?K I GI(?K I ) + GI(X) ?I(X) = 1 PROOF WE NOTE THE FOLLOWING FACTS SINCE F ? I(X) = KI XKI?1, WE HAVE, FOR A PRIMITIVE KI-TH ROOT OF 1 AND 1 ?L ?N ?1, F ? I ?L I = KI ?L(KI?1) I = KI ??L I THE THEORY OF BERNOULLI AND ALLIED POLYNOMIALS FI(X) X ??K I ?I ?K I F ? I ?K I + KI ?I(1) GI(X) (1 45) IS THE LAGRANGE INTERPOLATION POLYNOMIAL FOR ?I(X) IN VIEW OF GI(1) = KI FROM THE ABOVE DATA, F ? I ?K I = KI ??K I AND ?I(?K I ) = F(X) X ??I ?R I F ?(?I) AND ALL SUMMANDS ON THE RIGHT ARE CONJUGATE ONE ANOTHER, SO THAT THE RIGHT- HAND SIDE IS THE TRACE THIS IS ESSENTIALLY USED TO ?ND THE DUAL BASIS IN A ?NITE EXTENSION OF Q THIS PAGE INTENTIONALLY LEFT BLANK THIS PAGE INTENTIONALLY LEFT BLANK CHAPTER GAMMA FUNCTION FIRST, WE DEVELOP THE THEORY OF THE GAMMA FUNCTION DE?NED BY (2 1), ONE OF EQUIVALENT CONDITIONS TO BE DISCUSSED IN CHAPTER 5 THE GAMMA FUNCTION, BEING THE MELLIN TRANSFORM OF E?X TO BE MENTIONED IN §7 4, IS DE?NED BY THE EULERIAN INTEGRAL OF THE SECOND KIND ?(S) = Z ? E?XXS?1 DX (2 1) FOR ? > 0 THIS IMPROPER INTEGRAL IS ABSOLUTELY AND UNIFORMLY CONVERGENT IN THE WIDE SENSE IN ? > 0, WHENCE IT FOLLOWS THAT ?(S) IS ANALYTIC IN THE RIGHT HALF-PLANE ? > 0 THE GAMMA FUNCTION MAY BE CONTINUED MEROMORPHICALLY OVER THE WHOLE PLANE BY THE DI?ERENCE RELATION ?(S) = ? (S + N + 1) S (S + 1) (S + N), N ?N ?{0} (2 2) EULER INDEED, FOR ? > 0, WE HAVE BY INTEGRATION BY PARTS, ?(S) = S E?XXS ? S Z ? E?XXS DX = S ?(S + 1), (2 3) WHOSE RIGHT-HAND SIDE IS ANALYTIC FOR ? > ?1 FROM (2 2) WE SEE THAT S = ?N, N ?N ?{0} ARE SIMPLE POLES AND THAT THE RESIDUES AT THESE POLES ARE RES S=?N ?(S) = (?1)N N! (2 4) ALSO IF WE PUT S = N ?N IN ?(S + 1) = S ?(S), (2 5) THEN WE HAVE ?(N + 1) = N!, WHICH MEANS THAT THE GAMMA FUNCTION IS A FUNCTION WHICH INTERPOLATES THE FACTORIAL N! THE THEORY OF THE GAMMA AND RELATED FUNCTIONS IN CONTRAST TO (?)R, ???R INDICATES THE POCHHAMMER SYMBOL (OR THE SHIFTED FACTORIAL, SINCE ?1?R = R!) DE?NED BY ???R := ?(? + R) ?(?) = ( (R = 0) ? (? + 1) (? + R ?1) (R ?N) (?1)N N! S + N + Z ? E?XXS DX (2 8) THE IMPROPER INTEGRAL IS ABSOLUTELY AND UNIFORMLY CONVERGENT IN ANY DISC |S| < R AND REPRESENTS AN INTEGRAL FUNCTION THE PROOF FOLLOWS ON DIVIDING THE INTERVAL OF INTEGRATION AND NOTING THAT Z (?1)N N! Z S + N LET THE BETA FUNCTION B(?, ?) BE DE?NED BY THE EULERIAN INTEGRAL OF THE ?RST KIND B(?, ?) = Z PROVE THE FORMULA ?(?) ?(?) = ?(? + ?) B(?, ?), (2 10) WHENCE IN PARTICULAR ? = ??, (2 11) OR THE VALUE OF THE PROBABILITY INTEGRAL R ? 