Tel Aviv University, 214/15 Analysis-III,IV 161 1 Improper integral 1a What is the problem................ 161 1b Positive integrands................. 162 1c Newton potential.................. 166 1d Special functions gamma and beta........ 169 1e Normed space of equivalence classes....... 173 1f Change of variables................. 177 1g Multidimensional beta integrals of Dirichlet.. 179 Riemann integral and Jordan measure are generalized to unbounded functions and sets. 1a What is the problem The n-dimensional unit ball in the l p metric, E = {(x 1,..., x n ) : x 1 p + + x n p 1}, is a Jordan measurable set, and its volume is a Riemann integral, v(e) = 1l E, R n of a bounded function with bounded support. In Sect. 1g we ll calculate it: v(e) = 2n Γ n( ) 1 p p n Γ ( n + 1) p where Γ is a function defined by Γ(t) = x t 1 e x dx for t > ; here the integrand has no bounded support; and for t = 1 < 1 it is also unbounded (near ). Thus we need a more general, so-called improper integral, p 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, k x t 1 e x dx = lim x t 1 e x dx. k 1/k
Tel Aviv University, 214/15 Analysis-III,IV 162 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. 1 1b Positive integrands We consider an open set R n and functions f : [, ) continuous almost everywhere. We do not assume that is bounded. We also do not assume that is Jordan measurable, even if it is bounded. 2 Continuous almost everywhere means that the set A of all discontinuity points of f satisfies m (A) =, recall Sect. 8f; but now A need not be bounded. For our purposes it is enough to know that m (A) = if and only if m (A 1 ) = for every bounded A 1 A (we may take this as the definition). We can use the function f 1l equal f on and on R n \, but must be careful: 1l and f 1l need not be continuous almost everywhere. We define { (1b1) f = sup g g : Rn R integrable, R n } g f on, g = on R n \ [, ]. The condition on g may be reformulated as g f 1l. 1b2 Exercise. (a) Without changing this supremum we may restrict ourselves to continuous g with bounded support; or, alternatively, to step functions g; (b) if f is bounded and is bounded, then f = f 1l R n, and in particular, 1 = v (); 3 (c) if f is bounded and is Jordan measurable, then the integral defined by (1b1) is equal to the integral defined by (6g16). Prove it. 1 Additional literature (for especially interested): M. Pascu (26) On the definition of multidimensional generalized Riemann integral, Bul. Univ. Petrol LVIII:2, 9 16. (Research level) D. Maharam (1988) Jordan fields and improper integrals, J. Math. Anal. Appl. 133, 163 194. 2 A bounded open set need not be Jordan measurable, even if it is diffeomorphic to a disk, as was noted in Sect. 8e (p. 138). 3 According to 8e, v () = e v() = m().
Tel Aviv University, 214/15 Analysis-III,IV 163 1b3 Exercise. Consider the case = R n, and let be a norm on R n. (a) Prove that f = lim min R n k x <k ( f(x), k ) dx. (b) For a locally bounded 1 f prove that f = lim f(x) dx. R n k x <k (c) Can it happen that f is locally bounded, not bounded, and R n f <? 1b4 Example (Poisson). Consider I = R 2 e x 2 dx. On one hand, by 1b3 for the Euclidean norm, I = lim k k x 2 +y 2 <k 2 e (x2+y2) dxdy = lim k r dr e r2 2π dθ = lim k π On the other hand, by 1b3 for (x, y) = max( x, y ), I = lim k x <k, y <k e (x2 +y 2) dxdy = lim k ( k k )( k e x2 dx and we obtain the celebrated Poisson formula: + e x2 dx = π. k k2 e u du = π. ) ( + e y2 dy = e x2 dx) 2, 1b5 Exercise. Consider I = x a y b e (x2 +y 2) dxdy [, ] x>,y> 1 That is, bounded on every bounded subset of R n.
Tel Aviv University, 214/15 Analysis-III,IV 164 for given a, b R. Prove that, on one hand, ( I = and on the other hand, ( I = )( π/2 ) r a+b+1 e r2 dr cos a θ sin b θ dθ, )( ) x a e x2 dx x b e x2 dx. 1b6 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 f = 2π g(r) r dr. R 2 (b) If g is continuous on (, ), then f = 2π g(r) r dr [, ]. R 2 Prove it. 1 1b7 Exercise. Consider f : R n [, ) of the form f(x) = g ( x ) for a given g : [, ) [, ) and a given norm on R n. (a) If g is integrable then f is integrable, and R n f = nv g(r) r n 1 dr where V is the volume of {x : x < 1}. (b) If g is continuous on (, ), then R n f = nv g(r) r n 1 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. 2 1b8 Example. e R x 2 dx = nv r n 1 e r2 dr; in particular, n e R x 2 dx = nv n n r n 1 e r2 dr where V n is the volume of the (usual) n-dimensional unit ball. On the other hand, e x 2 dx = ( R n R e x2 dx )n = π n/2. Therefore Not unexpectedly, V 2 = V n = π n/2 n r n 1 e r2 dr. π 2 = π. re r2 dr Clearly, cf = c f for c (, ). 1b9 Proposition. (f 1 + f 2 ) = f 1 + f 2 [, ] for all f 1, f 2 on, continuous almost everywhere. 1 Hint: (a) either polar coordinates, or 9g4; (b) use (a). 2 Hint: (a), (b) similar to 9g4, using also 9c3 and 6g12; (c) use (b).
