Chapter 2 Fractional Part Integrals

Transcription

1 Chapter Fractioal Part Itegrals If I had the theorems! The I should fid the proofs easily eough Berhard Riema Sigle Itegrals I have had my results for a log time: but I do ot yet ow how I am to arrive at them Carl Friedrich Gauss A de la Vallée Poussi itegral Calculate { d Prove that { d lπ γ Let A deote the Glaisher Kieli costat defied by the limit A lim / / / e / cotiued O Furdui, Limits, Series, ad Fractioal Part Itegrals: Problems i Mathematical Aalysis, Problem Boos i Mathematics, DOI 7/ , Spriger Sciece+Busiess Media New Yor 3 99

2 Fractioal Part Itegrals cotiued 3 A cubic itegral Prove that { 3 d γ + 3 lπ 6lA 4 Let beaitegerprovethat { d p ζp + +p p 5 a Let beaitegerprovethat { d l γ, where γ deotes the Euler Mascheroi costat b More geerally, if q is a positive real umber, the { q q γ lq if q, d q q {q + q γ lq + q+ q if q > 6 Let beaitegerprovethat { d lπ γ l l! 7 Let be a iteger Prove that { d where ζ deotes the Riema zeta fuctio 8 Let be a iteger Prove that ζ,

3 Sigle Itegrals { d ζ 9 Let be a iteger Prove that { d ζ A Euler s costat itegral Prove that { { d γ a A quadratic itegral ad Euler s costat Prove that { { d 4lπ 4γ 5 b A class of fractioal part itegrals Let beaitegerthe { { d + j ζ j + j m ζm + ζm + +, + m where ζ is the Riema zeta fuctio ad the first sum is missig whe Prove that { { d 5 γ lπ 3 Prove that { d + l π, where a deotes the greatest iteger ot eceedig a 4 a Prove that

4 Fractioal Part Itegrals { d π b More geerally, let α, β,adγ be positive real umbers Study the covergece of { β γ α d, where a deotes the greatest iteger ot eceedig a 5 Prove that { d γ 6 Let m > be a real umber Prove that {l m d e m+ m + e m+ + m 7 The first Stieltjes costat, γ, is the special costat defied by l γ lim l Prove that { ld γ + γ 8 Let be a positive real umber Fid the value of { lim d 9 Calculate L lim { d ad lim Let m > be a real umber Calculate { t m dt t { d L

5 Double Itegrals 3 Let m be a iteger Prove that { m m d ζ+ζ3+ + ζm + m + a Let m ad be positive real umbers Prove that { m d! m +! j m + j! ζm + j + + j! b A ew itegral formula for Euler s costat If m isaiteger,the m { m+ d H m+ γ m+ j where H m+ deotes the m + th harmoic umber ζ j, j Double Itegrals Nature laughs at the difficultiesof itegratio Pierre-Simo de Laplace a Calculate { ddy y b More geerally, if is a iteger, calculate { y ddy 4 Let m be a iteger Calculate { m ddy + y

6 4 Fractioal Part Itegrals 5 Let be a iteger Calculate ddy, y + where a deotes the greatest iteger ot eceedig a 6 Let be a iteger Calculate ddy, y + where a deotes the greatest iteger ot eceedig a 7 Calculate { ddy y 8 Let be a iteger Prove that { ddy + y + + l γ Let m ad be positive itegers such that m Prove that { m ddy m l m y + 3 γ 3 Let be a real umber Prove that { ddy + y + γ + 3 Prove that { ddy y 3 γ + lπ 3 Let be a iteger Prove that { y ddy { d + + p ζp p p

7 Double Itegrals 5 33 Let be a iteger Prove that { y y ddy ζ+ζ3+ + ζ Let be a iteger ad let p be a real umber such that p > Prove that { y y p ddy ζ+ζ3+ + ζ + p + + p + 35 Let adletm, p be oegative itegers such that m p > ad p > Prove that { y m y m p + p ddy! m +! m + j! j + j! ζm + j + + p + 36 A surprisig appearace of ζ Prove that { y { y ddy π 37 Let,m > be real umbers Prove that { { y m y ddy y m ζ + ζm + m + + m + 38 Let > be a real umber Prove that { { y y ζ + ddy y Let beaitegeradlet I { { y ddy y

