tailieunhanh - Analytic Number Theory A Tribute to Gauss and Dirichlet Part 6

Tham khảo tài liệu 'analytic number theory a tribute to gauss and dirichlet part 6', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 92 ADRIAN DIACONU AND DORIAN GOLDFELD which is valid for p K v 1 one can write the first integral in as 23 2k e ent e 1 ni 1 K w 2z 4ite r 2 K 2it r 1 K w 2z r 1 2ite w 2z F 2 K w 2z 2 K 2ite 1 w 2z 2ite 1 ni i K w 2z r 2 K 2itc r 1 K w 2z r 1 2ite w 2z F 2 K w 2z 2 K 2ite 1 w 2z 2ite 1 If we replace the ớ-integral on the right hand side of by the above expression it follows that nw 2 K t w ww sin 2 J 1 t r ị it 2 1 w cos nw cos 22k 2nK 1 cos 52 e entr 2 K 2ite e 1 i 1 9a w 1 r 1 z r w z r z 2ni J r z w 1 i 1 9a w ni 1 K w 2z 4ite r 1 K w 2z e 2 r 1 2t w 2z F 2 K w 2z 2 K 2ite 1 w 2z eni 1 K w 2z r 1 K w 2z 2 r 1 2ite w 2z 2ite F 2 K w 2z 2 K 2ite 1 w 2z 2ite 1 1 dz nw 2 O e eQ w . To complete the proof of Proposition . we require the following Lemma. Lemma . Fix K 12. Let 1 K w 2 0 t 3 w 2 e K z e with Ề e small positive numbers and 3 z 2 3 w . Then we have the following estimates F 2 K w 2z 2 K 2ite 1 w 2z 2ite 1 c . min 1 2t 3 w 2z F 2 K w 2z 2 K 2ite 1 w 2z 2ite 1 min 1 2t 3 w 2z . Proof. We shall make use of the following well-known identity of Kummer F a b c 1 2c a bF c a c b c 1 . It follows that 4 . F 2 K w 2z 2 K 2ite 1 w 2z 2ite 1 22k 3F k 1 2ite K 1 w 2z 1 w 2z 2ite 1 and 4 . F 2 K w 2z 2 K 2ite 1 w 2z 2ite 1 22k 3F k 1 2ite K 1 w 2z 1 w 2z 2ite 1 . SECOND MOMENTS OF GLĩ AUTOMORPHIC L-FUNCTIONS 93 Now we represent the hypergeometric function on the right hand side of as ờ i . 17 F r c 1 r a e r b e r e dP F a b c 1 r a r b 2ni y r c e de S itt with a K 1 2ite b K 1 w 2z c 1 w 2z 2ite. This integral representation is valid if for instance 1 Ỗ 0. We may also shift the line of integration to 0 Ỗ 1 which crosses a simple pole with residue 1. Clearly the main contribution comes from small values of the imaginary part of e. If for example we use Stirling s formula r s ĩn 2 e 3n t tlog t 1 Ĩ Tĩĩ ơ 3 1 O t 1 where s Ơ it 0 Ơ 1 t 0 we have r a e r b e r c r e r a r b r c e eĩ w í 2t w Ị Ị Ị 2t t3 K W3 K W e ĩ k S e .