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Suppose that g is monotonic.Then a sufficient condition for convergence of the improper integral ω(f · g)(x) dx(6.79)ais that the one of the following pairs of conditions hold:ωα1 ) the integral a f (x) dx converges,β1 ) the function g bounded on [a, ω[,orbα2 ) the function F(b) = a f (x) dx is bounded on [a, ω[,β2 ) the function g(x) tends to zero as x → ω, x ∈ [a, ω[.4046IntegrationProof For any b1 and b2 in [a, ω[ we have, by the second mean-value theorem,b2ξ(f · g)(x) dx = g(b1 )b1f (x) dx + g(b2 )b2f (x) dx,ξb1where ξ is a point lying between b1 and b2 . Hence by the Cauchy convergencecriterion (Proposition 2), we conclude that the integral (6.79) does indeed convergeif either of the two pairs of conditions holds.6.5.3 Improper Integrals with More than One SingularityUp to now we have spoken only of improper integrals with one singularity causedeither by the unboundedness of the function at one of the endpoints of the intervalof integration or by an infinite limit of integration.
In this subsection we shall showin what sense other possible variants of an improper integral can be taken.If both limits of integration are singularities of either of these two types, then bydefinition ω2 c ω2f (x) dx :=f (x) dx +f (x) dx,(6.80)ω1ω1cwhere c is an arbitrary point of the open interval ]ω1 , ω2 [.It is assumed here that each of the improper integrals on the right-hand side of(6.80) converges.
Otherwise we say that the integral on the left-hand side of (6.80)diverges.By Remark 2 and the additive property of the improper integral, the definition(6.80) is unambiguous in the sense that it is independent of the choice of the pointc ∈ ]ω1 , ω2 [.Example 131−1√ 1dxdx+=√√221−x1−x1 − x2−10011= arcsin x −1 + arcsin x 0 = arcsin x −1 = π.dx=0Example 14 The integral+∞−∞e−x dx2is called the Euler–Poisson integral, and sometimes the Gaussian integral. It obvi√ously converges in the sense given above. It will be shown later that its value is π .6.5 Improper Integrals405Example 15 The integral+∞0dxxαdiverges, since for every α at least one of the two integrals +∞ 1dxdxandααxx01diverges.Example 16 The integral+∞0converges if each of the integrals 1sin xdxxα0sin xdxxα+∞and1sin xdxxαconverges.
The first of these integrals converges if α < 2, sincesin x1∼ α−1αxxas x → +0. The second integral converges if α > 0, as one can verify directlythrough an integration by parts similar to the one shown in Example 12, or by citingthe Abel–Dirichlet test. Thus the original integral has a meaning for 0 < α < 2.In the case when the integrand is not bounded in a neighborhood of one of theinterior points ω of the closed interval of integration [a, b], we set b ω bf (x) dx :=f (x) dx +f (x) dx,(6.81)aaωrequiring that both of the integrals on the right-hand side exist.Example 17 In the sense of the convention (6.81) 1dx√ = 4.|x|−11Example 18 The integral −1 dxx is not defined.Besides (6.81), there is a second convention about computing the integral of afunction that is unbounded in a neighborhood of an interior point ω of a closedinterval of integration.
To be specific, we set ω−δ b bPVf (x) dx := limf (x) dx +f (x) dx ,(6.82)aδ→+0aω+δ4066Integrationif the integral on the right-hand side exists. This limit is called, following Cauchy,the principal value of the integral, and, to distinguish the definitions (6.81) and(6.82), we put the letters PV in front of the second to indicate that it is the principalvalue.In accordance with this convention we haveExample 19PV1−1dx= 0.xWe also adopt the following definition: +∞f (x) dx := limPVRR→+∞ −R−∞Example 20PV+∞−∞f (x) dx.(6.83)x dx = 0.Finally, if there are several (finitely many) singularities of one kind or another onthe interval of integration, at interior points or endpoints, then the nonsingular pointsof the interval are divided into a finite number of such intervals, each containingonly one singularity, and the integral is computed as the sum of the integrals overthe closed intervals of the partition.It can be verified that the result of such a computation is not affected by thearbitrariness in the choice of a partition.Example 21 The precise definition of the logarithmic integral can now be writtenas! x dtif 0 < x < 1,t,li x = 0 ln x dtPV 0 ln t , if 1 < x.In the last case the symbol PV refers to the only interior singularity on the interval]0, x[, which is located at 1.
We remark that in the sense of the definition in formula(6.81) this integral is not convergent.6.5.4 Problems and Exercises1. Show that the following functions have the stated properties.xa) Si(x) = 0 sint t dt (the sine integral) is defined on all of R, is an odd function,and has a limit asx → +∞.∞b) si(x) = − x sint t dt is defined on all of R and differs from Si x only by aconstant;6.5 Improper Integrals407∞c) Ci x = − x cost t dt (the cosine integral) can be computed for sufficientlylarge values of x by the approximate formula Ci x ≈ sinx x ; estimate the region ofvalues where the absolute error of this approximation is less than 10−4 .2.
