Question

question 3 and 4

(3) Prove that nen! T($) = lim n70s(s+1)...(s+n) (4) (a) Prove Wallis's formula T_2.2 4.4 2m. 2m 2" 1.3 3.5 (2m – 1)(2m + 1) *** (b) Use it to prove the identity t(s)[(s +1/2) = V721–25T(28)

Answer 1

3 we know that r(s) = po fsilent at Re(s) you 1. is called gamma function For n a positive integer and Resyo, we con function In (s) = 1" (1- + h) est dit we show that - In (S) = neni s(s+) --(sth) limit formula obtain Euler's for the gamma and we function lim rn(s) = m(s). we write r' (s) - Puls) = 1, (s) +1268) + 53(3), where 1, (s) = Som ett slat [z(S) foto (24-(-)") esaldt 13(S) = 5" (e-11 ) ") est dk Obviously so(s) fensd to o as nyo. For Iz and I3 we have ostcn, and by the Taylor expansion of the logarithm we obtain lang (1- "En log(-) = -2, where L=th ( ² + 1 2 3 4 + 2 2 2 2 2 ton) It follows that Onet-line-etekent set, this [z(3) tends to o as nyo. For fg we have t/n < ² , which implies that Lee Won there is consequently, for tu al e we have

oset-(1144)" = et (1-e-t) & et le eteen Plugging this estimate into fg, we deduce that 153 (33) sep (Res +2), which tends to o as noooo .; r(s) - rm (5) ►o ay na 5) lim in (s) = res). - Next, in indegrating by pasts give rh(s) = ing en het ons emist me tats***dt 5 (5+1)(5+2) --- (sth) —133 n > Enns From ☺ and , r(s) = lim - hsn! n75 (+1) (5+2) --- (5+) a we have to prove en amu) (end 1) *** - myo let in be defined 1/2 by sinn ge du for n=1,2,or. Then - 2 sin ange du = 13.5.--(2n-1) and 2.. 2.4.6.- .leni 2n) Hoxha,2,... 2 Fanati - J% sin 2rl mom - 2,4,6.-=(20) - 3.5.7... (2n+1) fox hal,,... formulas in The key hollowing to proving reduction the above formula:

3) x = 10h Sinkeda = k l goth sin k-2 du = H (x-2 This formula is proven using integration by parts. Soth sin kuda = th sink"x. sinn du 3-com sinkt auf 2 + 5 "(k-1)} sink @osºudu = 1/2 (k-1) sin k-2 Alda - J V2 (K-1) sin Kada * Ek = (K-1) EK-2 - (K-1) Ik. → SK = KL IK prove ok (31) For equation (), when nai, 12 = f* sina xdu Jive 1-ci (2n) on - and the right hand side of (1) with has becomes (E) I , which verifies that is true for mal. Assume now that (1) is true for n. The following lines show it is true hos ntl. Tag+ 1) = ( 2 lanten tus) San 2(6+) -1 13.5... (2n- u n 2(n+1) 2.4.6.-..en 2 1-3.5.-:(2n-1) (2n+1) 2.4.6.... (2n) (2n+2) Ź ! Hence 1 is true Similarly (2) is true, as 1 I = ( 2 ) t = 2 / 8th sinndu = 2 / 2 to e is true for nas ] Now forom (3) , Iantz = ant! E20 2112

For xe [0, %], we know that of sinusl, then Sin 2n+2 na sinantln & Sinann. Thus, 2 Sin 2n+2 dn c 2n+2 m < 1/2 sin anti a / sir entre c /2 sin anda e » [anth < Ianto < [an - From (5) 2n4/7 Iante a [anti si Iantz = ant 2] Iznt2 Ian * lim | Soh - - - - Now, 2.4.6.... (2n) 3.5.7. - 2n41) 1 2 Inti = 1 / 1.3.5 (2n-1) 1 / 2 Ian -}. . . ne znatno 21 2n +1 + 12+ um lim 7 din î everything dime † i tiše 2- Can no } .l = lim §. memang meminta 2m. 2m Temy. (zmt es from wallis formula we have, 2.2.4.4... 2m. 2m . | V | 3 - - - -(2-1) 2 + 1) = lim _2.4. -- 2m moto 1.3.5. (2m-1) (2m4132 ng! Since (S) - - S (5+1) --- (Sun) n2stk 2t no 2 rm (8) 6. (s+4) = (25) (25+1) ---- (25+2n+i)

(25) = (zvy - (алу. (S) (25 +1) ---- (25+ 2n+1) 6 г.(s) ги(st: Гr (25) n“ (n!)ъ. 22" 23-1 (2n)! уп с ино (by walles product) and n2ny 2, 22 (2) B - Lim n-a (s) Гn(3+1) - по 1-25 Гn (23) ә г(9) г(+ 1) - , 1-25 г(25) - г1) г( 34 ) – уп 21-22 1(23) .