Handbook of mathematics for engineers and scienteists part 49 doc

304 INTEGRALS Example 3. Let us show that the improper integral  ∞ a sin x x λ dx is convergent for a > 0 and λ > 0. Set f(x)=sinx and g(x)=x –λ and verify conditions (i) and (ii) of Theorem 8. We have (i)      A a sin xdx     = |cos a –cosA| ≤ 2; (ii) since λ > 0, the function x –λ is monotonically decreasing and goes to zero as x →∞. So both conditions of Theorem 8 are met, and therefore the given improper integral is convergent. 7.2.7-3. Some remarks. 1 ◦ . If an improper integral is convergent and the integrand function tends to a limit as x →∞, then this limit can only be zero (it is such situations that were dealt with in Examples 1–3). However, the property lim x→∞ f(x)=0 is not a necessary condition for convergence of the integral (7.2.7.1). An integral can also be convergent if the integrand function does not have a limit as x →∞. For example, this is the case for Fresnel’s integrals:  ∞ 0 sin(x 2 ) dx =  ∞ 0 cos(x 2 ) dx = 1 2  π 2 . Furthermore, it can be shown that the integral  x 1 + x 6 sin 2 x dx is convergent regard- less of the fact that the integrand function, being everywhere positive, is not even bounded (f(πk)=πk, k = 1, 2, ). The graph of this function has infinitely many spikes with heights increasing indefinitely and base widths vanishing. At the points lying outside the spike bases, the function rapidly goes to zero. 2 ◦ .Iff(x) is a monotonic function for x ≥ 0 and the improper integral  ∞ 0 f(x) dx is convergent, then the following limiting relation holds:  ∞ 0 f(x) dx = lim ε→0 ε ∞  n=1 f(εn). 7.2.8. General Reduction Formulas for the Calculation of Improper Integrals Below are some general formulas, involving arbitrary functions and parameters, that may facilitate the calculation of improper integrals. 7.2.8-1. Improper integrals involving power functions.  ∞ 0 f  a + bx 1 + x  dx (1 + x) 2 = 1 b – a  b a f(x) dx;  ∞ 0 f(ax)–f(bx) x dx =  f(0)–f(∞)  ln b a if a > 0, b > 0, f (x) is continuous on [0, ∞), and f(∞) = lim x→∞ f(x)isafinite quantity; 7.2. DEFINITE INTEGRAL 305  ∞ 0 f(ax)–f(bx) x dx = f (0)ln b a if a > 0, b > 0, f (x) is continuous on [0, ∞ ), and the integral  ∞ c f(x) x dx exists; c > 0;  ∞ 0 f      ax – b x      dx = 1 a  ∞ 0 f(|x|) dx if a > 0, b > 0;  ∞ 0 x 2 f      ax – b x      dx = 1 a 3  ∞ 0 (x 2 + ab)f(|x|) dx if a > 0, b > 0;  ∞ 0 f      ax – b x      dx x 2 = 1 b  ∞ 0 f(|x|) dx if a > 0, b > 0;  ∞ 0 f  x, 1 x  dx x = 2  1 0 f  x, 1 x  dx x if f(x, y)=f(y, x);  ∞ 0 f  x, a x  dx x = 0 if f (x, y)=–f (y, x), a > 0 (the integral is assumed to exist). 7.2.8-2. Improper integrals involving logarithmic functions.  ∞ 0 f  x a + a x  ln x x dx =lna  ∞ 0 f  x a + a x  dx x if a > 0;  ∞ 0 f  x p a + a x p  ln x x dx = ln a p  ∞ 0 f  x p a + a x p  dx x if a > 0, p > 0;  ∞ 0 f(x a + x –a ) ln x 1 + x 2 dx = 0 (a special case of the integral below);  ∞ 0 f  x, 1 x  ln x 1 + x 2 dx = 0 if f(x, y)=f (y, x) (the integral is assumed to exist);  ∞ 0 f  x, 1 x  ln x x dx = 0 if f(x, y)=f(y, x) (the integral is assumed to exist). 