The Riemann-Stieltjes Integral

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1 Chpter 6 The Riemnn-Stieltjes Integrl 6.1. Definition nd Eistene of the Integrl Definition 6.1. Let, b R nd < b. ( A prtition P of intervl [, b] is finite set of points P = { 0, 1,..., n } suh tht = 0 < 1 < 2 < < n 1 < n = b. We write i = i i 1 for i = 1, 2,, n, nd define the norm P of P by P = m 1 i n i. (b A refinement of prtition P is prtition Q suh tht P Q. In this se, we lso sy tht Q is finer thn P. Given two prtitions P 1 nd P 2 of [, b], we ll the union P 1 P 2 their ommon refinement. Definition 6.2. Let α be monotonilly inresing funtion on [, b], f be bounded funtion on [, b] nd P = { 0,..., n } be prtition of [, b]. Define nd M j (f = We ll the numbers α j = α( j α( j 1 (1 j n sup f(, m j (f = inf f( [ j 1, j ] [ j 1, j ] U(P, f, α = M j (f α j, L(P, f, α = (1 j n. m j (f α j the upper (Riemnn-Stieltjes sum nd, respetively, the lower (Riemnn-Stieltjes sum of f with prtition P over [, b] with respet to α. Note tht, if m f( M for ll [, b], then m m j (f M j (f M for eh j = 1, 2,, n, nd hene m(α(b α( L(P, f, α U(P, f, α M(α(b α(. 1

2 2 6. The Riemnn-Stieltjes Integrl So, the sets {U(P, f, α P is prtition of [, b]} nd {L(P, f, α P is prtition of [, b]} re bounded sets in R. nd Define fdα = inf{u(p, f, α P is prtition of [, b]} fdα = sup{l(p, f, α P is prtition of [, b]} to be the upper Riemnn-Stieltjes integrl nd, respetively, the lower Riemnn- Stieltjes integrl of f over [, b] with respet to α. We sy tht f is Riemnn-Stieltjes integrble on [, b] with respet to α, nd write f R(α[, b], provided tht (6.1 fdα = fdα. In this se, the ommon vlue of the upper nd lower Riemnn-Stieltjes integrls in (6.1 is lled the Riemnn-Stieltjes integrl of f over [, b] with respet to α nd denoted by Sometimes, we lso write fdα = fdα. f(dα(; of ourse, the dummy vrible n be repled by ny other letters (eept for the letters f, α or d, to void obvious onfusion. Definition 6.3. When the funtion α is the identity funtion, i.e., α( =, we define the nottions U(P, f, L(P, f, to be the nottions, respetively, U(P, f, α, L(P, f, α, fd, fdα, fd, fdα, fd, R[, b] fdα, R(α[, b]. In this se, if f R[, b], then we sy f is Riemnn integrble on [, b] or, simply, integrble on [, b]. Theorem 6.1. If P, Q re prtitions of [, b] nd Q is finer thn P, then L(P, f, α L(Q, f, α U(Q, f, α U(P, f, α. Tht is, L(P, f, α inreses with P nd U(P, f, α dereses with P. Proof. Sine Q is obtined from P by dding finitely mny points, by indution, we only need to prove the se when Q is obtined from P by dding one etr point. So let P = { 0, 1,..., k 1, k,..., n }, Q = { 0, 1,..., k 1, y, k,..., n }, where k 1 < y < k. Then L(P, f, α = m j (f α j, where m j (f = inf [ j 1, j ]

