1. On some properties of definite integrals. We prove


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1 This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments. We only wish to dd few detils tht re not provided in [M]. We prove. On some properties of definite integrls Proposition.. Let f, g R[, b] be given. We hve (i) Let m f(x) M for x [, b], nd let Φ : [m, M] R be continuous, then Φ f R[, b]; (ii) fg R[, b]; (iii) f R[, b] nd it holds tht f Proof. (i) Let ε > be fixed. By the uniform continuity of Φ, there exists δ > such tht s t < δ implies Φ(s) Φ(t) < ε. Without loss of generlity, we my ssume tht δ < ε. Let P be prtition of [, b] such tht f. S(f, P ) s(f, P ) < δ 2. We divide the prtition P, tht is, the indices,..., n} into two subprtitions, A nd B so tht M i m i < δ if i A nd M i m i δ if i B. Set Mi = sup (Φ f)(t), nd m i = inf (Φ f)(t). t [x i,x i ] t [x i,x i ] Then, becuse of our choice of ε nd δ, for i A, we hve M i m i < ε. For i B we hve δ (x i x i ) i m i )(x i x i ) S(f, P ) s(f, P ) < δ i B i B(M 2, () tht is, Setting (x i x i ) < δ. i B K = sup Φ(t), t [,b] Non ftt in clsse, e non richiest per l esme
2 2 we then hve S(Φ f, P ) s(φ f, P ) = (Mi m i )(x i x i ) + i i A i B(M m i )(x i x i ) ε (x i x i ) + 2K i x i ) i A i A(x ε(b ) + 2Kδ ε(b + 2K). Since ε > ws rbitrry, conclusion (i) follows. (ii) Using prt (i), (f + g) 2, (f g) 2 R[, b]. Hence, 4fg = (f + g) 2 (f g) 2 R[, b]. (iii) With Φ(t) = t we obtin tht f R[, b]. If c = ± is such tht c f, then f = c f = cf f, since cf f. 2. On the Fundmentl Theorem of Clculus In this prt we wish to prove the Fundmentl Theorem of Clculus in more generl form thn the one presented in [M]. We recll tht function F : I R such tht F = f on I is clled n ntiderivtive of f on I, nd tht, in this cse, f is sid to dmit n ntiderivtive on I. Essentilly, the corollry sys tht if function f is Riemnnintegrble nd dmits n ntiderivtive, then the ntiderivtives must be the integrl function F (x) = f(t) dt + c. Theorem 2.. Let f R[, b] nd suppose tht there exists F : [, b] R differentible nd such tht F (x) = f(x), for x [, b]. Then f(t) dt = F (b) F (). Corollry 2.2. Let f R[, b] nd suppose tht there exists F : [, b] R differentible nd such tht F (x) = f(x). Then for some c R. F (x) = f(t) dt + c We need lemm. Given prtition P = = < x < < x n < x n = b} of [, b], we set l(p ) = supδ > : x i x i, i =,..., n}. Lemm 2.3. Let f R[, b]. Then f(t i )(x i x i ) = lim l(p ) for ny choice of t i [x i, x i ]. f(t) dt,
3 Proof. We know tht f R[, b] if nd only if for every ε > there exists prtition P such tht S(f, P ) s(f, P ) < ε. Since we hve m i = Notice tht, in prticulr, inf f(x) f(t i) sup f(x) = M i x [x i,x i ] x [x i,x i ] s(f, P ) f(t i )(x i x i ) S(f, P ). f(t i )(x i x i ) Moreover, if P is ny refinement of P, l(p ) l(p ) nd so tht lso s(f, P ) s(f, P ) S(f, P ) S(f, P ), S(f, P ) s(f, P ) < ε. f(t) dt < ε. (2) This implies tht lim l(p ) n f(t i)(x i x i ) exists nd the conclusion follows. 3 Proof of Thm. 2.. Let ε > nd P prtition of [, b] such tht S(f, P ) s(f, P ) < ε. Since F is differentible, by the men vlue theorem there exists t i [x i, x i ] such tht Therefore, nd by (2), F (x i ) F (x i ) = F (t i )(x i, x i ) = f(t i )(x i, x i ). f(t i )(x i x i ) = for every ε >. Hence, This proves the theorem. F (x i ) F (x i ) = F (b) F () F (b) F () F (b) F () = f(t) dt < ε, f(t) dt. Proof of Cor Applying the previous theorem to f, F on the intervl [, x] we obtin thus, the conclusion. F (x) = f(t) dt + F (),
4 4 Exmple 2.4. Consider the function f α (x) = sin x x α x =. Then there exists F differentible nd such tht F = f α if nd only if α =. Proof. Let α R nd F (x) = sin t dt. Then F (x) = sin x for x, by the Fundmentl Theorem of Clculus. On the other hnd, using n obvious chnge of vribles nd integrtion by prts we hve F (x) F () lim = lim x x x x sin t dt /y = lim y sin y + t dt = lim y sin s ds y + y s2 = lim y y + s 2 cos s) s=+ s=y cos y + = lim y y + y 2 2 =, y 2 y cos s ds s3 } cos s ds s3 } since y y s 3 cos s ds y y s 3 ds = 2y s y +. Hence, F is differentible t x = nd F () =. Consequentely, f dmits ntiderivtive. Next, let α nd suppose f α dmits ntiderivtive G. Then F G is differentible nd (F G) if x (x) = α if x =. Thus, (F G) hs discontinuity of first kind nd this is not possible (unless α = ). For x > we define 3. Euler s Gmm function Γ(x) = t x e t dt. Since t x e t t x s t +, the improper integrl tx e t dt converges exctly when x >. Moreover, since t x e t c x e t/2, for t, sy, the improper integrl t x e t dt converges. Hence, Γ(x) is well defined on (, + ).