7.2.8-3. Improper integrals involving trigonometric functions.  ∞ 0 f(x) sin x x dx =  π/2 0 f(x) dx if f(x)=f(–x)andf(x + π)=f(x);  ∞ 0 f(x) sin x x dx =  π/2 0 f(x)cosxdx if f(x)=f(–x)andf(x + π)=–f (x);  ∞ 0 f(sin x) x dx =  π/2 0 f(sin x) sin x dx if f(–x)=–f(x);  ∞ 0 f(sin x) x 2 dx =  π/2 0 f(sin x) sin 2 x dx if f(x)=f(–x);  ∞ 0 f(sin x) x cos xdx=  π/2 0 f(sin x) sin x cos 2 xdx if f(–x)=–f (x); 306 INTEGRALS  ∞ 0 f(sin x) x tan xdx=  π/2 0 f(sin x) dx if f(–x)=f(x);  ∞ 0 f(sin x) x 2 + a 2 dx = sinh(2a) 2a  π/2 0 f(sin x) dx cosh 2 a –cos 2 x if f(–x)=f(x);  ∞ 0 f  x + 1 x  arctan x x dx = π 4  ∞ 0 f  x + 1 x  dx x . 7.2.8-4. Calculation of improper integrals using analytic functions. Suppose F (z)=f (r, x)+ig(r, x), z = r(cos x + i sin x), i 2 =–1, where F (z) is a function analytic in a circle of radius r. Then the following formulas hold:  ∞ 0 f(x, r) x 2 + a 2 dx = π 2a F (re –a );  ∞ 0 xg(x, r) x 2 + a 2 dx = π 2 [F (re –a )–F (0)];  ∞ 0 g(x, r) x dx = π 2 [F (r)–F (0)];  ∞ 0 g(x, r) x(x 2 + a 2 ) dx = π 2a 2 [F (r)–F (re –a )].  Paragraph 10.1.2-8 presents a method for the calculation of improper integrals using the theory of functions of a complex variable. 7.2.8-5. Calculation of improper integrals using the Laplace transform. The following classes of improper integrals may be evaluated using the Laplace transform:  ∞ 0 f(x) x dx =  ∞ 0  f(p) dp,  ∞ 0 x n f(x) dx =(–1) n+1  d n dp n  f(p)  ∞ 0 , n = 1, 2, , (7.2.8.1) where  f(p)istheLaplace transform of the function f(x), which is defined as  f(p)=  ∞ 0 e –px f(x) dx. Short notation for the Laplace transform:  f(p)=L  f(x)  . Section 11.2 presents properties and methods for determining the Laplace transform, and Section T3.1 gives tables of the Laplace transforms of various functions. 7.2. DEFINITE INTEGRAL 307 Example 1. Evaluate the integral  ∞ 0 sin(ax) x dx. Using Table 11.2 from Subsection 11.2.2 (or the table from Subsection T3.1.6), we find the Laplace transform of the function sin(ax): L  sin(ax)  = a a 2 + p 2 . Substitute this expression into the first formula in (7.2.8.1) and integrate to obtain  ∞ 0 sin(ax) x dx =  ∞ 0 adp a 2 + p 2 =arctan p a     ∞ 0 = π 2 . Example 2. Evaluate Frullani’s integral  ∞ 0 e –ax – e –bx x dx,wherea > 0 and b > 0. Using the first formula in (7.2.8.1) and Table 11.2 from Subsection 11.2.2 (or the table from Subsec- tion T3.1.3), we obtain L  e –ax  = 1 p + a . Integrating yields  ∞ 0 e –ax – e –bx x dx =  ∞ 0  1 p + a – 1 p + b  dp =ln p + a p + b     ∞ 0 =–ln a b =ln b a . 7.2.9. General Asymptotic Formulas for the Calculation of Improper Integrals Below are some general formulas, involving arbitrary functions and parameters, that may be useful for determining the asymptotic behavior of improper integrals. 7.2.9-1. Asymptotic formulas for some improper integrals with parameter. 1 ◦ . For asymptotics of improper Laplace integrals I(λ)=  ∞ a f(x)exp[λg(x)] dx as λ →∞, see Remark 1 in Paragraph 7.2.4-3. 