Tel Aviv University, 26 Analysis-III 9 9 Improper integral 9a Introduction....................... 9 9b Positive integrands................... 9c Special functions gamma and beta......... 4 9d Change of variables................... 6 9e Iterated integral..................... 7 9f Multidimensional beta integrals of Dirichlet... 2 9g Non-positive (signed) integrands.......... 22 Riemann integral and volume are generalized to unbounded functions and sets. 9a Introduction The n-dimensional unit ball in the l p metric, E = {(x,..., x n ) x p + + x n p }, is an admissible set, and its volume is a Riemann integral, v(e) = R n l E, of a bounded function with bounded support. In Sect. 9f we ll calculate it: where Γ is a function defined by Γ(t) = v(e) = 2n Γ n ( p ) p n Γ( n p + ) x t e x dx for t > ; here the integrand has no bounded support; and for t = p < it is also unbounded (near ). Thus we need a more general, so-called improper integral, even for calculating the volume of a bounded body! In relatively simple cases the improper integral may be treated via ad hoc limiting procedure adapted to the given function; for example, x t e x dx = lim / x t e x dx.
Tel Aviv University, 26 Analysis-III In more complicated cases it is better to have a theory able to integrate rather general functions on rather general n-dimensional sets. Different functions may tend to infinity on different subsets (points, lines, surfaces), and still, we expect (af + bg) = a f + b g (linearity) to hold, as well as change of variables. 9b Positive integrands We consider an open set G R n and functions f G [, ) continuous almost everywhere. 2 We do not assume that G is bounded. We also do not assume that G is admissible, even if it is bounded. 3 Continuous almost everywhere means that the set A G of all discontinuity points of f has measure (recall Sect. 6d). We can use the function f l G equal f on G and on R n G, but must be careful: l G and f l G need not be continuous almost everywhere. We define (9b) G f = sup { R n g g R n R integrable, g f on G, g = on R n G} [, ]. The condition on g may be reformulated as g f l G. If f l G is integrable (on R n ), then clearly G f = R n f l G, which generalizes 4d5. This happens if and only if f l G is bounded, with bounded support, and f(x) f(x ) = as G x x for almost all x G (thin, why). (Void if G has measure.) 9b2 Exercise. (a) Without changing the supremum in (9b) we may restrict ourselves to continuous g with bounded support; or, alternatively, to step functions g; and moreover, in both cases, WLOG, g has a compact support inside G; Additional literature (for especially interested): M. Pascu (26) On the definition of multidimensional generalized Riemann integral, Bul. Univ. Petrol LVIII:2, 9 6. (Research level) D. Maharam (988) Jordan fields and improper integrals, J. Math. Anal. Appl. 33, 63 94. 2 This condition will be used in 9b9. 3 A bounded open set need not be admissible, even if it is diffeomorphic to a dis.
Tel Aviv University, 26 Analysis-III (b) if f is bounded (not necessarily a.e. continuous) and G is bounded, then G f = R n f l G, and in particular, G = v (G); (c) if f is bounded and G is admissible, then the integral defined by (9b) is equal to the integral defined by 4d5. Prove it. There are many ways to treat the improper integral as the limit of (proper) Riemann integrals; here are some ways. 9b3 Exercise. Consider the case G = R n, and let be a norm on R n of the form 2 x = ( x p + + x n p ) /p for x = (x,..., x n ); here p [, ] is a parameter (and x = max( x,..., x n ) if p = ). (a) Prove that R n f = lim x < min(f(x), ) dx. (b) For a locally bounded 3 f prove that R n f = lim x < f(x) dx. (c) Can it happen that f is locally bounded, not bounded, and R n f <? 9b4 Example (Poisson). Consider I = e x 2 dx. R 2 On one hand, by 9b3 for the Euclidean norm (p = 2), I = lim e (x2+y2) dxdy = lim x 2 +y 2 < 2 r dr e r2 2π dθ = lim π On the other hand, by 9b3 for (x, y) = max( x, y ) (p = ), I = lim e (x2 +y 2) dxdy = lim x <, y < ( e x2 dx)( and we obtain the celebrated Poisson formula: + e x2 dx = π. In fact, v (G) is Lebesgue s measure of G. 2 But in fact, the same holds for arbitrary norm. 3 That is, bounded on every bounded subset of R n. 2 e y2 dy) = ( e u du = π. + e x2 dx) 2,