0 E?X2 DX = ?? 2 SOLUTION FIRST, IN (2 9), PUT T = SIN2 ? TO OBTAIN B(?, ?) = Z ? SIN2??1 ? COS2??1 ? D? (2 12) IF IN (2 1), WE PUT T = X2, THEN ?(?) = Z ? X2??1E?X2 DX, THE THEORY OF THE GAMMA AND RELATED FUNCTIONS WHENCE FOR RE ? > 0, RE ? > 0, ?(?) ?(?) = X2??1E?X2 DX Z ? Y2??1E?Y2 DY = 4 LIM X?? Z X X2??1E?X2 DX Z X Y2??1E?Y2 DY ! = 4 LIM X?? Z X Z X X2??1Y2??1 E?(X2+Y2) DX DY = 4 LIM X?? ZZ D X2??1Y2??1 E?(X2+Y2) DX DY, (2 13) WHERE D = NX Y 0 ? P X2 + Y2 ?X O BY THE CHANGE OF VARIABLE X = R COS ?, Y = R SIN ?, WE HAVE THE CORRESPONDENCE D ?˜D = R ? 0 ?R ?X, 0 ?? ?? , WHERE THE ABSOLUTE VALUE OF THE JACOBIAN OF THIS TRANSFORMATION IS ?(X,Y) ?(R,?) = R HENCE ?(?) ?(?) = 4 LIM X?? ZZ ˜ D R2?+2??2 E?R2 SIN2??1 ? COS2??1 ? R DRD? = R2?+2??1E?R2 DR SIN2??1 ? COS2??1 ? D? = ?(? + ?) B(?, ?) BY (2 12) ABOVE THIS COMPLETES THE PROOF OF (2 10) PUTTING NOW ? = ? = 2, WE OBTAIN ? = ?(1) 2 DX = ? 2?, WHICH IS USED TO NORMALIZE THE DISTRIBUTION FUNCTION OF THE GAUSSIAN (OR NORMAL) DISTRIBUTION AN ORDINARY PROCEDURE FOR PROVING (2 11) IS TO USE (2 13) FOR ? = ? = E?X2 DX = 4 LIM R?? Z R R E?R2 DR Z ? D? ! = 2 E?R2? USE THE FORMULA IN COROLLARY A 4 TO PROVE THAT B(S, 1 ?S) = ? SIN ?S (2 14) OR ?(S) ?(1 ?S) = ? SIN ?S, (2 15) THE RECIPROCITY RELATION FOR THE GAMMA FUNCTION, WHERE 0 < S < 1 IN THE ?RST PLACE AND THEN FOR ALL S EXCEPT FOR INTEGER VALUES BY ANALYTIC CONTINUATION SOLUTION MAKE THE SUBSTITUTION T = X+1 IN (2 9) TO DEDUCE THAT B(S, 1 ?S) = Z ? DETERMINE THE VALUE OF THE PROBABILITY INTEGRAL ? BY CONSIDERING TWO FUNCTIONS F(X) = Z X E?T2 DT AND G(X) = Z T2 + 1 E?X2(T2+1) DT SOLUTION RECALLING THE FUNDAMENTAL THEOREM OF CALCULUS, WE OBTAIN F ?(X) = 2 E?X2 Z X E?T2 DT WE MAY DI?ERENTIATE G(X) UNDER THE INTEGRAL SIGN TO GET G?(X) = Z T2 + D DX E?X2(T2+1) DT = ?2X Z E?X2(T2+1) DT = ?2 E?X2 Z X E?U2 DU BY THE CHANGE OF VARIABLE U = XT THE THEORY OF THE GAMMA AND RELATED FUNCTIONS HENCE WE CONCLUDE THAT F ?(X) + G?(X) = 0, WHENCE BY THE NEWTON-LEIBNIZ PRINCIPLE (CF LEMMA B 1) THAT F(X) + G(X) = F(0) + G(0) = Z 0 = ? 4 NOW LETTING X ??, THEREBY NOTING THAT LIM X??G(X) = Z LIM X??E?X2(T2+1) DT = 0, WE CONCLUDE THAT LIM X??F(X) = Z ? = ? 4 , I E Z ? USE FORMULA (2 10) IN THE FORM B(?, ?) = ?(?) ?(?) ?(? + ?) (2 16) AND THE VALUE OF THE PROBABILITY INTEGRAL ? = ?? TO DEDUCE WALLIS’ FORMULA ? 2 = LIM N?? 2 4 (2N) 1 3 (2N ?1) 2N + 1 (2 17) SOLUTION FIRST, BY (2 16) AND (2 12) ? ? N + ?