8 6 Fractioal Part Itegrals Prove that a I {{/d b I 49/6 π /3 l c Ope problem Fid a eplicit formula for I whe 3 4 Let m be a iteger Prove that { y m { y m ddy ζ+ζ3+ + ζm + m + 4 Let m, be two itegers Prove that { y m { y ddy! m +! + m! +! p p m + p! ζm + p + + p! + p! ζ + p + m + p! 4 Let a > be a real umberad let be a positive iteger Prove that a+ a {d Let beaitegeradleta,,a be positive itegers Calculate a a { d d d The Stieltjes costats, γ m, are the special costats defied by l m γ m lim lm+ m + 43 A multiple itegral i terms of Stieltjes costats Let m bea iteger Prove that { d d d m m m γ!

9 3 Quicies 7 3 Quicies I my opiio, a mathematicia, i so far as he is a mathematicia, eed ot preoccupy himself with philosophy, a opiio, moreover, which has bee epressed by may philosophers Heri Lebesgue Let > be a real umber ad let beaitegerprovethat { d + 45 Calculate { d 46 Calculate { { { d 47 Let > adletm bereal umbers Calculate { m m d 48 Let be a positive iteger ad let be a atural umber Prove that {d! +! 49 Let f ad g be fuctios o [,] with g itegrable ad g g Prove that { f f gd where {a deotes the fractioal part of a 5 Calculate the double itegral 5 Prove that { y ddy + y { yddy gd, { + y ddy y

10 8 Fractioal Part Itegrals 5 Let > be a real umber Prove that { { { ddy y y γ 53 Let > be a real umber Fid { ddy y 54 Let ad m be positive real umbers Calculate { y m y m ddy y 55 Let be a iteger ad let m > be a real umber Calculate { m y m ddy + y 56 Let be a iteger ad let m > be a real umber Calculate { m y m ddy y 57 Let a >, let be a real umber, ad let g : [,a] [,a] R be a itegrable ad symmetric fuctio i ad y Prove that a a { y g,yddy a a g, yddy, where { deotes the fractioal part of 4 Ope Problems I could ever resist a itegral G H Hardy Itegratig a product of fractioal parts Let 3beaiteger Calculate { { { d d d 3

11 5 Hits 9 59 A power itegral Let 3adm beitegers Calculate, i closed form, the itegral { m d d Hits To myself I am oly a child playig o the beach, while vast oceas of truth lie udiscovered before me Sir Isaac Newto Sigle Itegrals It is ot certai that everythig is ucertai Blaise Pascal Mae the substitutio / t 3 Mae the substitutio / t ad the itegral reducesto the calculatio of the series 3 l + / + 3/ 3 / + The, calculate the th partial sum of the series by usig the defiitio of Glaisher Kieli costat 5 ad 6 Mae the substitutio / t 7 Mae the substitutio /y 8 Calculate the itegral by usig the substitutio /y 9 Mae the substitutio / y Mae the substitutio /t ad show the itegral reduces to the calculatio of the series l l Mae the substitutio / y ad observe that if y is a positive real umber, which is ot a iteger, oe has { y {y 4, 5,ad7 Mae the substitutio /y 8 Substitute / t ad apply Stolz Cesàro lemma the / case 9 Mae the substitutio / t ad prove that {/ d verifies the iequalities π + 6 < < π + 6