Tel Aviv University, 26 Analysis-III 2 9b5 Exercise. Consider I = x>,y> x a y b e (x2 +y 2) dxdy [, ] for given a, b R. Prove that, on one hand, I = ( and on the other hand, I = ( r a+b+ e r2 dr)( x a e x2 dx)( π/2 cos a θ sin b θ dθ), x b e x2 dx). 9b6 Exercise. Consider f R 2 [, ) of the form f(x) = g( x ) for a given g [, ) [, ). (a) If g is integrable, then f is integrable and R 2 f = 2π g(r) r dr. (b) If g is continuous on (, ), then R 2 f = 2π g(r) r dr [, ]. Prove it. 9b7 Exercise. Let be as in 9b3. 2 Consider f R n [, ) of the form f(x) = g( x ) for a given g [, ) [, ). (a) If g is integrable, then f is integrable, and R n f = nv g(r) r n dr where V is the volume of {x x < }. (b) If g is continuous on (, ), then R n f = nv g(r) r n dr [, ]. c) Let g be continuous on (, ) and satisfy g(r) r a for r +, g(r) r b for r +. Then f < if and only if b < n < a. Prove it. 3 9b8 Example. R n e x 2 dx = nv r n e r2 dr; in particular, R n e x 2 dx = nv n r n e r2 dr where V n is the volume of the (usual) n-dimensional unit ball. On the other hand, R n e x 2 dx = ( R e x2 dx) n = π n/2. Therefore Not unexpectedly, V 2 = V n = π 2 re r2 dr = π. π n/2 n r n e r2 dr. Hint: (a) polar coordinates; (b) use (a). 2 But in fact, the same holds for arbitrary norm. 3 Hint: (a) first, g = l [,a], second, a step function g, and third, sandwich; also, (a) (b) (c).
Tel Aviv University, 26 Analysis-III 3 Clearly, G cf = c G f for c (, ). 9b9 Proposition. G (f + f 2 ) = G f + G f 2 [, ] for all f, f 2 on G, continuous almost everywhere. Proof. The easy part: G (f + f 2 ) G f + G f 2. Given integrable g, g 2 such that g f l G and g 2 f 2 l G, we have g + g 2 = (g +g 2 ) G (f + f 2 ), since g + g 2 is integrable and g + g 2 (f + f 2 ) l G. The supremum in g, g 2 gives the claim. The hard part: G (f + f 2 ) G f + G f 2, that is, g G f + G f 2 for every integrable g such that g (f +f 2 ) l G. We introduce g = min(f, g), g 2 = min(f 2, g) (pointwise minimum on G; and on R n G) and prove that they are continuous almost everywhere (on R n, not just on G). For almost every x G, both f and g are continuous at x and therefore g is continuous at x. For almost every x G, g is continuous at x, which ensures continuity of g at x (irrespective of continuity of f ), since g(x) = (x G). Thus, g is continuous almost everywhere; the same holds for g 2. By Lebesgue s criterion 6d2, the functions g, g 2 are integrable. We have g +g 2 min(f +f 2, g) = g, since generally, min(a, c)+min(b, c) min(a+b, c) for all a, b, c [, ) (thin, why). Thus, g (g + g 2 ) = g + g 2 G f + G f 2, since g f l G, g 2 f 2 l G. 9b Proposition (exhaustion). For open sets G, G, G 2, R n, G G G f G f [, ] for all f G [, ) continuous almost everywhere. Proof. First of all, G f G+ f (since g f l G implies g f l G+ ), and similarly, G f G f, thus G f and lim G f G f. We have to prove that G f lim G f. We tae an integrable g, compactly supported inside G (recall 9b2(a)), such that g f on G. By compactness, there exists such that g f l G. Then g G f lim G f. The supremum in g proves the claim., 2 9b Corollary (monotone convergence for volume). For open sets G, G G 2, R n, G G v (G ) v (G). 9b2 Remar. Let G, G 2, R n be (pairwise) disjoint open balls. Then v (G G 2... ) = v(g ) + v ( G 2 ) +... even if the union is dense in R n (which can happen; thin, why). Compare it with 4c7: (f + g) f + g. 2 Really, this is easy to prove without 9b (try it).