3 6.1. Definition nd Eistene of the Integrl 3 Note tht k 1 L(Q, f, α = ( m j (f α j + inf f( [ k 1,y] ( + inf f( (α( k α(y + [y, k ] inf f( inf f(, [ k 1,y] [ k 1, k ] j=k+1 (α(y α( k 1 m j (f α j. inf f( inf f(. [y, k ] [ k 1, k ] Hene, sine α is inresing, we hve tht ( ( inf [ k 1,y] (α(y α( k 1 + inf [y, k ] (α( k α(y ( inf [ k 1, k ] (α(y α( k 1 + α( k α(y = m k (f α k. Consequently, k 1 L(Q, f, α m j (f α j + m k (f α k + j=k+1 m j (f α j = L(P, f, α. The proof of U(Q, f, α U(P, f, α is similr. Theorem 6.2. fdα fdα. Proof. Let P, Q be rbitrry two prtitions of [, b]. Then, sine P Q is refinement of both P nd Q, by the previous theorem, Hene L(P, f, α L(P Q, f, α U(P Q, f, α U(Q, f, α. fdα = sup{l(p, f, α} inf{u(q, f, α} = P Q fdα. Theorem 6.3 (Criterion for Integrbility. A bounded funtion f is in R(α[, b] if nd only if for eh ε > 0 there eists prtition P of [, b] suh tht (6.2 U(P, f, α L(P, f, α < ε. Proof. (Suffiieny for Integrbility. Let ε > 0. Assume tht there eists prtition P of [, b] suh tht U(P, f, α L(P, f, α < ε. Then U(P, f, α < L(P, f, α + ε, nd thus fdα U(P, f, α < L(P, f, α + ε fdα + ε. Sine ε > 0 is rbitrry, this proves tht fdα fdα, nd hene fdα = fdα; so f R(α[, b]. (Neessity for Integrbility. Assume f R(α[, b]; nmely, fdα = fdα. Let ε > 0. Then there eist prtitions P 1, P 2 of [, b] suh tht U(P 1, f, α < fdα + ε/2, L(P 2, f, α > fdα ε/2.

4 4 6. The Riemnn-Stieltjes Integrl Let P = P 1 P 2 be the ommon refinement of P 1 nd P 2. Then P is prtition of [, b], nd, using fdα = fdα, we hve ( U(P, f, α L(P, f, α U(P 1, f, α L(P 2, f, α < fdα + ε ( fdα ε = ε. 2 2 Theorem 6.4. Suppose tht (6.2 holds for prtition P = { 0, 1,, n }. ( Then (6.2 holds with P repled by ny refinement of P. (b If s j, t j re points in [ j 1, j ] for eh j, then f(s j f(t j α j < ε. ( If f R(α[, b], then f(t j α j for ll t j [ j 1, j ] with j = 1, 2,, n. fdα < ε Proof. ( follows esily sine L(P, f, α inreses with P nd U(P, f, α dereses with P. (b follows sine both f(s j, f(t j re between m j (f nd M j (f nd hene f(s j f(t j M j (f m j (f. The obvious inequlities L(P, f, α f(t j α j U(P, f, α, L(P, f, α fdα U(P, f, α prove (. Theorem 6.5. If f is ontinuous on [, b], then f R(α[, b]. Proof. Let ε > 0 be given, nd hoose η > 0 so tht (α(b α(η < ε. Sine f is uniformly ontinuous on [, b], there eists δ > 0 suh tht f( f(y < η for ll, y [, b] with y < δ. Let P = { 0,..., n } be ny prtition of [, b] with norm P < δ. Sine f is ontinuous on eh subintervl [ j 1, j ], by the Etreme Vlue Theorem, there eist j, d j [ j 1, j ] suh tht M j (f = f( j, m j (f = f(d j. Sine j d j j P < δ, we hve Therefore, U(P, f, α L(P, f, α = M j (f m j (f = f( j f(d j < η (M j (f m j (f α j η So, by the Criterion for Integrbility, f R(α[, b]. (1 j n. α j = η(α(b α( < ε. Theorem 6.6. If α is ontinuous on [, b], then every monotoni funtion on [, b] belongs to R(α[, b].