Show that +∞ x +∞a) the integrals 1 sinx α dx, 1lutely only for α > 1;b) the Fresnel integrals1C(x) = √2√x2cos xxαcos t dt,0dx converge only for α > 0, and abso-1S(x) = √2√xsin t 2 dt0are infinitely differentiable functions on the interval ]0, +∞[, and both have a limitas x → +∞.3. Show thata) the elliptic integral of first kind sin ϕdt#F (k, ϕ) =0(1 − t 2 )(1 − k 2 t 2 )is defined for 0 ≤ k < 1, 0 ≤ ϕ ≤π2and can be brought into the form ϕdψ;F (k, ϕ) =01 − k 2 sin2 ψb) the complete elliptic integral of first kind π/2dψK(k) =01 − k 2 sin2 ψincreases without bound as k → 1 − 0.4. Show thatx ta) the exponential integral Ei(x) = −∞ et dt is defined and infinitely differentiable for x < 0;−xb) − Ei(−x) = e x (1 − x1 + x2!2 − · · · + (−1)n xn!n + o( x1n )) as x → +∞;n n!c) the series ∞n=0 (−1) x n does not converge for any value of x ∈ R;xd) li(x) ∼ ln x as x → +0. (For the definition of the logarithmic integral li(x)see Example 21.)5.
Show thatx2a) the function Φ(x) = √1π −x e−t dt, called the error function and often denoted erf(x), is defined, even, and infinitely differentiable on R and has a limit asx → +∞;4086Integrationb) if the limit in a) is equal to 1 (and it is), then x22erf(x) = √e−t dt =π 0 121·31·3·5112= 1 − √ e−x− 2 3 + 3 5 − 4 7 +o 72x 2 xx2 x2 xπas x → +∞.6.
Prove the following statements.a) If a heavy particle slides under gravitational attraction along a curve givenin parametric form as x = x(θ ), y = y(θ ), and at time t = 0 the particle had zerovelocity and was located at the point x0 = x(θ0 ), y0 = y(θ0 ), then the followingrelation holds between the parameter θ defining a point on the curve and the time tat which the particle passes this point (see formula (6.63) of Sect.
6.4) θ' (x (θ ))2 + (y (θ ))2dθ,t =±2g(y0 − y(θ ))θ0in which the improper integral necessarily converges if y (θ0 ) = 0. (The ambiguoussign is chosen positive or negative according as t and θ have the same kind ofmonotonicity or the opposite kind; that is, if an increasing θ corresponds to anincreasing t, then one must obviously choose the positive sign.)b) The period of oscillation of a particle in a well having cross section in theshape of a cycloidx = R(θ + π + sin θ ),y = −R(1 + cos θ ),|θ | ≤ π,is independent√ of the level y0 = −R(1 + cos θ0 ) from which it begins to slide and isequal to 4π R/g (see Problem 4 of Sect.
6.4).Chapter 7Functions of Several Variables: Their Limitsand ContinuityUp to now we have considered almost exclusively numerical-valued functions x →f (x) in which the number f (x) was determined by giving a single number x fromthe domain of definition of the function.However, many quantities of interest depend on not just one, but many factors,and if the quantity itself and each of the factors that determine it can be characterized by some number, then this dependence reduces to the fact that a valuey = f (x 1 , . . .
, x n ) of the quantity in question is made to correspond to an orderedset (x 1 , . . . , x n ) of numbers, each of which describes the state of the correspondingfactor. The quantity assumes this value when the factors determining this quantityare in these states.For example, the area of a rectangle is the product of the lengths of its sides. Thevolume of a given quantity of gas is computed by the formulaV =RmT,pwhere R is a constant, m is the mass, T is the absolute temperature, and p is thepressure of the gas. Thus the value of V depends on a variable ordered triple ofnumbers (m, T , p), or, as we say, V is a function of the three variables m, T , and p.Our goal is to learn how to study functions of several variables just as we learnedhow to study functions of one variable.As in the case of functions of one variable, the study of functions of severalnumerical variables begins by describing their domains of definition.7.1 The Space Rm and the Most Important Classes of Its Subsets7.1.1 The Set Rm and the Distance in ItWe make the convention that Rm denotes the set of ordered m-tuples (x 1 , .
. . , x m )of real numbers x i ∈ R (i = 1, . . . , m).© Springer-Verlag Berlin Heidelberg 2015V.A. Zorich, Mathematical Analysis I, Universitext,DOI 10.1007/978-3-662-48792-1_74094107 Functions of Several Variables: Their Limits and ContinuityEach such m-tuple will be denoted by a single letter x = (x 1 , . . . , x m ) and, inaccordance with convenient geometric terminology, will be called a point of Rm .The number x i in the set (x 1 , . . . , x m ) will be called the ith coordinate of the pointx = (x 1 , .
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