Tel Aviv University, 26 Analysis-III 4 9c Special functions gamma and beta The Euler gamma function Γ is defined by (9c) Γ(t) = x t e x dx for t (, ). This integral is not proper for two reasons. First, the integrand is bounded near for t [, ) but unbounded for t (, ). Second, the integrand has no bounded support. In every case, using 9b, Γ(t) = lim / x t e x dx <, since the integrand (for a given t) is continuous on (, ), is O(x t ) as x, and (say) O(e x/2 ) as x. Thus, Γ (, ) (, ). Clearly, Γ() =. Integration by parts gives (9c2) / x t e x dx = x t e x x=/ + t / Γ(t + ) = tγ(t) for t (, ). x t e x dx ; In particular, (9c3) Γ(n + ) = n! for n =,, 2,... We note that (9c4) x a e x2 dx = 2 Γ(a + ) for a (, ), 2 since x a e x2 dx = u a/2 e u du 2. For a = the Poisson formula (recall u 9b4) gives (9c5) Γ( 2 ) = π. Thus, (9c6) Γ( 2n + 2 ) = 2 3 2 2n π. 2 The volume V n of the n-dimensional unit ball (recall 9b8) is thus calculated: (9c7) V n = πn/2 n 2 Γ( n 2 ). This is rather Γ (,).
Tel Aviv University, 26 Analysis-III 5 Not unexpectedly, V 3 = π3/2 3 By 9b5, 2 that is, (9c8) = π3/2 2 Γ( 3 2 ) 3 2 2 π = 4 3 π. Γ( a+b+2 2 ) π/2 cos a θ sin b θ dθ = 2 In particular, π/2 Γ( a+ 2 ) 2 b+ Γ( ) 2 for a, b (, ); cos α θ sin β θ dθ = Γ( α 2 )Γ( β 2 ) 2 Γ( α+β 2 ) for α, β (, ). (9c9) π/2 sin α θ dθ = π/2 cos α θ dθ = π 2 Γ( α) 2 Γ( α+). 2 The trigonometric functions can be eliminated: π/2 cos α θ sin β θ dθ = cos α 2 θ sin β 2 θ 2 sin θ cos θ dθ = 2 α 2 ( u) 2 u β 2 2 du; thus, 2 π/2 (9c) where (9c) x α ( x) β dx = B(α, β) for α, β (, ), B(α, β) = Γ(α)Γ(β) Γ(α + β) for α, β (, ) is another special function, the beta function. 9c2 Exercise. Chec that B(x, x) = 2 2x B(x, 2 ). 9c3 Exercise. Chec the duplication formula: 2 Γ(2x) = 22x π Γ(x) Γ(x + 2 ). 9c4 Exercise. Calculate x4 x 2 dx. π Answer: 32. 9c5 Exercise. Calculate x m e xn dx. m+ Answer: nγ( ). n 9c6 Exercise. Calculate xm (ln x) n dx. Answer: ( ) n n! (m+) n+. Hint: π/2 2 sin θ cos θ ( ) 2x dθ. 2 2 Hint: use 9c2.
Tel Aviv University, 26 Analysis-III 6 9c7 Exercise. Calculate π/2 Answer: Γ2 (/4) 2 2π. dx cos x. 9c8 Exercise. Chec that Γ(t)Γ( t) = We mention without proof another useful formula x t + x dx = π sin πt x t +x dx for < t <. for < t <. There is a simple proof that uses the residues theorem from the complex analysis course. This formula yields that Γ(t)Γ( t) = for < t <. π sin πt Is the function Γ continuous? For every compact interval [t, t ] (, ) the given function of two variables (t, x) x t e x is continuous on [t, t ] [, ], therefore its integral in x is continuous in t on [t, t ] (recall 4e6(a)). Also, / x t e x dx Γ(t) uniformly on [t, t ], since / x t e x dx / x t dx as and x t e x dx x t e x dx as. It follows that Γ is continuous on arbitrary [t, t ], therefore, on the whole (, ). In particular, tγ(t) = Γ(t + ) Γ() = as t +; that is, 9d Change of variables Γ(t) = t + o( t ) as t +. 9d Theorem (change of variables). Let U, V R n be open sets, ϕ U V a diffeomorphism, and f V [, ). Then (a) (f is continuous almost everywhere on V ) (f ϕ is continuous almost everywhere on U) ((f ϕ) det Dϕ is continuous almost everywhere on U); (b) if they are continuous almost everywhere, then V f = U (f ϕ) det Dϕ [, ]. Item (a) follows easily from 8c (similarly to the proof of 8a(a) in Sect. 8c but simpler: 8c4 is not needed now). Hint: change x to y via ( + x)( y) =.