12 Fractioal Part Itegrals ad π + 6 < < π + 6 ad Use the substitutio / y ad the improper itegral formula /a /Γ e a d where a > 5 Double Itegrals As for everythig else, so for a mathematical theory; beauty ca be perceived but ot eplaied Arthur Cayley I geeral, these double itegrals ca be calculated by writig the double itegral f,yddy as a iterated itegral f,ydyd, maig a particular chage of variables i the ier itegral, ad the itegratig by parts 3 Write the itegral i the form { y dy d; mae the substitutio y t, i the ier itegral; ad itegrate by parts 4 Write the itegral as + { m t dt d ad itegrate by parts 6 Use that dy d /y + / dt d ad itegrate by /t + parts 8 Substitute /y t, i the ier itegral, to get that {/ydy d {t/t dt d ad itegrate by parts 9 Write { t m m y y dy d m m {t dt t 3 { y dy d t /y {t dt t 3 35 Use the substitutio /y t aditegrateby parts d + / t {t { t dt d ad i- 38 Write { { y y y tegrate by parts dy d ad itegrate by parts d ad itegrate by parts 4 ad 4 Use the substitutio /y t, i the ier itegral, to get that {/ym {y/ dy d { {tm dt t d ad itegrate by parts t 4 Let m a be the floor of a ad calculate the itegral o itervals of the form [ j, j + ] Write the multiple itegral as a iterated itegral ad use part of the problem

13 6 Solutios 53 Quicies Nature ot oly suggests to us problems, she suggests their solutio Heri Poicaré These itegrals are solved by usig either symmetry or the followig idetity ivolvig the fractioal part fuctio A fractioal part idetity If R \ Z, the { + { 44 Mae the substitutio y ad 49 Mae the substitutio y ad use the fractioal part idetity 48 Observe that if is a positive iteger, the { y {y, forall y [,] ecept for, /,/,, /, The, use the substitutio y ad the defiitio of the Beta fuctio 5 54 ad 57 Use symmetry ad the fractioal part idetity 55 ad 56 Observe that if is a positive iteger, the { y {y, for all y [,] ecept for, /,/,, /, The, use symmetry 6 Solutios Everythig you add to the truth subtracts from the truth Aleader Solzheitsy 98 8 This sectio cotais the solutios to the problems from the first three sectios of the chapter 6 Sigle Itegrals The itegral equals γ We have Sir, I have foud you a argumet I am ot obliged to fid you a uderstadig Samuel Johso

14 Fractioal Part Itegrals { {t + d t dt t t dt l + + γ Remar This itegral, which is well ow i the mathematical literature [43, p 3], [65, pp 9 ], is related to a surprisig result, due to de la Vallée Poussi [5], which states that lim { γ I words, if a large iteger is divided by each iteger, the the average fractio by which the quotiet / falls short of the et iteger is ot /, but γ! See the solutio of Problem 6 3 Usig the substitutio / t the itegralbecomes {t 3 + t dt t 3 t dt 3 l Let S 3 l + / + 3/ 3 / + be the th partial sum of the series A calculatio shows that S l 3 l + 3 O the other had, l + l + +! l 3 3 l + ad 3 l + 3 l + + 3! 3 l e 3 Let + 3! e e ! 3

15 6 Solutios 3 The first fractio coverges to /A 6, ad for calculatig the limit of the secod fractio, we have, based o Stirlig s formula,! π/e,that + 3! 3 π π 3 e 3 e Thus, π 3 e 3 /A 6, which implies that lim S γ + 3 lπ 6lA 4 See the solutio of Problem 5 a Usig the substitutio / t, we get that { {t l+ d t dt l l l t l t dt l l l + l l + A calculatio shows that S l l+ l l l l +, + ad this implies that lim S + / + + / l γ b Usig the substitutio q/y, we obtai that { q {y I d q q y dy We distiguish here two cases Case q We have I q q q l q + + y dy + q γ lq Case q > A calculatio shows that q + I q q y y dy l + + y q {y y dy + q + y dy q q {q γ lq + + q q + q

16 4 Fractioal Part Itegrals Remar If is a iteger, the d l , ad this geeralizes a itegral from Pólya ad Szegösee[4, p 43] 6 Usig the substitutio / t, the itegral becomes l { d {t t dt l Let S be the th partial sum of the series We have l l l + + l l + l l l + l l l + + l + l! l! Sice l! lπ+ + l, it follows that S is approimated by +lπ+l l! l l ad hece, lim S lπ γ l l! 7 We have, based o the substitutio /y,that { {y d y + dy m m m+ m+ {y m y + dy m m y + m+ + +m y m+ m m m m + + m m m + m m + m + m ζ m y m dy y+ m + m