Tel Aviv University, 214/15 Analysis-III,IV 165 Proof. First we prove that (f 1 + f 2 ) f 1 + f 2. 1 iven integrable g 1, g 2 such that g 1 f 1 1l and g 2 f 2 1l, we have g 1 + g 2 = (g1 + g 2 ) (f 1 + f 2 ), since g 1 + g 2 is integrable and g 1 + g 2 (f 1 + f 2 ) 1l. The supremum in g 1, g 2 gives the claim. It remains to prove that (f 1 + f 2 ) f 1 + f 2, that is, g f 1 + f 2 for every integrable g such that g (f 1 + f 2 ) 1l. We introduce g 1 = min(f 1, g), g 2 = min(f 2, g) (pointwise minimum on ; and on R n \ ) and prove that they are continuous almost everywhere (on R n, not just on ). For almost every x, both f 1 and g are continuous at x and therefore g 1 is continuous at x. For almost every x, g is continuous at x, which ensures continuity of g 1 at x (irrespective of continuity of f 1 ), since g(x) = (x / ). Thus, g 1 is continuous almost everywhere; the same holds for g 2. By Theorem 8f1, the functions g 1, g 2 are integrable. We have g 1 + g 2 min(f 1 + f 2, g) = g, since generally, min(a, c) + min(b, c) min(a + b, c) for all a, b, c [, ) (think, why). Thus, g (g 1 + g 2 ) = g 1 + g 2 f 1 + f 2, since g 1 f 1 1l, g 2 f 2 1l. 1b1 Proposition (exhaustion). For open sets, 1, 2, R n, k = f f [, ] k for all f : [, ) continuous almost everywhere. Proof. First of all, k f k+1 f (since g f 1l k implies g f 1l k+1 ), and similarly, k f f, thus k f and lim k k f f. We have to prove that f lim k k f. Let a step function g : R n R satisfy g f 1l ; we have to prove that g lim k k f, but we ll prove that moreover, g lim k k g. By linearity, WLO, g = 1l C for a box C, C. By 1b2, k g = 1l R n C k = v (C k ) and g = v(c); by 8e9, v (C k ) v (C ) = v(c ). 1b11 Exercise. Let 1 2 R n be two open sets, and f : 2 [, ) continuous almost everywhere. If f = on 2 \ 1, then 2 f = 1 f. Prove it. 2 1b12 Exercise. The following four conditions on a function f : [, ) continuous almost everywhere are equivalent: 1 Compare it with (6d1). 2 Hint: just (1b1).
Tel Aviv University, 214/15 Analysis-III,IV 166 (a) f = ; (b) f(x) = for every continuity point x of f; (c) f(x) = for almost all x ; (d) the set {x : f(x) = } is dense in. Prove it. 1 1c Newton potential By the celebrated Newton s law of universal gravitation, the gravitational force exerted by a particle of mass m at point ξ on a particle of mass m at point x is m mg ξ (x), and mg ξ ( ) is the gravitational field generated by m, (1c1) g ξ (x) = g (x ξ) = x ξ x ξ 3 = U (x ξ) ; here the function U : x 1 is proportional to the gravitational potential x (energy), and is the gravitational constant. 2 The reason to replace the force by the potential is simple: it is easier to work with scalar functions than with the vector ones. 3 What happens if we have a system of point masses µ 1,..., µ k at points ξ 1,..., ξ k? The forces are to be added, and the corresponding potential is U(x) = k j=1 µ j x ξ j. A continuously distributed mass is described in physics by its density ρ. Mathematically it means that the density is a point function, the mass is an additive box function, and these two functions are related according to Sect. 6a (and 8c): the mass within a box B is ρ. enerally, ρ is not quite B integrable but improperly integrable; and still, the mass within a box B is assumed to be ρ (improper integral) for evident physical reasons; and the B total mass is ρ. R 3 1 Hint: (b)= (c)= (d): easy; (d)= (a): use 1b2(a); (a)= (b): otherwise f( ) ε on some neighborhood of x. 2 6.674 1 11 N(m/kg) 2 ; that is, if m = µ = 1 kg and x ξ = 1 m then the force is 6.674 1 11 newtons. 3 Knowing the force F one can write down the differential equations of motion of the particle (Newton s second law) m ẍ = F, or ẍ = U (note that m does not matter). Then one hopes to integrate these equations, thus finding out where is the particle at time t.