Tel Aviv University, 26 Analysis-III 7 9d2 Lemma. Let U, V, ϕ, f be as in Th. 9d, and in addition, f be compactly supported within V. Then 9d(b) holds. Proof. This is basically Prop. 8d; there U, V are admissible, since otherwise the integrals over U and V are not defined by 4d5. Now they are defined (see the paragraph after (9b)): V f = R n f l V (and similarly for U), and the proof of 8d given in Sect. 8d applies (chec it). Proof of Th. 9d(b). First, we prove that (9d3) V f U (f ϕ) det Dϕ. Assume the contrary. By 9b2(a) there exists integrable g, compactly supported within V, such that g f on V and V g > U (f ϕ) det Dϕ. By 9d2, V g = U (g ϕ) det Dϕ U (f ϕ) det Dϕ ; this contradiction proves (9d3). Second, we apply (9d3) to ϕ = ϕ V U and f = (f ϕ) det Dϕ U [, ): U f V (f ϕ ) det Dϕ. By the chain rule, ϕ ϕ = id V implies ((Dϕ) ϕ )(Dϕ ) = id, thus ((det Dϕ) ϕ )(det Dϕ ) =. We get f ϕ = (f ϕ ϕ ) (det Dϕ) ϕ = (f ϕ ) det Dϕ = f; U (f ϕ) det Dϕ = U f V f. 9e Iterated integral f det Dϕ ; We consider an open set G R m+n and functions f G [, ) continuous almost everywhere. Similarly to Sect. 5d, the section f(x, ) of f need not be continuous almost everywhere on the section G x = {y (x, y) G} of G; thus, Gx f(x, ) is generally ill-defined. Similarly to Th. 5d we need the lower integral (but no upper integral this time). We define the lower integral by (9b) again, but this time f G [, ) is arbitrary (rather than continuous almost everywhere). That is, for open G R n (rather than R m+n, for now) (9e) f = sup { g g R n R integrable, G R n g f on G, g = on R n G} [, ].
Tel Aviv University, 26 Analysis-III 8 In particular, if f is continuous almost everywhere on G, then G f = G f. As before, the condition on g may be reformulated as g f l G. Still, 9b2(a) applies (chec it). And 9b2(b) becomes: if f is bounded and G is bounded, then G f = R n f l G, the latter integral being proper, that is, defined in Sect. 4c. Similarly to 9b, for open sets G, G, G 2, R n, (9e2) G G G f G f [, ] for arbitrary f G [, ). Similarly to 9b3(a), G G G min(f, ) G f [, ]. If, in addition, G are bounded, then we may rewrite it as (9e3) R n min(f, )l G G f, the left-hand side integral being proper. An increasing sequence of integrable functions can converge to a function that is not almost everywhere continuous (and moreover, is discontinuous everywhere). Nevertheless, a limiting procedure is possible, as follows. 9e4 Proposition. If g, g 2, R n [, ) are integrable and g f R n [, ), then R n g R n f. 2 This claim follows easily from an important theorem (to be proved in Appendix). 9e5 Theorem (monotone convergence for Riemann integral). If g, g, g 2, R n R are integrable and g g, then R n g R n g. Proof that Th. 9e5 implies Prop. 9e4. Clearly, g f implies lim g f; we have to prove that lim g f. Given an integrable g f, we have min(g, g) min(f, g) = g and, by 9e5, min(g, g) g. Thus, g lim g ; supremum in g gives f lim g. We return to an open set G R m+n and its sections G x R n for x R m. Pointwise, not uniformly. 2 Do you thin that g f for arbitrary (not integrable) g? No, this is wrong. Recall f of 4e7 and consider f.