5 Lemm 3.. For ny x > it holds tht Moreover, for ny positive integer n we hve Γ(x + ) = xγ(x). Γ(n + ) = n!. Notice tht this lst identity sys tht the Gmm function interpoltes the fctorils. Euler obtined the expression of the Gmm function by serching function tht interpoltes the fctorils using the theory of infinite products nd then proving the integrl representtion tht now is used s definition of the Gmm function. Proof. The proof just relies on n integrtion by prts nd then simple induction rgument. We hve, Γ(x + ) = since the term t x e t t=+ vnishes. t= t x e t dt = t x e t t=+ + x = xγ(x), t= t x e t dt Next, clerly Γ() = =!. Now we proceed by inductio. Assuming tht Γ(n) = (n )! we see tht Γ(n + ) = nγ(n) = n!. This proves the lemm Locl Existence nd Uniqueness of Solutions of Cuchy Problems Let (, y ) R 2 e set I = [, + ], J = [y b, y + b]. nd set Ω = I J R 2. Let f : I J R be continuous on Ω, nd Lipschitz in y uniformly in x, tht is, there exists L > such tht f(x, y ) f(x, y 2 ) L y y 2 (3) for ll x I nd y, y 2 J. Theorem 4.. Let f, Ω be s bove, ( y ) I J. Then, there exists δ > such tht the Cuchy problem y = f(x, y) y( ) = y (4) dmits unique solution defined on the intervl I δ = [ δ, + δ]. Proof. By the continuity of f we hve tht is well defined nd finite. Let M = mx f(x, y) (x,y) I J δ < min, b M, }. (5) L
6 6 Consider the metric spce C(I δ ) = C R (I δ ), the complete metric spce of continuous functions defined on I δ, hving rel vlues, with the uniform norm. Let } B = y C(I δ ) : y y C(Iδ ) b be the closed bll in C(I δ ), with center in y nd rdius b (where y represents the function tht is constnt nd equl to y ). Then B is itself complete metric spce. Consider the mpping F : B B given by F (y) = z, where z is the function z(x) = y + f(t, y(t)) dt, nd x I δ. Let us show tht in fct F tkes vlues in B, tht is, if y B, then F (y) B. We wnt to show tht F (y) y C(Iδ ) b. (6) We observe tht, for every x I δ, we hve F (y)(x) y = z(x) y = f(t, y(t)) dt f(t, y(t)) dt M dt Mδ b. Hence, (6) holds true. We now show tht F : B B is ctully contrction, so tht by the fixed point theorem, it would hve unique fixed point. Using the Lipschitz condition (3) on f we hve, F (y ) F (y 2 ) = L f(t, y (t)) f(t, y 2 (t)) dt f(t, y (t)) f(t, y 2 (t)) dt y (t) y 2 (t) dt L ( supt Iδ y (t) y 2 (t) ) dt Lδ y y 2 C(Iδ ), where k = Lδ < by ssumption (5). Therefore, T is contrction on the complete metric C(I δ ) nd it dmits unique fixed point tht in turns is the unique solution of the Volterr eqution nd therefore of the Cuchy problem (4). Diprtimento di Mtemtic F. Enriques, Università degli Studi di Milno, Vi C. Sldini 5, I233 Milno Emil ddress: