Abstract:Last, but not the least, I would like to thank my institute 'The Institute of Mathematical Sciences' for providing me a vibrant research environment. I have enjoyed excellent computer facility, a well managed library, clean and well furnished hostel and office rooms, and catering services at my institute. I would like to express my special gratitude to the library and administrative staffs of my institute for their efficient service.
“…Using a pole on the line Re(s) = 1/4, Landau's method gives ∆(x) = Ω ± (x 1/4 ). In [6], we show that…”
Section: Introductionmentioning
confidence: 69%
“…Under the assumption(6), there exists T 0 with T ≤ T 0 ≤ 2T such that ∆(T 0 )e −T0/y T α )e −u/y u α du ≪ log T.Proof.The assumption (6) implies thatlog 2 T ≥ 2T T |∆(u)| 2 u 2α+1 e −u/y du = 2T T |∆(u)| 2 u 2α e −2u/y e u/y u du ≥ min T ≤u≤2T |∆(u)| u α e −u/ywhich proves the first assertion. To prove the second assertion, we use the previous assertion and Cauchy-Schwartz inequality along with assumption (6) to get 2α+1 e −u/y du ∞ T0 ue −u/y du ≪ y 2 log 2 T.…”
“…Using a pole on the line Re(s) = 1/4, Landau's method gives ∆(x) = Ω ± (x 1/4 ). In [6], we show that…”
Section: Introductionmentioning
confidence: 69%
“…Under the assumption(6), there exists T 0 with T ≤ T 0 ≤ 2T such that ∆(T 0 )e −T0/y T α )e −u/y u α du ≪ log T.Proof.The assumption (6) implies thatlog 2 T ≥ 2T T |∆(u)| 2 u 2α+1 e −u/y du = 2T T |∆(u)| 2 u 2α e −2u/y e u/y u du ≥ min T ≤u≤2T |∆(u)| u α e −u/ywhich proves the first assertion. To prove the second assertion, we use the previous assertion and Cauchy-Schwartz inequality along with assumption (6) to get 2α+1 e −u/y du ∞ T0 ue −u/y du ≪ y 2 log 2 T.…”
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