Tel Aviv University, 214/15 Analysis-III,IV 167 ρ(ξ) x ξ dξ; Similarly, the potential is assumed to be U ρ where U ρ (x) = R 3 this integral is improper (in general) and must be finite. 1 Let us compute the potential of the homogeneous mass distribution, of density 1, within the ball of radius R centered at the origin: U R (x) = ξ <R dξ x ξ. Due to rotation invariance (Theorem 9c1), U R is a radial function, that is, depends only on x. Thus, it suffices to compute U R (x) at the point x = (,, a), a [, ). The integral is proper for a (R, ) and improper for a [, R]. First, consider the proper integral, for a > R. Using the spherical coordinates ξ = (r cos ϕ sin θ, r sin ϕ sin θ, r cos θ) (recall 9b3) we have U R (x) = R π dr 2π r 2 sin θ dθ (a r cos θ)2 + r 2 sin 2 θ = = R π dr 2π r 2 sin θ dθ a2 2ar cos θ + r 2 } {{ } V a(r) Intuitively, the under-braced expression V a (r) is the potential of the homogeneous sphere of radius r; but rigorously, integration over spheres and other surfaces will be treated much later. We compute V a (r) using the variable t = a 2 2ar cos θ + r 2. Then a r < t < a + r, and t dt = ar sin θ dθ. We get V a (r) = 2πr 2 a+r a r t dt art = 2πr a r2 2r = 4π a. 1 Mathematical rigorosity is of little interest to physicists, and still, the distinction between proper and improper integrals may be physically sound. Imagine a material ball of mass M and radius R, consisting of a large number of uniformy distributed particles that are balls of mass m and radius r. Outside the (large) ball, near its surface, the gravitational field is M/R 2 in a good approximation. Inside the ball, near the surface of a particle, the gravitational field of this single particle is m/r 2. Let M = 1 kg, R =.1 m, m = 1 25 kg, r = 1 14 m; then M/R 2 = 1 kg/m 2 while m/r 2 = 1 kg/m 2. Here, a single particle generates a field 1 times stronger than the improper integral that will be calculated! Do not think that such parameters are physically unrealistic; these m, r are the parameters of a typical atomic nucleus..
Tel Aviv University, 214/15 Analysis-III,IV 168 Now we easily find U R (x) by integration: U R (x) = R R r 2 4πR3 V a (r) dr = 4π dr = a 3a = 4πR3 3 x for x > R. We turn to the case a < R, and treat the improper integral by exhaustion: ( U R (x) = lim ε + = lim ε + ξ <a ε dξ x ξ + ( a ε V a (r) dr + ) dξ = x ξ ) R V a (r) dr = V a (r) dr [, ], a+ε< ξ <R R a+ε the latter integral being improper, since V a need not be bounded near a. For r < a we have V a (r) = 4π r2 as before. For r > a we still use t = a a2 2ar cos θ + r 2, and t is still strictly increasing in θ (, π), but now a2 2ar + r 2 = r a, thus r a < t < r + a, and we get V a (r) = 2πr 2 r+a r a t dt art = 2πr 2a = 4πr. a A surprise: V a appears to be bounded near a, and extends by continuity to (, R), thus the one-dimensional integral may be treated as proper. We have a R R U R (x) = V a (r) dr = 4π r2 a dr + 4πr dr = a ( ) a 2 = 4π 3 + R2 2 a2 = 2π 2 3 (3R2 a 2 ) = 2π 3 (3R2 x 2 ) for x < R. The case a = R is easy: U R (x) = R V a(r) dr = R 4π r2 R3 dr = 4π a 3a for x = R. The function U R appears to be continuous. Finally, = 4πR2 3 U R (x) = { 4πR 3 3 x for x R, 2π 3 (3R2 x 2 ) for x R. Observe that 4πR 3 /3 is exactly the total mass of the ball. That is, together with Newton, we arrived at the conclusion that the gravitational potential, and hence the gravitational force exerted by the homogeneous ball on a particle is the same as if the whole mass of the ball were concentrated at its center, as long as the point is outside the ball. Of course, you heard about this already in the high-school.