(N + 1) = B 2, N + COS2N ? D? = S2N, SAY, AND ? ? (N + 1) ?(N + 2) = B COS2N+1 ? D? = S2N+1 = N + N = 1 + = ? N + 2 ? N (N ?1) = (2N ?1)(2N ?3) (2N)(2N ?1) LOG N ? N + LOG X ?X + C + O(1) (2 21) AND DETERMINE THE VALUE OF C THROUGH THE ASYMPTOTIC FORMULA FOR LOG 2N N = LOG ?(2N + 1) ?(N + 1)2 THE THEORY OF THE GAMMA AND RELATED FUNCTIONS FIRST, FROM (2 21), LOG ?(2N + 1) ?2 LOG ?(N + 1) = 2N + LOG(2N + 1) ?(2N + 1) + C ?(2N + 1) LOG(N + 1) + 2(N + 1) ?2C + O(1) = (2N + 1) LOG 2N + 2 LOG(2N + 1) + 1 ?C + O(1) SECONDLY, SINCE THE ?RST TERM ON THE RIGHT IS (2N + 1) LOG 2N + N + 1 = (2N + 1) LOG 2 ?(2N + 1) LOG 2N + 2N + = (2N + 1) LOG 2 ?(2N + 1) LOG 1 + = LOG 22N + LOG 2 ?1 + O N , WE OBTAIN LOG ?(2N + 1) ?(N + 1)2 = LOG 22N ?LOG ? 2N + 1 + LOG 2 ?C + O(1), (2 22) WHENCE LOG 2N N ?2N + 22N = LOG 2 ?C + O(1) (2 23) SINCE 2 4 2N , WHOSE LIMIT IS LOG Q ? BY WALLIS’ FORMULA HENCE C = LOG 2 ?LOG R LOG X ?X + LOG ? 2? + O X (2 24) LET ?(X) DENOTE THE NUMBER OF PRIMES ?X THEN USING A SPECIAL CASE OF STIRLING’S FORMULA (EXERCISE 2 6, (2 24)) LOG 2N N = LOG 22N ?2N + 1 + LOG R ? + O X LOG X SOLUTION FIRST WE NOTE THE INEQUALITIES I + 1 (2 27) THE ?RST ONE IS TRIVIAL WHILE THE SECOND ONE FOLLOWS FROM X 2 ?1 < X OR X < 2 X N PL , R = LOG 2N LOG P (2 28) THE THEORY OF THE GAMMA AND RELATED FUNCTIONS BEARING (2 27) IN MIND, WE SEE THAT EACH SUMMAND IS 0 OR 1, AND SO ? ? R ?LOG 2N LOG P HENCE 2N N = Y P?2N PR < (2N)?(2N)R LOG P < (2N)?(2N) LOG 2N NOW BY (2 25), 2N N ?2N, SO THAT LOG 2N ?LOG 2N N ??(2N) LOG 2N HENCE N LOG 2N ??(2N), SO THAT FOR X ?2, USING (2 27), WE OBTAIN ?(X) ?? I ? X LOG 2 X ? X LOG X TO PROVE THE OTHER INEQUALITY IN (2 26), WE RECALL THE REMARK AFTER (2 28) TO THE E?ECT THAT THOSE PRIME P, N < P ?2N DO NOT DIVIDE THE DENOMINATOR, AND THEREFORE 2N N MUST BE DIVISIBLE BY Q N<P?2N P HENCE N?(2N)??(N) ? Y N<P?2N P ? 2N N WE DEDUCE FROM THIS, ON USING (2 25) IN THE FORM LOG 2N N ?N, THAT (?(2N) ??(N)) LOG N ?N, WHENCE THAT ?(2N) LOG 2N ??(N) LOG N = (?(2N) ??(N)) LOG N + ?(2N) LOG 2 ?N IN VIEW OF ?(X) < X HENCE WE HAVE ?(X) LOG X ?? X LOG X 2 ?X (2 29) REPLACING X IN (2 29) BY X 2L , L = 0, , R, R = H LOG X LOG 2L X (2 30) THE INTEGRAL REPRESENTATION (2 1) AND EULER’S INTERPOLA- TION FORMULA (TELESCOPING SERIES) LIM N?? ?(N + Z) ?(N) NZ = 1, 0 < RE Z < (2 31) ARE EQUIVALENT PROOF ?(S) BEING MEROMORPHIC OVER THE WHOLE PLANE, IT SU?CES TO CON- SIDER THE CASE 0 < Z = X < 1 FIRST WE SHOW THAT (2 1) IMPLIES (2 31) IN (2 1) WITH S = N ?N PUT T = NU TO OBTAIN Z ? E?NUUN?1 DU = ?