17 6 Solutios 5 8 Usig the substitutio /y we obtai that { d + {y m+ y m dy + y+ dy m m y+ + y + m+ m + +m y m+ m m + m m m + + m m + m m + m m + m m + ζ ζ 9 The substitutio / y implies that { d {y dy y+ y + ζ ζ We used that see the solutio of Problem 7 {y dy y+ ζ The substitutio /t implies that I { { {t t dt + { t t d y y + dy + {y dy y+ + ζ + { {t t t t t t dt { t t dt

18 6 Fractioal Part Itegrals We calculate the itegral ad the sum separately We have that + { t t + t t t dt t t t + t dt t t dt l + + l O the other had, the substitutio t u implies that { {t t { u u + t dt t u + du u {t t + t dt l + + l + + u u + t t + t dt Puttig all these together we get that I l + + l { du u Let S be the th partial sum of the series A calculatio shows that S l + + l l + + l , + ad this implies lim S γ Remar It is worth metioig that the followig itegral formulae hold { { d For a solutio of this problem, see [45] We eed the followig lemma: Lemma The followig formulae hold: lim + l + + { { d γ

19 6 Solutios d d + + l The lemma ca be proved by elemetary calculatios Usig the substitutio /t, the itegralbecomes I {t { t t dt t {t { t t dt + t l + l + +, + l, > + t We calculate the itegral ad the sum separately We have + t S { t t dt t + t t t t dt + t dt t t Usig part 3 of Lemma we get that S l + l + + l + + l O the other had, we have based o the substitutio t u that {t { t {u + J t dt t u + { u + u t du { t dt t u u + The substitutio /u t, combied with part of Lemma,shows that J {t t + t dt t t + t dt l + l + + { du u 3

20 8 Fractioal Part Itegrals Combiig 3, we get that { { d Let S be the th partial sum of the series, ie, S A calculatio shows Thus, + l + + l + + l l + + l l + + l + + l + + l + + l l + l! + l l + + S + l + l! + l l + l l + l! + l l + + Sice l! lπ/ + + /l + O/, we get that S + l + lπ+ + l l + + ad this implies that lim S 5/ lπ γ + O,

21 6 Solutios 9 Remar We have, due to symmetry reasos, that the followig evaluatios hold { { d 5 lπ γ { { { { + d 5 lπ γ { { { { { { + + { { d 5 lπ γ d 5 lπ γ 3 Usig the substitutio / y, the itegral becomes I { y y /y dy Sice for a positive real umber y, which is ot a iteger, oe has { y {y, we get that { y y { + I y dy y y { + y dy y y dy p { y p p p p p y dy + {y y dy + p+ p p p p+ p p y p p p y dy + p p p p + l p {y y dy y p y dy p ly + p y p y p y dy + l p + p p l p + p p p p + p + l p + p p + l π, p + p p p + p+ l p + p p p + where the last equality follows from the Wallis product formula

22 Fractioal Part Itegrals 4 a Usig the substitutio / t the itegralbecomes { {t t + t I d t 3 dt t 3 dt + t t 3 dt + π 6 b The itegral coverges if ad oly if α + > γ Maig the substitutio / t, the itegral equals { β γ Iα,β,γ α {t β t γ d t α+ dt + t β γ t α+ dt t u γ u β du + u α+ We have, sice α+ < + u α+ < + α+,that γ β + + α+ < Iα,β,γ < β + α+ Thus, the itegral coverges if ad oly if α + > γ Remar The covergece of the itegral is idepedet of the parameter β, which ca be chose to be ay real umber strictly greater tha 5 We have I { { d d γ { d Let J { d Usig the substitutio / y we obtai that {y m+ J y + dy {y m+ m m y + dy m m m m m+ m + + m m y m dy y+ y + m y + dy m y + m m+ + y + m m m + + m + m m [ m + ] m m + +