5 6.1. Definition nd Eistene of the Integrl 5 Proof. Without loss of generlity, we ssume f is monotonilly inresing funtion on [, b]. Let ε > 0 be given, nd hoose η > 0 so tht (f(b f(η < ε. Sine α is uniformly ontinuous on [, b], there eists δ > 0 suh tht α( α(y < η for ll, y [, b] with y < δ. Let P = { 0,..., n } be ny prtition of [, b] with norm P < δ. Sine f is monotonilly inresing on eh subintervl [ j 1, j ], we hve Sine j j 1 = j P < δ, we hve m j (f = f( j 1, M j (f = f( j. α j = α( j α( j 1 < η (1 j n. Therefore, U(P, f, α L(P, f, α = (f( j f( j 1 α j η (f( j f( j 1 = η(f(b f( < ε. So, by the Criterion for Integrbility, f R(α[, b]. Theorem 6.7. Suppose f is bounded on [, b], f hs only finitely mny points of disontinuity on [, b], nd α is ontinuous t every point t whih f is disontinuous. Then f R(α[, b]. Proof. Let ε > 0 be given. Put M = sup [,b] f( nd let E be the set of points in [, b] t whih f is disontinuous. Sine E is finite nd α is ontinuous t every point of E, we n over E by finitely mny disjoint intervls [u j, v j ] in [, b] suh tht (α(v j α(u j < ε. Furthermore, we n hoose these intervls in suh wy tht every point of E (, b lies in the interior of some [u j, v j ]. If E, we ssume [, v 0 ] is one of these intervls; if b E, ssume [u 0, b] is one of these intervls. Remove ll open intervls (u j, v j, nd possibly [, v 0 or (u 0, b] if or b is in E. The remining set K is then ompt, nd f is ontinuous on K. Hene f is uniformly ontinuous on K, nd there eists δ > 0 suh tht f(s f(t < ε if s, t K nd s t < δ. Define prtition P = { 0, 1,, n } of [, b] s follows. All u j, v j belong to P. No points in (u j, v j belong to P, nd no points in [, v 0 or (u 0, b] belong to P. If i 1 is not one of u j, then i < δ. Note tht M i (f m i (f 2M for every i, nd tht M i (f m i (f ε unless i 1 is one of u j. Hene U(P, f, α L(P, f, α = (M i (f m i (f α i = i 1 =u j (M i (f m i (f α i + i=1 Sine ε > 0 is rbitrry, this proves f R(α[, b]. i 1 u j (M i (f m i (f α i 2Mε + ε(α(b α(. Theorem 6.8. Suppose f R(α[, b], m f M, φ is ontinuous on [m, M], nd h( = φ(f( on [, b]. Then h R(α[, b].

6 6 6. The Riemnn-Stieltjes Integrl Proof. Let ε > 0 be given. Sine φ is uniformly ontinuous on [m, M], there eists δ (0, ε suh tht φ(s φ(t < ε if s t δ nd t, s [m, M]. Sine f R(α[, b], there is prtition P = { 0, 1,, n } of [, b] suh tht (6.3 U(P, f, α L(P, f, α < δ 2. Let M j (f, m j (f be defined for f s bove; similrly, let M j (h, m j (h be defined for funtion h. Divide the numbers i = 1, 2,, n into two lsses: i A if M i (f m i (f < δ; i B if M i (f m i (f δ. For i A, we hve f( f(y M i (f m i (f < δ for ll, y [ i 1, i ]; hene, our hoie of δ shows tht M i (h m i (h = sup φ(f( [ i 1, i ] inf φ(f(y = y [ i 1, i ] For i B, we hve M i (h m i (h 2K, where K = δ i B sup (φ(f( φ(f(y ε.,y [ i 1, i ] sup φ(t < +. By (6.3, t [m,m] α i i B(M i (f m i (f α i U(P, f, α L(P, f, α < δ 2, so tht i B α i < δ. Thus it follows tht U(P, h, α L(P, h, α = i A (M i (h m i (h α i + i B(M i (h m i (h α i ε i A α i + 2K i B α i ε[α(b α(] + 2Kδ < ε[α(b α( + 2K]. Sine ε > 0 is rbitrry, by the Criterion of integrbility, this proves h R(α[, b] Further Properties of the Integrl Theorem 6.9. The Riemnn-Stieltjes integrl hs the following properties. (6.4 ( (Liner property If f 1, f 2 R(α[, b], then 1 f f 2 R(α[, b] for ll rel numbers 1, 2, nd ( 1 f f 2 dα = 1 f 1 dα + 2 f 2 dα. (b (Order property If f 1, f 2 R(α[, b] nd f 1 ( f 2 ( on [, b], then f 1 dα f 2 dα. ( (Additivity If f R(α[, b] nd < < b, then f R(α[, ] nd f R(α[, b]; moreover, fdα = fdα + fdα. Conversely, if < < b nd if f R(α[, ] nd f R(α[, b], then f R(α[, b], nd (6.4 holds. (d (Positive ombintion If f R(α 1 [, b] nd f R(α 2 [, b], nd k 1, k 2 re nonnegtive onstnts, then f R(k 1 α 1 + k 2 α 2 [, b], nd fd(k 1 α 1 + k 2 α 2 = k 1 fdα 1 + k 2 fdα 2.