Tel Aviv University, 214/15 Analysis-III,IV 169 Another important conclusion is that the potential of the homogeneous sphere does not depend on the point inside the sphere! 1 Hence, the gravitational force is zero inside the sphere. The same is true for the homogeneous shell {ξ : a < ξ < b}: there is no gravitational force inside the shell. 1c2 Exercise. Check that all the conclusions are true when the mass distribution ρ is radial: ρ(ξ) = ρ(ξ ) whenever ξ = ξ. 1c3 Exercise. Find the potential of the homogeneous solid ellipsoid (x 2 + y 2 )/b 2 + z 2 /c 2 < 1 at its center. 1c4 Exercise. Find the potential of the homogeneous solid cone of height h and radius of the base r at its vertex. 1c5 Problem. Show that at sufficiently large distances the potential of a solid is approximated by the potential of a point with the same total mass located at the center of mass of the solid with an error less than a constant divided by the square of the distance. The potential itself decays as the distance, so the approximation is good: its relative error is small. 2 1d Special functions gamma and beta Integrating a function of two variables in one variable we get a function of the other variable. An interesting example was seen in 7e2: the function F (t) = π/2 ln(t 2 sin 2 x) dx appeared to be the elementary function F (t) = π ln t+ t 2 1. But generally it is not elementary. Here is a much more 2 important example. The Euler gamma function Γ is defined by 3 (1d1) Γ(t) = x t 1 e x dx for t (, ). This integral is not proper for two reasons. First, the integrand is bounded near for t [1, ) but unbounded for t (, 1). Second, the integrand has no bounded support. In every case, using 1b1, k Γ(t) = lim x t 1 e x dx <, k 1/k since the integrand (for a given t) is continuous on (, ), is O(x t 1 ) as x, and (say) O(e x/2 ) as x. Thus, Γ : (, ) (, ). 1 Since V a (r) does not depend on a for a < r. 2 This estimate is rather straightforward. A more accurate argument shows that the error is of order constant divided by the cube of the distance. 3 This is rather Γ (, ).
Tel Aviv University, 214/15 Analysis-III,IV 17 Clearly, Γ(1) = 1. Integration by parts gives (1d2) k 1/k x t e x dx = x t e x k x=1/k + t k 1/k Γ(t + 1) = tγ(t) for t (, ). x t 1 e x dx ; In particular, (1d3) Γ(n + 1) = n! for n =, 1, 2,... We note that (1d4) x a e x2 dx = 1 ( a + 1 ) 2 Γ 2 for a ( 1, ), since x a e x2 dx = u a/2 e 1b4) gives (1d5) Thus, (1d6) ( 2n + 1 ) Γ 2 u du 2 u. For a = the Poisson formula (recall ( 1 Γ = 2) π. = 1 2 3 2 2n 1 π. 2 The volume V n of the n-dimensional unit ball (recall 1b8) is thus calculated: (1d7) V n = πn/2 n 2 Γ( n 2 ). Not unexpectedly, V 3 = π3/2 3 ) π/2 1 By 1b5, ( 1, ); that is, (1d8) In particular, (1d9) Γ( a+b+2 2 2 π/2 = π3/2 2 Γ( 3 2 ) 3 2 1 2 π = 4 3 π. cos a θ sin b θ dθ = 1Γ( a+1 2 2 ) 1 Γ( ) b+1 2 2 for a, b cos α 1 θ sin β 1 θ dθ = 1 Γ( α)γ( β ) 2 2 2 Γ( α+β ) for α, β (, ). 2 π/2 sin α 1 θ dθ = π/2 cos α 1 θ dθ = π 2 Γ ( ) α 2 Γ ( α+1 2 ).
Tel Aviv University, 214/15 Analysis-III,IV 171 π/2 The trigonometric functions can be eliminated: cos α 1 θ sin β 1 θ dθ = 1 π/2 cos α 2 θ sin β 2 θ 2 sin θ cos θ dθ = 1 1 α 2 (1 u) 2 u β 2 2 du; thus, 2 2 (1d1) where (1d11) 1 x α 1 (1 x) β 1 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. 1d12 Exercise. Check that B(x, x) = 2 1 2x B(x, 1). 2 Hint: π/2( 2 sin θ cos θ ) 2x 1 dθ. 2 1d13 Exercise. Check the duplication formula: ( Γ(2x) = 22x 1 Γ(x) Γ x + 1 ). π 2 Hint: use 1d12. 1d14 Exercise. Calculate 1 x4 1 x 2 dx. π Answer:. 32 1d15 Exercise. Calculate x m e xn dx. 1 Answer: Γ( ) m+1 n n. 1d16 Exercise. Calculate 1 xm (ln x) n dx. Answer: ( 1) n n! (m+1) n+1. 1d17 Exercise. Calculate π/2 Answer: Γ2 (1/4) 2 2π. dx cos x. 1d18 Exercise. Check that Γ(p)Γ(1 p) = Hint: change x to t via (1 + x)(1 t) = 1. We mention without proof another useful formula x p 1 1 + x dx = π sin πp x p 1 1+x dx. for < p < 1. There is a simple proof that that uses the residues theorem from the complex analysis course. This formula yields that Γ(t)Γ(1 t) = π. sin πt
