= 1 2 x (x 1) + 1 {x} (1 {x}). [t] dt = 1 x (x 1) + O (1), [t] dt = 1 2 x2 + O (x), (where the error is not now zero when x is an integer.

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1 Problem Sheet,. i) Draw the graphs for [] and {}. ii) Show that for α R, α+ α [t] dt = α and α+ α {t} dt =. Hint Split these integrals at the integer which must lie in any interval of length, such as [α, α + ]. iii) Prove that for > [t] dt = [] ([] ) + {} [] = ( ) + {} ( {}). This is often written as [t] dt = ( ) + O (), though the error term is zero when Z. The result will be used in the course in the form [t] dt = + O (), (where the error is not now zero when is an integer.). How fast does the logarithm grow?. i) Recall the fundamental idea from Chapter, ( ) ( ) inf f (t) ( y) f (t)dt sup f (t) ( y). (5) t [y,] t [y,] y Use (5) to show that log y < y < y 30

2 for all real y >. ii) By an appropriate choice of y in part i show that for all n we have log < n /n (6) for >. iii) Deduce that for all ε > 0, we have log ε ε. This means that the logarithm of grows slower than any power of. Here log ε ε means there eists a constant C, though it does depend on ε, for which log C (ε) ε for all. iv) In the notes we make use of both of /3 log < C / and δ < C log l, for any constant C > 0, δ < and l. Prove these inequalities both hold for sufficiently large. 3. A result used in more advanced results on the Prime Number Theorem. For a function whose growth lies between that of δ for any δ < and / log l for any l, consider with constants C > 0 and α. Prove that a) If δ < then for any C > 0 ep ( C (log ) α ) δ ep ( C (log ) α ) for all sufficiently large as long as α <. b) If l then for any C > 0 ep ( C (log ) α ) log l for all sufficiently large as long as α > 0. 3

3 4. i) Estimates of integral found in error terms. Show that for α 0 and l we have. t α log l t dt +α l,α log l. Hint Split the integral at and use the fact that log t is an increasing function of t. ii) Show that for α > and l we have log l t t dt α l,α log l α. Hint Split the integral at and use (5). In the shorter interval again use that log t is an increasing function while in the longer interval use log t ε with some appropriately chosen ε. iii) A result used in more advanced results on the Prime Number Theorem. Show that with C > 0 for some C > 0. ep ( C (log t) α )dt ep ( C (log ) α ), 3

4 Generalising Euler s constant. Recall from lectures that a result for comparing sums with integrals is that if f has a continuous derivative, is non-negative and monotonic then f (n) = f (t)dt + O (ma (f (),f ())), (7) n for all real. In fact we deduced (7) from the Euler Summation: Let f have a continuous derivative > 0. Then n f (n) = for all real. f (t)dt + f () {} f () + n {t} f (t)dt As an application of (7) we showed that there eists a constant γ such that ( ) n = log + γ + O, (8) for. The essential idea here is that we come across a convergent integral {t} t dt. This could simply be bounded as O (), but for a better result it is completed to infinity and the tail end, the integral from to, is bounded (often using the results of Question 4). This is a standard method and is used in the following question in which we generalise (8) to a sum of (log n) l /n for any integer l. The result of Question 5 is used later when eamining the Laurent Epansion of ζ (s). 5. Prove that for all l there eists a constant C l such that log l n = ( log l ) n l + logl+ + C l + O, n for all real. 33

5 The result when l = 0 has C 0 = γ, Euler s constant. Notice how we have the best possible error term. More sums of logs 6. Generalise a result in lectures written (in a weakened form) as log n = log + O () n i) With integer l justify log l n = for all real. ii) Prove that n log l tdt + O ( log l ) (9) Deduce log l tdt = log l + O l ( log l ). ( log l n = log l + O l log l ). n for all real. Note the best error term here would be O ( log l ), far smaller than the one here. 7. Improve the result of Qu 6 to the best possible error term. Change the variable of integration in (9) to u = log t so i) Prove by induction that y 0 e u u d du = e y log l tdt = d r=0 log 0 e u u l du ( ) r d! (d r)! yd r ( ) d d! (30) 34

6 for all d 0. ii) Prove that for any integer l 0 we have where log l n = P l (log ) + O ( log l ), n P d (y) = d r=0 ( ) r d! (d r)! yd r, a polynomial of degree d. 35

7 Partial Summation Partial Summation is often no more than an eercise in interchanging Sums and Integrals. 8. Let be real, {a n } n a sequence of comple numbers and A () = n a n. a) Show that a n ( n) = n A (t)dt. b) Show that c) Show that n n a n log n = a n (e e n ) = A (t) dt = t e t A (t)dt = log 0 e Hint Write n, log (/n) and e e n as integrals. e A (e y )dy. A (log y)dy. 9. The last question concerned sums over n, this will be for sums over n >. For f with a continuous derivative for > 0 satisfying f () 0 as and f (t) dt <, then a n f (n) = A ()f () n> A (t)f (t)dt. for sequences {a n } for which the sum and integral converge. Hint Write f (n) as an integral from n to. 36

