Question: for 1, we have (1) since the integrand is nonnegative. Since N n Hence, (2) n p > === S log p < n

for 1, we have (1) since the integrand is nonnegative. Since N n Hence, (2) n p > === S log p < n arctan for n24. Also, R+L * L > >= 10 + 1 - S(in) log T R-L 37 R L cos R+L n R-L - arctan < L sin o R- L cos L sin odo n T RL cos o L (224) 3T R L2 3/2 = log do $). do, R+L R-L n R+L n log. - R-L 3T n R+L p==10g +1+0(1). -log R-L (-). tan

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