Tel Aviv University, 214/15 Analysis-III,IV 172 Is the function Γ continuous? For every compact interval [t, t 1 ] (, ) the given function of two variables (t, x) x t 1 e x is Lipschitz continuous on [t, t 1 ] [ 1, k], therefore k the integral is Lipschitz continuous on [t, t 1 ] (recall 7b). Also, k 1/k x t 1 e x dx Γ(t) uniformly on [t, t 1 ], since 1/k x t 1 e x dx 1/k x t 1 dx as k and x t 1 e x dx k x t1 1 e x dx as k. It follows that Γ is continuous on arbitrary k [t, t 1 ], therefore, on the whole (, ). In particular, tγ(t) = Γ(t + 1) Γ(1) = 1 as t +; that is, Γ(t) = 1 ( 1 ) t + o as t +. t Is the function Γ differentiable? By Theorem 7e1 the function t k 1/k xt 1 e x dx is continuously differentiable, and its derivative is t k 1/k xt 1 e x ln x dx; this relation results from application of Prop. 7b4 (iterated integral) to the function (t, x) t xt 1 e x = x t 1 e x ln x on [t, t 1 ] [ 1, k]. Regretfully, iterated improper k integral is not an easy matter. 1 Instead, we use exhaustion, as follows. As before, k x t 1 e x ln x dx x t 1 e x ln x dx uniformly on [t, t 1 ] 1/k (check it), therefore t1 k t1 dt x t 1 e x ln x dx dt t 1/k t On the other hand, k 1/k t1 dx dt x t 1 e x ln x = t = k 1/k k 1/k x t 1 1 e x dx ( x t 1 e x ) t 1 t=t dx = k 1/k x t 1 e x ln x dx. x t 1 e x dx Γ(t 1 ) Γ(t ). 1 If f : [, 1] [, 1] [, ) is improperly integrable, then f x : y f(x, y) is improperly integrable on [, 1] for almost every x; however, the function ϕ : x f x need not be improperly integrable. Rather, ϕ is equivalent to a function semicontinuous from below (possibly, unbounded on every interval), and ϕ = f.
Tel Aviv University, 214/15 Analysis-III,IV 173 Thus, Γ(t 1 ) Γ(t ) = t 1 t dt x t 1 e x ln x dx, which implies Γ (t) = x t 1 e x ln x dx. Similarly, Γ is differentiable; continuing this way we get 1e Γ (k) (t) = x t 1 e x (ln x) k dx for k = 1, 2,... Normed space of equivalence classes In order to integrate signed functions we reuse the simple trick of (8e1). We define (g h) = g h whenever g, h : [, ) are continuous almost everywhere and g <, h < ; this definition is correct, that is, g 1 h 1 = g 2 h 2 whenever g 1 h 1 = g 2 h 2 ; proof: (1e1) g 1 h 1 = g 2 h 2 = g 1 +h 2 = g 2 +h 1 = (g 1 +h 2 ) = (g 2 +h 1 ) = = g 1 + h 2 = g 2 + h 1 = g 1 h 1 = g 2 h 2. 1e2 Lemma. The following two conditions on a function f : R continuous almost everywhere are equivalent: (a) there exist g, h : [, ), continuous almost everywhere, such that g <, h < and f = g h; (b) f <. Proof. (a)= (b): g h ( ) g + h = g + h <. (b)= (a): we introduce the positive part f + and the negative part f of f, (1e3) f + (x) = max (, f(x) ), f (x) = max (, f(x) ) ; f = ( f) + ; f = f + f ; f = f + + f ; they are continuous almost everywhere (think, why); f + f <, f f < ; and f + f = f.
Tel Aviv University, 214/15 Analysis-III,IV 174 (1e4) We summarize: f = f + f whenever f : R is continuous almost everywhere and such that f <. Such functions will be called improperly integrable 1 (on ). 1e5 Exercise. Prove linearity: cf = c f for c R, and (f 1 + f 2 ) = f 1 + f 2. Similarly to Sect. 6e, a function f : R continuous almost everywhere will be called negligible if 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 : R are a vector space. On this space, the functional f f is a seminorm. The corresponding equivalence classes are a normed space (therefore also a metric space). Similarly to 6e3, the integral is a continuous linear functional on this space. If is Jordan measurable then the space of improperly integrable functions on is embedded into the space of improperly integrable functions on R n by f f 1l. 1e6 Lemma. Let 1 2 R n be two open sets, and f : 2 R continuous almost everywhere. If f = almost everywhere on 2 \ 1, then f 1l 1 is continuous almost everywhere on 2 and equivalent to f. Proof. The set A = {x 2 \ 1 : f(x) } is of Lebesgue measure. If f is continuous at x 2 while f 1l 1 is not, then clearly ( x 2 \ 1 ; and moreover, x A (since lim t x f(t) = implies lim t x f(t) 1l1 (t) ) = ). Thus, f 1l 1 is continuous almost everywhere on 2. Finally, f 1l 1 = f on 2 \ A. In particular, if 1 contains almost all points of 2 (that is, 2 \ 1 is of Lebesgue measure ), 2 then the condition f = almost everywhere on 2 \ 1 holds vacuously; in this case the values of f on 2 \ 1 do not influence the equivalence class of f. 1e7 Corollary. Let 1 2 R n be two open sets, and f : 2 R improperly integrable. If f = almost everywhere on 2 \ 1, then 2 f = 1 f. 1 In one dimension they are usually called absolutely (improperly) integrable. 2 Warning: this condition implies v ( 1 ) = v ( 2 ) and is implied by v ( 1 ) = v ( 2 ) <, but is not implied by v ( 1 ) = v ( 2 ) =.