8 0. a) Use Partial Summation to prove that if f has a continuous derivative on [, ] then for a sum over primes we have f (p) = p π (t)f (t)dt + π ()f (). (3) b) Recalling that θ () = p log p, deduce that ( ) θ () = π () log + O. log Compare this with Theorem.3. Hint for b. You may have to use Chebyshev s bound π () = O (/ log ) in the integral that arises along with Question 4 37

9 Deductions from Merten s Theorem.. Prove that n n log n = log log + O (). This last result could be compared with Merten s result = log log + O (). p p This may lead you to think that the n-th prime is of size n log n.. i) Prove that converges for all α > 0. n= n (log n) +α We saw in the previous question that the series diverges when α = 0. Hint Consider the partial sum over n and let. ii) After Question it might be thought that the n-th prime p n is approimately of size n log n, which we might write as p n n log n. In which case log p n may be thought of as of size log p n log(n log n) = log n + log log n log n. Thus p n (log p n ) α would be approimately of the size p n (log p n ) α (n log n)(log n) α = n(log n) +α. Hence part i might then suggest that the sum over primes p (log p) α. converges for all α > 0. Prove that this is so. p Hint for Part ii. Use Partial Summation to remove the / (log p) α so you can apply Merten s Theorem, Theorem.9, and in particular (8). 38

10 3. Prove that for a fied c >, <n c Λ (n) n, i.e. this sum is bounded for all >. (What is more difficult, and thus of more interest, is to show that this sum is bounded below and thus non-zero. For this would then show the eistence of n [, c] for which Λ (n) 0. This n would be a power of a prime, and since powers greater or equal to are rare, this would lead to the eistence of a prime in [, c] for any c >.) Hint Use Merten s result (6) (twice). 4. Prove that ψ (u) u du = log + O (). Hint Interchange the integral and the summation within the definition of ψ, use Merten s Theorem and Chebyshev s bound ψ (). 5. (Hard) Prove that for a fied constant c >, ψ (cu) ψ (u) u du = (c ) log + O (). Note, Since for sufficiently large the right hand side is greater than 0 we must have that the integrand is non-zero, thus ψ (cu) ψ (u) > 0, and in particular, there is a prime in [u, cu], for some values of u. Unfortunately this does not tell us for which u these intervals contain a prime (it is, in fact, for all u sufficiently large) and how many primes are in these intervals. 39

11 Prime Number Theorem 6. (Tricky) Assume the Prime Number Theorem in the form π () / log as. Prove that the n-th prime p n satisfies p n n log n as n, i.e. lim n p n n log n =. Hint Note that π (p n ) = n, apply the Prime Number Theorem in the form given in the question and take logarithms. Stirling s Theorem Stirling s Theorem relates to any asymptotic result for N! as N. In Problem Sheet you were asked to show e ( ) N N N! en e ( ) N N,. (3) e by eamining log N! = n N log n. In Chapter we show that log = log + O (log ), (33) n for any real >. Choosing = N, an integer, and then eponentiating gives N! = e N N N e O(log N). But what does e O(log N) means? It represents e f(n) for some function f (N) satisfying f (N) = O (log N), i.e. C log N < f (N) < C log N, for some C. Then N C = e C log N < e f(n) < e C log N = N C, 40

12 which we epress as e O(log N) = N O(). Thus we have a very weak form, N! = e N N N+O(), of Stirling s Theorem. To improve this we need to improve (33), but only for integer. 7. The function {t} is periodic, period. The result of Question α+ α {t} dt =, can be interpreted as saying that {t} has average value /. Define P () = ( {t} ) dt. Prove that P () is periodic, period and P (n) = 0 for all n Z. 8. Euler Summation, Proposition.8, is for sums over n with real. Improved results can be given when is an integer. i) Prove log n = N log N N + + N n N {t} dt, (34) t for integers N. ii) Prove, by integrating by parts, log n = N log N N + N log N + + n N P (t) t dt, (35) for integers N, where P (t) is the function seen in Question 7. iii) Prove that there eists a constant C such that n N log n = N log N N + log N + C + O ( N ), (36) for integers N. 4

13 Hint Use an idea seen in the proof of Theorem.9, replacing any integral over [, ] of an integrable function by an integral over [, ) and then estimating the tail end integral over (, ). This could be compared with Eample., log n = log + O (log ), for real. n iv) Deduce Stirling s formula in the form N! = A for some constant A. It can be shown that A = π. ( ) N ( ( )) N N + O, e N This shows that the true result for N! lies midway between the bounds in (3). 4