23 6 Solutios + ζ + + ζ + + Thus, I ζ ζ + + γ, sice ζ + / + γ [, Etry 35, p 73] For a alterative solutio, see [3] 6 The solutio is give i [5] 7 For a solutio see [5] 8 The limit equals /+Leta {/ d Usig the substitutio / t, we get that {t {t a t dt dt t b, c where b {t /t dt ad c /, ad we ote that b coverges to sice b {t /t dt < /t dt / For calculatig lim a,weusestolz Cesàro lemma the / case ad we get that lim a b + b lim lim c + c + + t lim + t + lim y + y dy {t dt t + dt lim + y dy + + lim + y + y dy {t t dt For a alterative solutio, see Problem 7 9 We prove that L Let {/ d Usig the substitutio / t, we get that {t + t dt y + y dy {t + t dt y + y t dt dy t

24 Fractioal Part Itegrals Sice / + < / + y < /, for positive itegers, ad y,, it follows that + dy < < dy y Thus, π + 6 which implies that lim y < < π + 6, The secod limit equals We have, based o the precedig iequalities, that π + 6 < < π + 6, ad the result follows sice lim π /6 / The itegral equals /m ζm + /m + See the calculatio of J from the solutio of Problem 5 ad Let { V,m m {t p+ d t m+ dt {t p+ p p t m+ dt t p p p t m+ dt t py y p p + y m+ dy y p p + y m+ dy 4 Sice we have that p + y +m e p+yu u m+ du, m +! p p + y m+ m +! e p+yu u m+ du p u m+ e yu e du pu m +! p m +! u m+ e yu e u du 5

25 6 Solutios 3 Combiig 4ad5, oe has that V,m m +! m +! y u m+ e yu e u du dy y e yu dy du u m+ e u Let J y e yu dy Itegratig by parts we get the recurrece formula J e u /u +/uj Let a J u /! ad we ote that a e u u u! + a This implies that a e u u u! + u! + + u + e u! u e u e u u u u u!!! Thus, ad it follows that O the other had, e u u V,m j u + j + j! J!e u + j!, m +! j j! + j! u j u m+ j e u e u du u m+ j e u u m+ j e u e u du p e pu du u m+ j e +pu du p p Γm + j + + p m+ j+ m + j!ζm + j +

26 4 Fractioal Part Itegrals Hece, V,m Whe m, oe has that V m,m m + j! m +! j ζm + j + m + j! ζm + j + + j! ζ+ζ3+ + ζm +, m + sice j ζ j + Part b of Problem follows from the calculatio of V,m combied with the formula j ζ j + / j + γ 6 Double Itegrals But just as much as it is easy to fid the differetial of a give quatity, so it is difficult to fid the itegral of a give differetial Moreover, sometimes we caot say with certaity whether the itegral of a give quatity ca be foud or ot Joha Beroulli a The itegral equals π We have { I ddy y { dy d y Usig the substitutio y t, i the ier itegral, we obtai that { I dt d t Itegratig by parts, with { f dt, t { f, g, g,

27 6 Solutios 5 we obtai that I { { t dt dt t { { d ζ, d { { d d where the last equality follows from Problem whe m b The itegral equals j j+ ζ j + j j j + Use the same techique as i the solutio of part a of the problem 4 Let I m deote the value of the double itegral We have { m I m dy d +yt + { m dt d + y t We calculate the itegral by parts, with + { m f dt, f t g adg, ad we get that + { m I m dt t { m { m dt t dt + t t t m dt t m dt + { m + { { m + { m d m d { m, { m d