Tel Aviv University, 214/15 Analysis-III,IV 175 Proof. First, 2 f = 2 f 1l 1 since [f] = [f 1l 1 ] by 1e6. Second, 2 f + 1l 1 = 1 f + by 1b11; the same holds for f, and therefore for f + f = f. Once again, if 1 contains almost all points of 2, then we get 2 f = 1 f for all f improperly integrable on 2. We may admit a function f partially defined on, provided that for almost every x, f is defined near x. 1 2 In other words: f : \ A R, and the (relative) closure of A in is of Lebesgue measure. In this case almost all points of belong to ( \ A). Such partially defined functions may be used as well as functions defined on the whole, whenever only equivalence classes matter. Thus, we need not hesitate saying that, for instance, 1 dx = 2 for 1 x α 1 α α < 1, even though the integrand is undefined at. 1e8 Proposition (Exhaustion). Let open sets 1 2 R n be such that k k contains almost all points of. Then f f as k k for all f improperly integrable on. Proof. First, the open set = k k contains almost all points of, therefore f = f. Second, k ; 1b1 gives k f = k f + k f f + f = f. In particular, if k are also Jordan measurable and such that f is defined and bounded on each k, then k f is the proper (Riemann) integral, and we obtain the improper integral f as the limit of proper integrals. 1e9 Proposition. Let R n be an open set, and f an improperly integrable function on. 3 Then there exist Jordan measurable open sets 1 2... such that k, k k contains almost all points of, and f is defined and bounded on every k. 1 Not just at x! 2 In fact, for every set A of Lebesgue measure (even if dense in ), every function f : \ A R continuous almost everywhere can be extended to a function R continuous almost everywhere (and all such extentions evidently are mutually eqiuvalent). Hint: lim inf t x,t \A f(t) f(x) lim sup t x,t \A f(t) for every x A such that f is bounded near x. Such f is continuous at every continuity point of f. 3 We admit partially defined f, as explained above.
Tel Aviv University, 214/15 Analysis-III,IV 176 Proof. We ll prove that for every box B R n and every ε > there exists a Jordan measurable open set B,ε B such that f is defined and bounded on B,ε and v( B,ε ) v ( B) ε. This is sufficient, since we may take B k R n and ε k, and then the sets k = B1,ε 1 Bk,ε k fit. We take a set A of Lebesgue measure such that for each x \ A, f is defined near x and continuous at x. 1 iven B and ε, we take an open U such that A B U and v (U B ) ε/2. We also take a compact set K B such that v (K) v (B ) ε/2. Then the compact set K \ U satisfies v (K \ U) v (K) v (U B ) v (B ) ε, and every point of K \ U has a Jordan measurable open neighborhood (just a ball, or a box) on which f is defined and bounded. We choose a finite subcovering; the union of the chosen neighborhoods (intersected with B ) is B,ε. The normed space of equivalence classes, introduced above, does not admit an inner product. 2 Now we turn to improperly square integrable functions; these are functions f : R continuous almost everywhere and such that f 2 <. If [f] = [g] then f 2 = g 2 (check it via 1b12), thus, square integrability applies to equivalence classes. We denote the set of all square integrable equivalence classes by L 2 (), 3 and often write f L 2 () instead of [f] L 2 (). This set is a vector space (since (f + g) 2 (f + g) 2 + (f g) 2 = 2f 2 + 2g 2 ). If f, g L 2 () then their pointwise product fg is improperly integrable (since f 2 2 fg + g 2 ), and we define the inner product (1e1) [f], [g] = fg and the corresponding norm (1e11) [f] 2 = [f], [f], that is, f 2 = f 2 satisfying [f] [] = [f] 2 > (check it via 1b12). We often write f, g and f 2 instead of [f], [g] and [f] 2. Every 2-dimensional subspace of L 2 () is a Euclidean plane, which ensures the triangle inequality (1e12) f + g 2 f 2 + g 2 1 Not continuous near x! 2 Its 2-dimensional subspace of step functions is not the Euclidean plane; you may check it similarly to the paragraph before 1e3. 3 The widely used notation L 2 is reserved for the corresponding notion in the framework of Lebesgue integration.
