[I71, Lemmas 8, 9 (lower bound), 10] #
- Lemma 8:
š(s-1) < 4 s² š(s)fors ā„ 3. - Lemma 10:
log t š(t) < 2 t/(t-1)² š(t-1)fort ā„ 2; henceš(t) t log t / š(t-1) < 9/2fort ā„ 3, and the CorollaryG_{r+1}(t) (t-1)² log t / (G_r(t-1) t) < 4fort ā„ 2. - Lemma 9 (lower bound):
G_r(s) ā„ exp(-s log s - s log log s - c s)fors ā„ 3.
[I71] proves Lemma 10 assuming that š(t-1)/š(t) is increasing; we instead use a first
crossing argument for the inequality itself (cf. [Opera de Cribro, Lemma 11.7]). For Lemma 9 we
compare š with the explicit subsolution exp(-(s log s + s log log s + 2s)) of (3.12),
instead of quoting the asymptotics of Hua's book.
Lemma 8 #
Lemma 10 #
The numerical comparison in the main range of Lemma 10. With q² = t - 1, the
logarithmic bound makes the integrated exponential estimate smaller than 2t/(t-1)².
Corollary to [I71, Lemma 10]: G_{r+1}(t) (t-1)² log t / (G_r(t-1) t) < 4 for t ℠2
(and all r; [I71] states it for t > (5 + (-1)^r)/2).
Lemma 9 (lower bound) #
The exponential lower bound, integrated and multiplied by the factor in (3.12),
has coefficient at least one. The key estimate is l (s-1)² ⤠s (exp l - 1).