Tel Aviv University, 26 Analysis-III 9 9e6 Theorem (iterated improper integral). If a function f G [, ) is continuous almost everywhere, then R m dx G x dy f(x, y) = G f(x, y) dxdy [, ]. Unlie Th. 5d, both integrals in the left-hand side are lower integrals. The function x Gx dy f(x, y) need not be almost everywhere continuous, even if G = R 2 and f is continuous. Moreover, it can happen that x R f(x, ) is unbounded on every interval, even if f R 2 [, ) is bounded, continuously differentiable, R 2 f <, and f(x, y), f(x, y) as x 2 + y 2. (Can you find a counterexample? Hint: construct separately f R [2,2 + ] for each.) It is easy to see (try it!) that G f does not exceed the iterated integral; but the equality needs more effort. Proof. We tae admissible open sets G R m+n such that G G, and introduce f = min(f, )l G, that is, f(x, y), if (x, y) G and f(x, y), f (x, y) =, if (x, y) G and f(x, y),, if (x, y) G. By Lebesgue s criterion 6d2, each f is integrable. By (9e3), R m+n f G f, the left-hand side integral being proper. Given x R m, we apply the same argument to the sections f (x, ), f(x, ), (G ) x, G x, taing into account that f (x, ) need not be integrable, and we get R n f (x, ) = R n min(f(x, ), )l G (x, ) Gx f(x, ). By Th. 5d (applied to f ), the function x R n f (x, ) is integrable, and its integral is equal to R m+n f. Applying Prop. 9e4 to these functions we get f = R (x f (x, )) m+n R m R n (x f(x, )) ; R m G x but on the other hand, R m+n f G f. 9e7 Corollary. The volume 2 of an open set G R m+n is equal to the lower integral of the volume of G x (even if G is not admissible). For example, we may use the interior of the union of all N-pixels contained in G [ N, N] n. 2 That is, v (G) if G is bounded; and G (in fact, the Lebesgue measure of G) in general.
Tel Aviv University, 26 Analysis-III 2 9f Multidimensional beta integrals of Dirichlet 9f Proposition. x,...x n>, x + +x n< for all p,... p n >. For the proof, we denote x p... x pn n dx... dx n = I(p,..., p n ) = x,...x n>, x + +x n< This integral is improper, unless p,..., p n. Γ(p )... Γ(p n ) Γ(p + + p n + ) x p... x pn n dx... dx n. 9f2 Lemma. I(p,..., p n ) = B(p n, p + + p n + )I(p,..., p n ). Proof. The change of variables ξ = ax (that is, ξ = ax,..., ξ n = ax n ) gives (by Theorem 9d ) ξ,...ξ n>, ξ + +ξ n<a ξ p... ξ pn n dξ... dξ n = a p + +p n I(p,..., p n ) for a >. Thus, using 9e6 and (9c), I(p,..., p n ) = = dx n x pn n x,...x n >, x + +x n < x n p x... x p n n dx... dx n = x pn n ( x n ) p + +p n I(p,..., p n ) dx n = = I(p,..., p n )B(p n, p + + p n + ). Proof of Prop. 9f. Induction in the dimension n. For n = the formula is obvious: x p dx = p = Γ(p ) Γ(p + ). But a linear change of variables does not really need 9d; it is a simple generalization of 7c or even (4h5).