Tel Aviv University, 214/15 Analysis-III,IV 177 and the Cauchy-Schwarz inequality (1e13) f 2 g 2 f, g f 2 g 2. More generally, for arbitrary p [1, ) we introduce the norm ( ) 1/p f p = f p on the vector space L p () of [f] such that f p < (two special cases p = 1 and p = 2 being already treated). The triangle inequality f + g p f p + g p follows from convexity of the ball {f : f p 1} (recall 1e14); convexity of the ball follows from convexity of the functional f f p (recall 1e13); and convexity of this functional follows from convexity of the function t t p (similarly to 1e15). The triangle inequality ensures that L p () is a vector space. The Hölder inequality fg f p g q for f L p (), g L q (), 1 p + 1 q = 1, is obtained similarly to 6d15(b) (but harder); first, ab ap + bq for a, b p q [, ); second, ( fg 1 min c> p cf p p + 1 1 ) q c g q = f p g q. q 1f Change of variables 1f1 Theorem. Let U, V R n be open sets, ϕ : U V a diffeomorphism, and f : V R. Then (a) f is improperly integrable on V if and only if (f ϕ) det Dϕ is improperly integrable on U; and (b) in this case f = (f ϕ) det Dϕ. V U Proof. We reuse the arguments from the proof of Theorem 9a1. There, U and V are assumed to be Jordan measurable, but this assumption is used only in the last paragraph of the proof. Before that we constructed Jordan measurable open sets V k V (denoted there by K i ) 1 such that V k V, 1 But notations U, V are swapped there; compare 9a1 and 9a2.
Tel Aviv University, 214/15 Analysis-III,IV 178 the sets ϕ 1 (V k ) = U k U are Jordan measurable, U k U, and we showed that the claim of the theorem (for Riemann integral) holds for every f whose support is contained in some V k, therefore, for every f with a compact support inside V. Now, given f : V R, we note that f is improperly integrable on V if and only if it is improperly integrable on each V k, and lim k V k f <, and in this case f = lim V k V k f. Similarly, (f ϕ) det Dϕ is improperly integrable on U if and only if it is improperly integrable on each U k, and lim k U k (f ϕ) det Dϕ <, and in this case (f ϕ) det Dϕ = lim U k U k (f ϕ) det Dϕ. Thus, in order to prove the theorem for arbitrary f it is sufficient to prove it for f whose support is contained in some V k. Theorem 9a1, applied to the diffeomorphism ϕ Uk : U k V k, gives the needed claim for proper integration, that is, for bounded f (boundedness of (f ϕ) det Dϕ follows, since the determinant is bounded on U k ). It remains to generalize this claim to unbounded f : V k R. Taking into account that f = f + f we may assume that f : V k [, ). We note that f is improperly integrable on V k if and only if each f l = min(f, l) is integrable on V k, and in this case V k f = lim l V k f l (since every integrable g such that g f 1l Vk satisfies g l for some l). Similarly, taking into account that det Dϕ is bounded away from on U k, we see that (f ϕ) det Dϕ is improperly integrable on U k if and only if each (f l ϕ) det Dϕ is improperly integrable on U k, and in this case U k (f ϕ) det Dϕ = lim l U k (f l ϕ) det Dϕ. The claim follows. 1f2 Exercise. Prove the equality (1d1) once again, avoiding (1b5) and triginometric functions; to this end, consider ( )( 1 ) u α+β 1 e u du x α 1 (1 x) β 1 dx