7 6.2. Further Properties of the Integrl 7 (e (Absolute integrbility If f R(α[, b], then f R(α[, b], nd fdα f dα. Proof. Let s prove ( nd (e only. ( Assume f R(α[, b]. Let [, d] be subintervl of [, b]. Let ε > 0. Choose prtition P of [, b] suh tht U(P, f, α L(P, f, α < ε. Let P = P {, d} nd P 1 = P [, d]. Then P is refinement of P on [, b] nd P 1 is prtition of [, d], whih is prt of prtition P of [, b]. Therefore, we hve U d (P 1, f, α L d (P 1, f, α U(P, f, α L(P, f, α U(P, f, α L(P, f, α < ε, where U d (P 1, f, α, L d (P 1, f, α re for f defined on [, d], with P 1 being prtition on [, d]. Hene, by the Criterion for integrbility, f R(α[, d]. Now ssume < < b nd f R(α[, ] nd f R(α[, b]. Let ε > 0. prtitions P 1 of [, ] nd P 2 of [, b] suh tht U (P 1, f, α L (P 1, f, α < ε/2, Let P = P 1 P 2. Then P is prtition of [, b], nd we hve U b (P 2, f, α L b (P 2, f, α < ε/2. Choose U(P, f, α L(P, f, α = [U (P 1, f, α + U b (P 2, f, α] [L (P 1, f, α + L b (P 2, f, α] < ε. Hene, f R(α[, b]. To verify the dditivity property (6.4, ssume P is prtition of [, b]. Let P 0 = P {}, P 1 = P 0 [, ], nd P 2 = P 0 [, b]. Then P 0 = P 1 P 2 nd Hene U(P, f, α U(P 0, f, α = U (P 1, f, α + U b (P 2, f, α L(P, f, α L(P 0, f, α = L (P 1, f, α + L b (P 2, f, α fdα fdα + but, fdα = fdα, nd this proves fdα = fdα, fdα + fdα fdα. fdα + fdα + fdα + fdα; fdα, fdα. (e Assume f R(α[, b]. Then, with φ(t = t in Theorem 6.8, we hve f R(α[, b]. Sine f( f( f( ( [, b], by prt ( nd (b, we hve whih proves fdα f dα. f dα fdα f dα,

8 8 6. The Riemnn-Stieltjes Integrl Remrk 6.4. The onverse of (e in the theorem is flse. Indeed, onsider the funtion { 1 Q f( = 1 / Q. Then f( = 1 is onstnt nd hene f R[, b]; however, it is esily seen tht nd hene f / R[, b]. fd = b > 0, Theorem If f, g R(α[, b], then fg R(α[, b]. fd = b < 0, Proof. Using φ(t = t 2 in Theorem 6.8, we hve f 2, g 2, (f + g 2 R(α[, b], nd hene fg = (f + g2 f 2 g 2 2 R(α[, b]. Theorem Assume α is monotonilly inresing nd differentible on [, b] nd α R[, b]. Let f be bounded on [, b]. Then f R(α[, b] if nd only if fα R[, b]. In this se, fdα = f(α (d. Proof. Let ε > 0. Sine α R[, b], there is prtition P = { 0,, n } of [, b] suh tht (6.5 U(P, α L(P, α < ε. By the MVT, for eh j = 1, 2,..., n, there eists t j ( j 1, j suh tht α j = α (t j j. Let s j [ j 1, j ] be rbitrry. Then, by (6.5 nd Theorem 6.4, we hve α (t j α (s j j < ε. Put M = sup [,b] f < +. Sine α j = α (t j j, it follows tht f(s j α j f(s j α (s j j Mε. In prtiulr, f(s j α j U(P, fα + Mε; f(s j α (s j j U(P, f, α + Mε. Sine s j [ j 1, j ] is rbitrry, these two inequlities imply tht nd hene tht (6.6 U(P, f, α U(P, fα + Mε; fdα U(P, fα + Mε, U(P, fα U(P, f, α + Mε, fα d U(P, f, α + Mε.

9 6.2. Further Properties of the Integrl 9 Now note tht (6.5 remins true if P is repled by the refinement P Q, where Q is ny given prtition of [, b]. Hene (6.6 lso remins true with P repled by P Q, whih yields tht (6.7 fdα U(Q, fα + Mε, fα d U(Q, f, α + Mε for ll prtitions Q of [, b]. Tking the infim over prtitions Q, we hve fdα Sine ε > 0 is rbitrry, this implies fα d + Mε, fdα = fα d fα d, fdα + Mε. whih is vlid for ll bounded funtions f. The equlity fdα = fα d follows in etly the sme mnner. Hene the theorem is proved. Theorem 6.12 (Chnge of Vribles. Suppose φ is stritly inresing ontinuous from n intervl [A, B] onto [, b]. Suppose α is monotonilly inresing on [, b] nd f R(α[, b]. Define β nd g on [A, B] by β(y = α(φ(y, g(y = f(φ(y. Then g R(β[A, B], nd (6.8 B A gdβ = fdα. Proof. To eh prtition P = { 0,, n } of [, b] orresponds unique prtition Q = {y 0,, y n } of [A, B] suh tht j = φ(y j, nd vie vers. Then α j = α( j α( j 1 = α(φ(y j α(φ(y j 1 = β(y j β(y j 1 = β j for eh j = 1, 2,..., n. Sine the vlues tken by f on [ j 1, j ] re etly the sme s those tken by g on [y j 1, y j ], we see tht U(Q, g, β = U(P, f, α, L(Q, g, β = L(P, f, α. If f R(α[, b], then, for every ε > 0, it follows tht U(P, f, α L(P, f, α < ε for some prtition P of [, b]; hene, with the orresponding prtition Q of [A, B], we hve U(Q, g, β L(Q, g, β < ε. Hene g R(β[A, B], nd (6.8 holds. Combing the previous two theorems, we obtin the following hnge of vrible theorem for Riemnn integrls. Theorem If f R[, b] nd if φ: [A, B] [, b] is stritly inresing nd differentible with φ R[A, B], then where = φ(a, b = φ(b. f( d = B A f(φ(yφ (y dy,

10 10 6. The Riemnn-Stieltjes Integrl 6.3. Integrtion nd Differentition Theorem Let f R[, b]. Define F ( = f(tdt ( b. Then F is ontinuous on [, b]; furthermore, if f is ontinuous t point [, b], then F is differentible t, with F ( = f(. Proof. Suppose M = sup [,b] f < +. If < y b, then y F (y F ( = f(t dt M(y. Hene F is uniformly ontinuous on [, b]. Now ssume f is ontinuous t point [, b]. Given ε > 0, hoose δ > 0 suh tht f(t f( < ε t [, b], t < δ. Then, for ny [, b] with < < + δ, F ( F ( f( = 1 (f(t f( dt 1 nd similrly, for for ny [, b] with δ < <, F ( F ( f( = 1 (f(t f( dt 1 Hene, whenever [, b] nd 0 < < δ, it follows tht F ( F ( f( < ε; so F ( = f(. f(t f( dt < ε, f(t f( dt < ε. Theorem 6.15 (Fundmentl Theorem of Clulus. If f R[, b] nd if there is differentible funtion F on [, b] suh tht F = f, then f( d = F (b F (. Proof. Let ε > 0 be given. Choose prtition P = { 0,..., n } of [, b] suh tht U(P, f L(P, f < ε. The MVT furnishes points t j ( j 1, j suh tht Thus But hene F ( j F ( j 1 = f(t j j f(t j j = L(P, f (j = 1, 2,..., n. (F ( j F ( j 1 = F (b F (. f(t j j U(P, f, L(P, f f(t dt U(P, f; b F (b F ( f(t dt U(P, f L(P, f < ε. Sine ε > 0 is rbitrry, this ompletes the proof.

11 6.5. Retifible Curves in R k 11 Theorem 6.16 (Integrtion by Prts. Suppose F, G re differentible on [, b], F = f R[, b], nd G = g R[, b]. Then F (g( d = F (bg(b F (G( f(g( d. Proof. Clerly F g, fg R[, b]. Let H( = F (G( on [, b]. Then H ( = f(g(+ g(f ( nd hene H R[, b]. So, by Theorem 6.15, H(b H( = proving the theorem. H ( d = 6.4. Integrtion of Vetor-Vlued Funtions f(g( d + g(f ( d, Definition 6.5. Let f : [, b] R k nd α: [, b] R be monotonilly inresing on [, b]. Let f( = (f 1 (,, f k (, whih eh oordinte funtion f j is rel vlued. We sy f R(α[, b] if eh of its oordinte funtions f j R(α[, b]. In this se, we define ( f dα = f 1 dα,, f k dα. Mny of the results on rel-vlued funtions lso hold for these vetor-vlued funtions. To illustrte, we stte the nlogue of the fundmentl theorem of lulus. Theorem If f nd F mp [, b] into R k, f R[, b], nd F = f on [, b], then f(d = F(b F(. Theorem If f R(α[, b], then f R(α[, b], nd f dα f dα. Proof. If f = (f 1,, f k then f = (f f k 21/2. Hene if eh f j R[, b], then f R[, b]. To show the inequlity on the integrls, we ssume y = f dα 0; otherwise there is nothing to prove. By linerity nd order properties nd the Cuhy- Shwrz inequlity, y 2 = y f dα = Cnelling y > 0 proves the inequlity Retifible Curves in R k y f dα y f dα = y f dα. Definition 6.6. A ontinuous funtion γ from n intervl [, b] into R k is lled urve in R k. To emphsize the prmeter intervl [, b], we my lso sy tht γ is urve on [, b]. If γ is one-to-one, then γ is lled n r. If γ( = γ(b then γ is sid to be losed urve. Note tht urve in R k is funtion, not the rnge of γ, whih is point set in R k ; different urves my hve the sme rnge.

12 12 6. The Riemnn-Stieltjes Integrl Let γ : [, b] R k be urve. We ssoite to eh prtition P = { 0,, n } of [, b] the number Λ(P, γ = γ( j γ( j 1. (This ould be defined for urves in ny metri spe X, with γ( j γ( j 1 repled by d(γ( j, γ( j 1. Define the length of γ to be the number (inluding + Λ(γ = sup Λ(P, γ, where the supremum is tken over ll prtitions P of [, b]. We sy tht γ is retifible if Λ(γ < +. Theorem If γ is ontinuous on [, b], then γ is retifible, nd Λ(γ = γ (t dt. Proof. If j 1 < j b, then j γ( j 1 γ( j = γ (t dt j 1 Hene Λ(P, γ j j 1 γ (t dt = j j 1 γ (t dt. γ (t dt for ll prtitions P of [, b]. Consequently, Λ(γ γ (t dt. To show the opposite inequlity, let ε > 0 be given. Sine γ is uniformly ontinuous on [, b], there eists δ > 0 suh tht γ (t γ (s < ε if s t < δ, s, t [, b]. Let P = { 0,, n } be prtition of [, b] with P < δ. If t [ j 1, j ] then γ (t γ ( j + ε. Hene, by Theorem 6.17, j j γ (t dt γ ( j j + ε j = [γ (t + γ ( j γ (t] dt j 1 j 1 + ε j j j γ (t dt j 1 + [γ ( j γ (t] dt j 1 + ε j j γ( j γ( j 1 + γ ( j γ (t dt + ε j j 1 γ( j γ( j 1 + 2ε j. Adding these inequlities, we obtin γ (t dt Λ(P, γ + 2ε(b. Sine ε > 0 is rbitrry, it follows tht γ (t dt Λ(P, γ. This ompletes the proof. Suggested Homework Problems Pges Problems: 1 5, 7 9, 15, 17, 19

13 6.6. Improper Riemnn Integrls* Improper Riemnn Integrls* Definition 6.7. Let (, b R, where < b +, nd f : (, b R. We sy tht f is lolly integrble on (, b if f R[, d] for eh finite losed subintervl [, d] of (, b. We sy tht f is improperly (Riemnn integrble on (, b if f is lolly integrble on (, b nd the it ( (6.9 f( d = f( d + d b eists nd is finite. In this se, this it is lled the improper (Riemnn integrl of f on (, b. Sometimes we lso use the nottion f( d = + f( d to distinguish the improper integrls from the Riemnn integrls defined erlier. Lemm The order of its in (6.9 does not mtter. In prtiulr, if the it in (6.9 eists nd is finite, then the it ( f( d eists nd equls the it in (6.9. d b Proof. Let 0 (, b. Then ( f( d + d b + ( 0 = f( d + f( d + d b 0 0 (6.10 = f( d + + d b f( d. 0 Sine, for eh, 0 b ( d b d b f(d eists, we hve f( d 0 [ = 0 b d b = d b [ ( 0 ] = f( d f( d 0 b d b f( d 0 ] f( d f( d 0 b 0 Therefore, in (6.10 letting 0 b, we obtin tht ( 0 ( 0 b + f( d = + f( d = 0. d b f( d.