Tel Aviv University, 26 Analysis-III 2 From n to n: using 9f2 (and (9c)), I(p,..., p n ) = Γ(p n)γ(p + + p n + ) Γ(p + + p n + ) Γ(p )... Γ(p n ) Γ(p + + p n + ) = = Γ(p )... Γ(p n ) Γ(p + + p n + ). A seemingly more general formula, x,...,x n>, x γ + +xγn n < x p... x pn n dx... dx n = γ... γ n Γ( p γ )... Γ( pn γ n ) Γ( p γ + + pn γ n + ), results from 9f by the (nonlinear!) change of variables y j = x γ j j. A special case: p = = p n =, γ = = γ n = p; x,...,x n> x p + +xp n< dx... dx n = Γ n ( p ) p n Γ( n p + ). We ve found the volume of the unit ball in the metric l p : v(b p ()) = 2n Γ n ( p ) p n Γ( n p + ). If p = 2, the formula gives us (again; see (9c7)) the volume of the standard unit ball: V n = v(b 2 ()) = 2πn/2 nγ( n 2 ). We also see that the volume of the unit ball in the l -metric equals 2n n!. Question: what does the formula give in the p limit? 9f3 Exercise. Show that x + +x n< x,...,x n> ϕ(x + + x n ) dx... dx n = (n )! ϕ(s)s n ds
Tel Aviv University, 26 Analysis-III 22 for every good function ϕ [, ] R and, more generally, x + +x n< x,...,x n> Hint: consider 9g We define ϕ(x + + x n )x p... x pn n dx... dx n = ds ϕ (s) x + +x n<s x,...,x n> = Γ(p )... Γ(p n ) Γ(p + + p n ) ϕ(u)u p +...p n du. x p... x pn n dx... dx n. Non-positive (signed) integrands G (g h) = G g G h whenever g, h G [, ) are continuous almost everywhere and G g <, G h < ; this definition is correct, that is, due to 9b9: G g G h = G g 2 G h 2 whenever g h = g 2 h 2, g h = g 2 h 2 g + h 2 = g 2 + h G (g + h 2 ) = G (g 2 + h ) G g + G h 2 = G g 2 + G h G g G h = G g 2 G h 2. 9g Lemma. The following two conditions on a function f G R continuous almost everywhere are equivalent: (a) there exist g, h G [, ), continuous almost everywhere, such that G g <, G h < and f = g h; (b) G f <. Proof. (a) (b): G g h G ( g + h ) = G g + G h <. (b) (a): we introduce the positive part f + and the negative part f of f, (9g2) f + (x) = max(, f(x)), f (x) = max(, f(x)) ; f = ( f) + ; f = f + f ; f = f + + f ; they are continuous almost everywhere (thin, why); G f + G f <, G f G f < ; and f + f = f.
Tel Aviv University, 26 Analysis-III 23 We summarize: (9g3) G f = G f + G f whenever f G R is continuous almost everywhere and such that G f <. Such functions will be called improperly integrable (on G). 9g4 Exercise. Prove linearity: G cf = c G f for c R, and G (f + f 2 ) = G f + G f 2. Similarly to Sect. 4e, a function f G R continuous almost everywhere will be called negligible if G f =. Functions f, g continuous almost everywhere and such that f g is negligible will be called equivalent. The equivalence class of f will be denoted [f]. Improperly integrable functions f G R are a vector space. On this space, the functional f G f is a seminorm. The corresponding equivalence classes are a normed space (therefore also a metric space). The integral is a continuous linear functional on this space. If G is admissible, then the space of improperly integrable functions on G is embedded into the space of improperly integrable functions on R n by f f l G. 9g5 Proposition (exhaustion). For open sets G, G, G 2, R n, G G G f G f R for all improperly integrable f G R. 9g6 Theorem (change of variables). Let U, V R n be open sets, ϕ U V a diffeomorphism, and f V R. Then (a) (f is continuous almost everywhere on V ) (f ϕ is continuous almost everywhere on U) ((f ϕ) det Dϕ is continuous almost everywhere on U); (b) if they are continuous almost everywhere, then V f = U (f ϕ) det Dϕ [, ] ; (c) and if the integrals in (b) are finite, then V f = U (f ϕ) det Dϕ R. In one dimension they are usually called absolutely (improperly) integrable.
Tel Aviv University, 26 Analysis-III 24 9g7 Exercise. Prove 9g5 and 9g6. 9g8 Exercise. If < t < t <, then the function (x, t) x t e x ln x is improperly integrable on (, ) (t, t ), and Prove it. t t dt dx x t e x ln x = Γ(t ) Γ(t ). 9g9 Exercise. (a) The function t x t e x ln x dx is continuous on (, ); (b) the gamma function is continuously differentiable on (, ), and Γ (t) = x t e x ln x dx for < t < ; (c) the gamma function is convex on (, ). Prove it. Index beta function, 5 change of variables, 6, 23 equivalent, 23 exhaustion, 3, 23 gamma function, 4 improper integral signed, 22, 23 unsigned, improperly integrable, 23 iterated improper integral, 9 linearity, 23 lower integral, 7 monotone convergence for integral, 8 for volume, 3 negligible, 23 Poisson formula, volume of ball, 4, 2 B, 5 [f], 23 f l G, f +, f, 22 Γ, 4 Hint: apply 9e6 twice, to f + and f.