(N) N?N (2 32) SIMILARLY, BY (2 1) WITH S = N + 1 AND (2 5), WE HAVE Z ? E?NUUN DU = ?(N) N?N (2 33) NOTE THAT Z E?NTTN?1 DT ? Z E?NTTN DT = N Z D DT(E?NTTN) DT = N E?N (2 34) SUBTRACTING (2 34) FROM (2 32), WE DEDUCE THAT Z E?NTTN DT + Z ? N E?N (2 35) ADDING (2 34) TO (2 33), WE OBTAIN Z E?NTTN?1 DT + Z ? E?NTTN DT = ?(N) N?N + N E?N (2 36) THE THEORY OF THE GAMMA AND RELATED FUNCTIONS HENCE, FOR 0 < X < NX+N ?(X + N) = Z + Z ? < Z E?NTTN?1 DT < ?(N) N?N + N E?N BY (2 36) SIMILARLY, N E?N < ?(X + N) NX N?N < ?(N) N?N + N E?N, OR 1 ?NN?1E?N ?(N) < ?(X + N) ?(N) NX < 1 + NN?1E?N ?(N) LETTING N ??, WE CONCLUDE (2 31) NOW WE TURN TO THE DEDUCTION OF (2 1) FROM (2 31) PUTTING ?N(X) = NXN! X(X + 1) (X + N), WE HAVE LIM N???N(X) = ?(X) (2 37) BY (2 31) EXPANDING (?1)K N K X + K , WHOSE RIGHT-HAND SIDE CAN BE EXPRESSED AS NX Z TX?1(1 ?T)N DT HENCE ?N(X) = Z N 1 ?T N N DT VISTAS OF SPECIAL FUNCTIONS NOW CHOOSE 0 < ? < X+2 AND DIVIDE THE INTERVAL [0, N] INTO [0, N?]?[N?, N] FOR 0 ?T ?N? WE HAVE N LOG 1 ?T N = ?T + O T2??1 , AND SO Z N? 1 ?T N N DT = Z N? , WHILE FOR N? ?T ?N, N LOG 1 ?T N ?N LOG 1 ?N??1 ??C N?, C > 0, SO THAT Z N N? = O N?E?CN? IT FOLLOWS THAT ?(X) = Z N? SUPPOSE F(X) IS CONTINUOUS AND INTEGRABLE ON R+ = (0, ?) LET F(X) = R X 0 F(T) DT THEN FOR EACH N ?N PROVE THAT Z ? F(X) E?NX DX = N Z ? F(X) E?NX DX (2 38) THE THEORY OF THE GAMMA AND RELATED FUNCTIONS AND THAT LIM N?? Z ? F(X) E?NX DX = 0 (2 39) SOLUTION BY ASSUMPTION, LIM X??F(X) EXISTS AND SO F(X) = O(1) UNIFORMLY ON R+ CONSTRUCT THE AUXILIARY FUNCTION G(X) = N Z X F(T) E?NT DT + F(X) E?NX THEN G(0) = F(0) = 0 AND G(?) = N Z ? F(T) E?NT DT ALSO G?(X) = N F(X) E?NX + F ?(X) E?NX ?N F(X) E?NX = F(X) E?NX BY THE FUNDAMENTAL THEOREM OF CALCULUS THEREFORE G(X) = Z X F(T) E?NT DT + G(0) = Z X F(T) E?NT DT HENCE, IN PARTICULAR Z ? F(T) E?NT DT = G(?) = N Z ? F(T) E?NT DT, WHICH IS (2 38) NOW, TO PROVE (2 39), WE NOTE THAT F(T) IS DI?ERENTIABLE ON R+ AND, A FORTIORI, CONTINUOUS IN THE RIGHT NEIGHBORHOOD OF 0 HENCE, FOR ANY ? > 0, THERE IS A ? > 0 SUCH THAT |F(T)| = |F(T) ?F(0)| < ?, 0 < T < ? NOW, DIVIDE THE INTERVAL (0, ?) INTO (0, ?) AND (?, ?) TO OBTAIN Z ? F(T) E?NT DT = O N Z ? |F(T)| E?NT DT ! + O N Z ? ? |F(T)| E?NT DT = O ?[?E?NT]? + O [?E?NT]? ? = O(?) + O(E?N?), WHICH IS O(?) IF N IS BIG ENOUGH THIS PROVES (2 39) DEDUCE FROM (2 39) THE INTEGRAL REPRESENTATION FOR EULER’S CONSTANT DE?NED BY (5 16): ? = ? Z ? F(T, 1) DT, (2 40) WHERE F(T, Z) = E?T T ? E?ZT 1 ?E?T (2 41) PROVE THE GAUSS’ INTEGRAL REPRESENTATION FOR THE DIGAMMA FUNCTION ? DE?NED BY (5 18) ?(Z) = Z ? F(T, Z) DT = Z ? E?T T ? E?ZT 1 ?E?T DT, RE Z > 0 (2 42) SOLUTION WE NOTICE THAT R = Z ? E?TR DT = Z ? 1 ?E?NT 1 ?E?T DT (2 43) BY EXERCISE A 1, (A 2), WE HAVE ALSO LOG(N + 1) = Z ? R ?LOG(N + 1) = ? Z ? F(T, 1) 1 ?E?NT DT ??? Z ? F(T, 1) DT, N ?? BY EXERCISE 2 9 HENCE, BY DE?NITION, (2 40) FOLLOWS LET F(Z) = Z ? F(T, Z) DT, THE INTEGRAL BEING (ABSOLUTELY) CONVERGENT FOR RE Z > 0 THEN F(Z) + ? = Z ? E?T ?E?ZT 1 ?E?T DT, (2 45) THE THEORY OF THE GAMMA AND RELATED FUNCTIONS WHENCE F(Z + 1) ?F(Z) = Z ? E?ZT DT = 1 + K ? Z + K + F(Z + N) ?F(1 + N) (2 48) BY (2 46) HENCE LIM N??(F(Z + N) ?F(1 + N)) = 1+K ? Z+K SINCE F(Z + N) ?F(1 + N) = Z ? PROVE LEGENDRE’S (INTEGRAL) REPRESENTATION ?(Z) + ? = Z 1 ?X DX, RE Z > 0 (2 50) VISTAS OF SPECIAL FUNCTIONS USE (2 50) TO DEDUCE ? SIN ?Z = Z ? U?Z 1 + U DU = Z ? 1 + V DV, (2 51) AND ? COS ?Z = Z ? 1 + U DU = 2 < RE Z < (2 52) (CF COROLLARY A 4) SOLUTION (2 50) FOLLOWS FROM (2 42) BY THE CHANGE OF VARIABLE T = ?LOG X USING (2 50), WE HAVE ?(Z) ?? = Z 1 ?X DX = ? Z ? 2 ?T?Z 1 ?T DT, RE Z > 0, (2 53) WHERE THE SECOND EXPRESSION FOLLOWS FROM THE ?RST BY THE CHANGE OF VARIABLE X = T?1 SIMILARLY, ?(1 ?Z) ?? 2 ?T?Z 1 ?T DT = ? Z ? 1 ?X DX, RE Z < 1 (2 54) SUBTRACTION OF (2 54) FROM (2 53) GIVES ?(Z)??(1?Z) = ? Z ? 2 ?T?Z 1 ?T DT = Z ? 1 ?X DX, 0 < RE Z < 1 IF WE ADMIT THE RELATION ?(Z) = (LOG ?(Z))?, THEN FROM (2 15), WE DEDUCE THAT ?(Z) ??(1 ?Z) = ?? COT ?Z, (2 55) AN IMPORTANT RELATION LINKING THE DIGAMMA AND THE TRIGONOMETRIC FUNCTIONS USE THIS TO WRITE THE ABOVE EQUALITY AS ? COT ?Z = Z ? 1 ?X DX, WHICH BECOMES, ON WRITING T = U2, X = V2, Z ?Z 2: ? 2 COT ? 2 Z = Z ? DU = ? Z ? 1 ?V DV, 0 < RE Z < 2 (2 56) THE THEORY OF THE GAMMA AND RELATED FUNCTIONS AND ? 2 TAN ? 2 Z = ? Z ? DU = Z ? DV, ?1 < RE Z < 1, (2 57) BY THE INTERCHANGE OF Z BY 1 ?Z ADDING (2 56) AND (2 57), WE DEDUCE (2 51), WHENCE (2 52) FOLLOWS BY CHANGING Z BY COS 2PK Q LOG 2 SIN K Q ? (2 58) PROOF WE MAKE THE CHANGE OF VARIABLE X = UK IN LEGENDRE’S FORMULA (2 50) WITH Z = P Q TO OBTAIN ? P Q + ? = Z (2 60) LET ? = E2?I AK U ??K , ?PK ?1 Z ?PK LOG 2I SIN 2? Q K ? K ?LOG Q, (2 64) THE THEORY OF THE GAMMA AND RELATED FUNCTIONS K SIN 2?P Q K (2 67) HENCE ADDING (2 66) AND (2 67), WE OBTAIN 2 ? P Q + 2? =