Tel Aviv University, 214/15 Analysis-III,IV 179 and change the variables u, x to t 1, t 2 as follows: { t 1 = ux { u = t 1 + t 2 t 2 = u(1 x) x = t 1 t 1 +t 2 1g Multidimensional beta integrals of Dirichlet 1g1 Proposition. x 1,...x n>, x 1 + +x n<1 for all p 1,... p n >. For the proof, we denote I(p 1,..., p n ) = x p 1 1 1... x pn 1 n dx 1... dx n = x 1,...x n>, x 1 + +x n<1 This integral is improper, unless p 1,..., p n 1. Γ(p 1 )... Γ(p n ) Γ(p 1 + + p n + 1) x p 1 1 1... x pn 1 n dx 1... dx n. 1g2 Lemma. I(p 1,..., p n ) = B(p n, p 1 + + p n 1 + 1)I(p 1,..., p n 1 ). Proof. We introduce proper integrals I ε (p 1,..., p n ) = x p 1 1 1... x pn 1 dx 1... dx n x 1,...x n>ε, x 1 + +x n<1 for ε >. 1 Clearly, I ε (p 1,..., p n ) I(p 1,..., p n ), and I ε (p 1,..., p n ) I(p 1,..., p n ) as ε +. The change of variables ξ = ax (that is, ξ 1 = ax 1,..., ξ n = ax n ) gives (by Theorem 1f1) ξ p 1 1 1... ξn pn 1 dξ 1... dξ n = a p 1+ +p n I ε (p 1,..., p n ) for a >. ξ 1,...x n>aε, x 1 + +x n<a 1 If nε 1 then I ε (p 1,..., p n ) =, of course. n
Tel Aviv University, 214/15 Analysis-III,IV 18 We use iterated integral (proper! ): I ε (p 1,..., p n ) = On one hand, 1 ε = dx n x pn 1 n 1 I ε (p 1,..., p n ) I(p 1,..., p n 1 ) ε x 1,...x n 1 >ε, x 1 + +x n 1 <1 x n x p 1 1 1... x p n 1 1 n 1 dx 1... dx n 1 = x pn 1 n (1 x n ) p 1+ +p n 1 I ε/(1 xn)(p 1,..., p n 1 ) dx n. 1 x pn 1 n (1 x n ) p 1+ +p n 1 dx n = = I(p 1,..., p n 1 )B(p n, p 1 + + p n 1 + 1) for all ε, therefore I(p 1,..., p n ) B(p n, p 1 + + p n 1 + 1)I(p 1,..., p n 1 ). On the other hand, for arbitrary δ >, I ε (p 1,..., p n ) for all ε, therefore 1 δ ε I(p 1,..., p n ) xn pn 1 (1 x n ) p 1+ +p n 1 I ε/(1 xn)(p 1,..., p n 1 ) dx n 1 δ ε 1 δ x pn 1 n (1 x n ) p 1+ +p n 1 I ε/δ (p 1,..., p n 1 ) dx n xn pn 1 (1 x n ) p 1+ +p n 1 I(p 1,..., p n 1 ) dx n for all δ, and finally, I(p 1,..., p n ) B(p n, p 1 + + p n 1 + 1)I(p 1,..., p n 1 ). Proof of Prop. 1g1. Induction in the dimension n. For n = 1 the formula is obvious: 1 From n 1 to n: using 1g2, x p 1 1 1 dx 1 = 1 p 1 = Γ(p 1) Γ(p 1 + 1). I(p 1,..., p n ) = Γ(p n)γ(p 1 + + p n 1 + 1) Γ(p 1 + + p n + 1) Γ(p 1 )... Γ(p n 1 ) Γ(p 1 + + p n 1 + 1) = = Γ(p 1 )... Γ(p n ) Γ(p 1 + + p n + 1).
Tel Aviv University, 214/15 Analysis-III,IV 181 There is a seemingly more general formula, x p 1 1 1... x pn 1 1 n dx 1... dx n = γ 1... γ n x 1,...,x n>, x γ 1 1 + +xγn n <1 Γ ( p 1 ) ( γ 1... Γ pn ) γ n Γ ( p 1 γ 1 + + pn γ n + 1 ), easily obtained from the previous one by the change of variables y j = x γ j j. A special case: p 1 = = p n = 1, γ 1 = = γ n = p; x 1,...,x n> x p 1 + +xp n<1 dx 1... dx n = Γn( 1 p ) p n Γ ( n p + 1). We ve found the volume of the unit ball in the metric l p : v ( B p (1) ) = 2n Γ n( ) 1 p p n Γ ( n + 1). p If p = 2, the formula gives us (again; see (1d7)) the volume of the standard unit ball: V n = v ( B 2 (1) ) = 2πn/2 nγ ( ). n 2 We also see that the volume of the unit ball in the l 1 -metric equals 2n n!. Question: what does the formula give in the p limit? 1g3 Exercise. Show that ϕ(x 1 + + x n ) dx 1... dx n = x 1 + +x n<1 x 1,...,x n> 1 (n 1)! 1 for every good function ϕ : [, 1] R and, more generally, ϕ(x 1 + + x n )x p 1 1 1... x pn 1 n dx 1... dx n = x 1 + +x n<1 x 1,...,x n> Hint: consider 1 ds ϕ (s) x 1 + +x n<s x 1,...,x n> = Γ(p 1)... Γ(p n ) Γ(p 1 + + p n ) 1 x p 1 1 1... x pn 1 n dx 1... dx n. ϕ(s)s n 1 ds ϕ(u)u p 1+...p n 1 du.
Tel Aviv University, 214/15 Analysis-III,IV 182 Index almost all points, 174 beta function, 171 Cauchy-Schwarz inequality, 177 continuous almost everywhere, 162 equivalent, 174 exhaustion, 165, 175 gamma function, 169 gravitational constant, 166 Hölder inequality, 177 improper integral signed, 173, 174 unsigned, 162 improperly integrable, 174 inner product, 176 Lebesgue measure zero, 162 linearity, 174 negligible, 174 Newton s law, 166 partially defined, 175 Poisson formula, 163 square integrable, 176 triangle inequality, 176, 177 volume of ball, 17, 181 B, 171 [f], 174 f 1l, 162 f +, f, 173, 162, 166 Γ, 169 L 2 (), 176 L p (), 177