Euler-Maclaurin Summation Formula 1

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1 Jnury 9, Euler-Mclurin Summtion Formul Suppose tht f nd its derivtive re continuous functions on the closed intervl [, b]. Let ψ(x) {x}, where {x} x [x] is the frctionl prt of x. Lemm : If < b nd, b Z, then <n b (f(x) + ψ(x)f (x)) dx + (f(b) f()). Proof: The proof proceeds long the lines of the Abel prtil summtion formul. (b )f(b) + f() (n )(f(n + ) f(n)) <n b n b (b )f(b) + f() (b )f(b) + f() (n ) n b n+ n b n (b )f(b) + f() + ( + ) bf(b) f() f(t)dt + f(t)dt + (t {t})f (t)dt {t}f (t)dt n+ n f (t)dt ([t] )f (t)dt f (t)dt ( {t} ) f (t)dt + (f(b) f()) (f(t) + ψ(t)f (t)) dt + (f(b) f()). [t]f (t)dt We define Bernoulli polynomils by the following three properties To determine the polynomils we introduce generting function F(t, x) B k (x) tk k!. B (x), () B k(x) kb k (x), k,,..., () B k (x)dx, k,,.... (3) These notes follow Anlytic Number Theory by H. Iwniec & E. Kowlski. Property () defines B k, k recursively up to constnt. The constnt is fixed by property (3). k

2 Observe F(t, x) x t k B k(x) tk k! kb k (x) tk k! t B k (x) tf(t, x). (k )! k k k Thus F(t, x) C(t)e tx (4) where C is some function of t. Now by the third defining property of B k (x), F(t, x)dx. Integrting both sides of (4) with respect to x over the intervl [, ] then implies nd hence; k C(t) t e t, B k (x) tk k! tetx e t. Using this generting function we cn find the first few Bernoulli polynomils: The Bernoulli numbers re defined by B (x), B (x) x, B (x) x x + 6, B 3 (x) x 3 3 x + x, B 4 (x) x 4 x 3 + x 3, B 5 (x) x 5 5 x x3 6 x, B 6 (x) x 6 3x x4 x + 4. B n B n (), tht is, the vlue of the Bernoulli polynomil t x. The generting function for Bernoulli numbers is clerly F(t) : t k B k k! t e t. n An esy clcultion shows F( t) F(t) + t; nd hence, F( t) F(t) t. This lst equlity implies tht B k+ for k,,.... Define ψ k (x) B k ({x}) (5) where {x} is the frctionl prt of x. Observe tht ψ(x) ψ (x) {x} which ppers in Lemm. Since {x} is periodic with period, so too re the functions ψ k (x) nd they hve generting function k ψ k (x) tk k! tet {x} e t

3 We now ssume tht f is twice continuously differentible in [, b]. We integrte by prts the term involving ψ in Lemm. First look t + f (x)ψ (x)dx f (t + )ψ (t + )dt since ψ is -periodic. In the intervl [, ], ψ (t) B (t). Thus x ψ (t)dt (ψ (x) B ) f (t + )ψ (t)dt (6) where we used defining property () of the Bernoulli polynomils. Note tht ψ (x)dx since ψ is -periodic. Integrting the lst integrl in (6) by prts gives + f (x)ψ (x)dx f (y + ) y ψ (t)dt y y f (y + ) (ψ (y) B ) + f (y + ) y ψ (t)dt f (y + )ψ (y)dy + B (f ( + ) f ()) f (x)ψ (x)dx + B (f ( + ) f ()) since ψ is -periodic. One notes tht the bove formul is vlid over ny intervl [ + n, + n]; nmely, +n f (x)ψ (x)dx +n f (x)ψ (x)dx+ +n +n B (f ( + n) f ( + n )), n,,..., b. Summing over n we obtin We hve proved f (x)ψ (x)dx f (x)ψ (x)dx + B (f (b) f ()). Lemm : Let f be twice continuously differentible on [, b] where < b nd, b Z. Then <n b { f(x) } ψ (x)f (x) dx + ( ) l ( f l (b) f l () ) B l. l (Recll B.) If one continues to integrte by prts one obtins the Theorem (Euler-Mclurin formul): Suppose f is k-times continuously differentible on the intervl [, b] with < b,, b Z. Then <n b } {f(x) ( )k ψ k (x)f (k) (x) dx + k! k ( ) l l ( ) f (l ) (b) f (l ) () B l. 3

4 Suppose f nd ll its derivtives go to zero s x. Then we obtin by letting b (nd dding f() to both sides) n f(x)dx + k f() ( ) l l f (l ) ()B l ( )k k! f (k) (x)ψ k (x)dx (7) Appliction of summtion formul to the Riemnn zet-function Let s σ + it where σ is the rel prt of s nd t is the imginry prt of s. Let σ > nd define the Riemnn zet-function ζ(s) ns, R(s) >. (8) n The series converges bsolutely nd uniformly in the hlf-plne σ R(s) + ε: First observe tht Now pply the Weiestrss M-test to the series n s n σ it n σ n ε n n +ε which is convergent for ll ε >. The series (8) clerly diverges t s. We now pply (7) with k to (8). Choose f(x) /x s. For R(s) >, The summtion formul then becomes ζ(s) s + s dx x s s. x s+ ψ (x)dx. This is derived under the ssumption tht σ >. Observe tht if we write ζ(s) s s x s+ ψ (x)dx then the right-hnd side of the bove eqution defines holomorphic function for σ > since the integrl x s+ ψ (x)dx dx < (9) x+σ (We used ψ (x) /.) We now use the right-hnd side (9) to define the left-hnd side of (9) for < σ. The two side gree for σ >. This is n exmple of nlytic continution. We hve mde sense out of the Riemnn zet-function for < R(s). We see tht it hs simple pole t s nd is holomorphic for ll other points R(s) >. If we pply (7) to (8) for rbitrry positive integer k we obtin fter some elementry computtions ζ(s) s + + k l B l ( )k s(s + ) (s + l ) k! 4 s(s + ) (s + k )x s k ψ k (x)dx

5 Since ψ k is -periodic nd equl to the polynomil B k (x) on [, ), ψ k (x) is bounded function on ll of R. Thus the integrl on the right-hnd side is convergent for ll σ+k > nd thus defines holomorphic function for σ > k. By repeting the bove rgument we see tht we hve nlyticlly continued the Riemnn zet-function to the right-hlf plne σ > k, for ll k,, 3,.... For exmple, it now mkes sense to sk for the vlue ζ ( ). 3 We summrize our findings in Theorem: The Riemnn zet-function ζ(s) defined by (8) for R(s) > cn be nlyticlly continued to C {} where it is holomorphic nd t s, ζ(s) hs simple pole. 3 Before we nlyticlly continued ζ(s) it clerly mkes no sense in (8) to sk for the derivtive t s since the series only converges for R(s) >. It is fct tht ζ ( )

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