28 6 Fractioal Part Itegrals Whe m, oe has that { I l + d l π, where the equality follows from Problem Whe m, oe has that { I l + d 5 π l γ, where the equality follows from part b of Problem Whe m 3, oe has that I m m 3 m m + m m m + m 3 m m + m m m + m! { m d j j +! ζ j +, m + j! where the equality follows from part a of Problem It is worth metioig that the case whe m was solved by a alterative method i [6] 5 We have, based o the substitutio y t, that I We itegrate by parts, with ddy y + / dt t + d / f t dt, + f, + g ad g /, ad we get that / dt I t + + d + d dt t t d + m m+ m dt t m d +

29 6 Solutios 7 m m ζ + m m + m m m + Let S m /m m + Sice m m+, oe has that m m m+ S ζ S This implies that S ζ+ S,adit follows that S + + j + j ζ j Thus, I + j + j ζ j Remar We metio that the case whe is due to Paolo Perfetti [99] 6 The itegral equals + ζ ζ3 ζ + Wehave,based o the substitutio y t, i the ier itegral, that dy / dt I /y + d /t + d We itegrate by parts, with / dt f /t +, f +, g ad g /, ad we get that / dt I /t + + dt /t + + d du u u + + m m m+ m du u m + + m + + m +, ad the result follows from Problem 38 7 See [88] or the solutios of Problems 8 or 9 d +

30 8 Fractioal Part Itegrals 8 We have, based o the substitutio /y t, that { I dy d y Itegratig by parts, with {t t dt d {t f t dt, f {, g ad g /, we get that I A calculatio shows that { d O the other had, {t t dt j {t t dt + { d {t t dt + {d {ydy j+ j j {ydy 6 j+ j j y jdy 7 l j + + j j l γ 8 Combiig 6 8, we get that the itegral is calculated ad the problem is solved Remar We also have, based o symmetry reasos, that 9 We have I { m y { y ddy ddy Itegratig by parts, with {t f m t dt, f m l γ + 4 { m m t y dy d m {t y m t dt d { m, g, g,

31 6 Solutios 9 we obtai, sice { m m for [,,that I m {t m t dt + { m m m {t m t dt + m m d m {t m t dt + m {t m t dt + {t t dt + m m t dt + γ + l m + 3 γ m d Remar It is worth metioig that, whe m >, the itegral equals m l m q γ + q + r + mr m, where q ad r are the itegers defied by m q + r with r < 3 Usig the substitutio t /y, i the ier itegral, we obtai that { { t y I ddy dy d {t y y t dt d Itegratig by parts, with {t f t dt, we obtai that + I + f { +, g, g + +, {t t dt + + {t + t dt + + d { d

32 3 Fractioal Part Itegrals γ γ + 3 ad 3 We have, based o the substitutio /y t, that { {t I dy d y t dt d We itegrate by parts, with {t f t dt, f {, g, g, ad we get that {t I t dt + { d {t t dt + + V, + + j ζ j j j +, 9 where the last equality follows from the calculatio of V,m see the solutios of Problems ad O the other had, we have, based o the substitutio / t, that { d {t dt, ad the equality t { ddy { d + y +, follows from the first lie of 9 The case this is Problem 3 { ddy 9 y p p ζp + p + p ζp + p + p ζp + + p γ + lπ, sice p ζp + /p + 3/ γ/ lπ/ [, p 3] ad ζp /p γ [, p 73] p

33 6 Solutios We have, based o the substitutio /y t, that { y m y p ddy { p y dy m d y m+ p {t dt d tm+ We itegrate by parts, with {t f dt, tm+ f { m+, g m+ p, g m+ p m + p, ad we get that { y m y p ddy m+ p m + p {t dt tm+ + m + p {t m + p t m+ dt + m + p p + V,m m + p + m + p p +! m + pm +! + j m p + p + Whe m this is Problem 34, we have that { y y p ddy p + Whe m p this is Problem 33, we have that { y y ddy 36 See [3] 37 See the solutio of Problem 38 p d m + j! ζm + j + + j! ζ+ζ3+ + ζ + + p + ζ+ζ3+ + ζ + +