14 14 6. The Riemnn-Stieltjes Integrl Remrk 6.8. (i If f is integrble on [, b] for ll (, b, then the improper Riemnn integrl of f on (, b is lso given by f( d = f( d := f( d. + + If this it eists nd is finite, we lso sy tht f is improperly integrble on (, b]. The similr sitution pplies t the endpoint b, in whih se we sy tht f is improperly integrble on [, b. (ii It is esily seen tht f is improperly integrble on (, b if nd only if f is improperly integrble on (, ] nd on [, b for ll (, b. In this se, we hve tht + f( d = + f( d + f( d. Theorem The funtion f( = 1/ p is improperly integrble on (0, 1] if nd only if p < 1, nd is improperly integrble on [1, + if nd only if p > 1. Proof. Eerise! Theorem 6.22 (Liner Property. If f, g re improperly integrble on (, b nd k, l R, then kf + lg is improperly integrble on (, b, nd (kf( + lg( d = k f( d + l g( d. Proof. Use the Liner Property of integrls on eh subintervl [, d] of (, b. Theorem 6.23 (Comprison Theorem for Improper Integrls. Suppose tht f, g re lolly integrble on (, b nd 0 f( g( for ll (, b. If g is improperly integrble on (, b, then f is lso improperly integrble on (, b nd f( d g( d. Proof. Fi (, b. Let F (d = f( d nd G(d = g(d for d [, b. Then by the Order Property, F (d G(d. Note tht F nd G re inresing on [, b nd G(b eists. Hene F is bounded bove by G(b nd so F (d eists nd is finite. This shows tht f is improperly integrble on [, b. By the similr rgument, we lso show tht f is improperly integrble on (, ]; thus f is improperly integrble on (, b. The order property f( d g( d. follows esily from the order property of the Riemnn integrls of f nd g on eh subintervl [, d] of (, b. Emple 6.1. Show tht f( = (sin / 3/2 is improperly integrble on (0, 1]. Proof. Sine 0 sin for ll [0, 1] (use elementry lulus to prove it!, it follows tht 0 f( 3/2 = 1/2 (0, 1]. Sine 1/2 is improperly integrble on (0, 1], by the theorem bove, f is improperly integrble on (0, 1]. Emple 6.2. Show tht f( = (ln / 5/2 is improperly integrble on [1, +.

15 6.6. Improper Riemnn Integrls* 15 Proof. Sine 0 ln for ll 1 (use elementry lulus to prove it!, it follows tht 0 f( 5/2 = 3/2 1. Sine 3/2 is improperly integrble on [1, +, by the theorem bove, f is improperly integrble on [1, +. Lemm If f is bounded nd lolly integrble on (, b nd g is improperly integrble on (, b, then fg is improperly integrble on (, b. Proof. Use 0 fg M g nd the Comprison Theorem bove. Definition 6.9. Let f : (, b R. We sy tht f is bsolutely integrble on (, b if f is lolly integrble on (, b nd f is improperly integrble on (, b. We sy tht f is onditionlly integrble on (, b if f is improperly integrble on (, b but f is not improperly integrble on (, b. Theorem If f is bsolutely integrble on (, b, then f is improperly integrble on (, b nd f( d f( d. Proof. Sine 0 f + f 2 f, by the Comprison Theorem, f + f is improperly integrble on (, b. Hene, by the Liner Property, f = ( f + f f is lso improperly integrble on (, b. Moreover, for ll < d in (, b, f( d f( d. We then omplete the proof by tking the it s + nd d b. The onverse of Theorem 6.25 is flse. Emple 6.3. Prove tht f( = sin is onditionlly integrble on [1, +. Proof. Integrting by prts, we hve for ll d > 1, sin = os d d os 2 Sine 1/ 2 is bsolutely integrble on [1, +, we hve (os / 2 is bsolutely integrble on [1, + ; hene (os / 2 is improperly integrble on [1, +. Tking the it s d + bove, we hve + sin + os d = os( d eists nd is finite. This proves tht (sin / is improperly integrble on [1, +. We now show tht sin / is not improperly integrble on [1, +, whih proves tht (sin / is onditionlly integrble on [1, +. Note tht if n N nd n 2 then nπ sin kπ sin 1 kπ d d sin d = 2 1 kπ π k. 1 k=2 (k 1π k=2 Hene nπ sin d = +. n 1 So sin / is not improperly integrble on [1, +. (k 1π d. k=2