34 3 Fractioal Part Itegrals 38 Recall that see the calculatio of J i the solutio of Problem 5ifα > is a real umber, the t α { t dt ζα + α α + We calculate the double itegral by maig the substitutio y t, i the ier itegral, ad we get that I y { y We itegrate by parts, with / f t {t { y / dy d + { t t {t dt, f { + { dt d t {, g + ad g + /+, ad we get, based o with α +, that + I { t {t t { t {t t { + d / ζ dt + dt { { { d 39 See [3] 4 ad 4 We have, based o the substitutio /y t, that { m { y { I m, dy d {t m y t We itegrate by parts, with { d dt t d { f {t m dt t t, f { m {, g ad g /, ad we get that

35 6 Solutios 33 I m, { {t m { t dt t + m d + { m d m { d V m, + V,m Recall that see the solutios of Problems ad Thus, { V,m m d I,m V m, + V,m +! m +! j! m +! m! +! j j Whe m this is Problem 4, oe has { m { y m ddy y m + m + j! ζm + j + + j! + j! ζ + j + m + j! m + j! ζm + j + + j! j ζm + j + ζ+ζ3+ + ζm +, m + sice p ζp + see[,p78] 4 Let m a be the floor of awehave a+ m+ +m j+ +a {d {d + {d + {d a a jm+ j +m m+ md + a +m jm+ j+ jd j +a + md +m

36 34 Fractioal Part Itegrals The itegral equals /a a LetI be the value of the itegral We have a a a I { d d d Usig the substitutio + + y, i the ier itegral, we get that a { d a, + + +a + + {ydy where the last equality follows based o the first part of the problem Thus, a a a I d d a a 43 See [35] 63 Quicies A mathematicia s reputatio rests o the umber of bad proofs he has give AS Besicovitch 44 We have { d y {y dy j+ y j dy j j u du j + 45 The itegral equals / Let I deote the value of the itegral We ote that if is a real umber ad is ot a iteger, the We have, based o the substitutio y,that { I y dy y { + { { y y dy I

37 6 Solutios The itegral equals γ / We have { I { { d y { y { { y y y { y { { y y y { { dy I, y y dy ad the result follows from Problem 47 The itegral equals m! /m +! We have { I m m d { y y m y m dy y y { y m y m dy y y y m y m dy I Bm +,m + I, dy where B deotes the Beta fuctio 48 First, we ote that if is a positive iteger, the { y {y,forall y [,] ecept for,,,,, Sice the Riema itegral does ot deped o sets of Lebesgue measure zero or sets with a fiite umber of elemets, we get, based o the substitutio y,that I {d y y { ydy y y {ydy y y dy I B +, + I! +! I Remar It is worth metioig that if f : [, ] R is a itegrable fuctio, the f {d f d

38 36 Fractioal Part Itegrals 49 Use the substitutio y ad idetity 5 The first solutio The itegral equals / We have I { ydy + { ydy d We ote that whe y,the y, ad hece { y y,adwhe y, the y, ad hece, { y y y y + It follows that I ydy + y + dy d The secod solutio By symmetry, Hece, I { yddy {y ddy I I + I { y + {y ddy ddy, because the set o which the itegrad is is a set of measure 5 We have, based o symmetry, that Thus, I I I + I { y + y ddy { y + + y { y ddy + y { y ddy + y ddy, because the set o which the itegrad is is a set of measure 5 Use symmetry ad idetity ad Problem 53 The itegral equals / Use symmetry ad idetity 54 The itegral equals /m + Use symmetry ad idetity 55 The itegral equals /m + First, we ote that if is a fied positive iteger, the { a {a, foralla [,] ecept for,,,,, We have, by symmetry reasos, that { { y I m y m ddy m y m ddy, + y + y

39 6 Solutios 37 ad hece, I I + I m y m ddy { + y { y + + y m d y m dy m y m ddy m + We used that { { y +, + y + y for all,y [,], ecept for the poits of a set A of Lebesgue area measure zero The set A is the set of poits for which y +y,,,,, ad hece, A turs out to be the uio of + lies through the origi 56 The itegral equals /m + See the solutio of Problem Use symmetry ad idetity

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