2 ◦ .Forλ →∞, the following asymptotic expansions of improper integrals involving trigonometric functions and a Bessel function hold:  ∞ 0 cos(λx)f (x) dx = n  k=1 (–1) k f (2k–1) (0)λ –2k + O(λ –2n–1 ),  ∞ 0 sin(λx)f(x) dx = n–1  k=0 (–1) k f (2k) (0)λ –2k–1 + O(λ –2n–1 ),  ∞ 0 J 0 (λx)f(x) dx = 1 √ π n–1  k=0 (–1) k k! Γ  k + 1 2  f (2k) (0)λ –2k–1 + O(λ –2n–1 ). The function f(x) is assumed to have 2n + 1 partial derivatives with respect to x for x ≥ 0 that monotonically go to zero as x →∞. 308 INTEGRALS 3 ◦ .Forλ →∞, the following asymptotic expansions hold:  ∞ 0 f(x)g  x λ  dx = n  k=0 (–1) k k! F (k) (0)g (k) (0)λ –k + O(λ –n–1 ),  ∞ 0 f(λx)g(x) dx = n  k=0 (–1) k k! F (k) (0)g (k) (0)λ –k–1 + O(λ –n–2 ), where F (t)=  ∞ 0 f(x)e –xt dx is the Laplace transform of the function f(x). 4 ◦ .Forλ →∞, the following asymptotic expansions hold:  ∞ 0 f(x) dx x + λ = n  k=0 F (k) (0) λ k+1 + O(λ –n–2 ), where F (t)=  ∞ 0 f(x)e –xt dx is the Laplace transform of the function f(x). 7.2.9-2. Behavior of integrals with variable limit of integration as x →∞. Let f(t) be a continuously differentiable function, let g(t) be a twice continuously differen- tiable function, and let the following conditions hold: f(t)>0, g  (t)>0; g(t) →∞as t →∞; f  (t)/f(t)=o  g  (t)  as t →∞; g  (t)=o  g 2 (t)  as t →∞. Then the following asymptotic relation holds as x →∞:  ∞ x f(t)exp[–g(t)] dt  f(x) g  (x) exp[–g(x)]. 7.2.9-3. π-related inequality. If f(x) ≥ 0, the inequality   ∞ 0 f(x) dx  4 ≤ π 2   ∞ 0 f 2 (x) dx   ∞ 0 x 2 f 2 (x) dx  holds, provided the integral on the left-hand side exists. The constant π 2 is best in the sense that there exist functions f(x) 0 for which the equality is attained. 7.2.10. Improper Integrals of Unbounded Functions 7.2.10-1. Basic definitions. 1 ◦ . Let a function f (x)bedefined and continuous for a ≤ x < b, but lim x→b–0 f(x)=∞.If there exists a finite limit lim λ→b–0  λ a f(x) dx, it is called the (convergent) improper integral of the unbounded function f (x) over the interval [a, b]. Thus, by definition  b a f(x) dx = lim λ→b–0  λ a f(x) dx.(7.2.10.1) 7.2. DEFINITE INTEGRAL 309 If no finite limit exists, the integral is called divergent. If lim x→a+0 f(x)=∞, then, by definition, it is assumed that  b a f(x) dx = lim γ→a+0  b γ f(x) dx. Finally, if f (x) is unbounded near a point c (a, b) and both integrals  c a f(x) dx and  b c f(x) dx are convergent, then, by definition,  b a f(x) dx =  c a f(x) dx +  b c f(x) dx. 2 ◦ . The geometric meaning of an improper integral of an unbounded function and also sufficient conditions for convergence of such integrals are similar to those for improper integrals with infinite limit(s). 7.2.10-2. Convergence tests for improper integrals of unbounded functions. Presented below are theorems for the case where the only singular point of the integrand function is the right endpoint of the interval [a, b]. T HEOREM 1(CAUCHY’S CONVERGENCE CRITERION). For the integral (7.2.10.1) to be convergent is it necessary and sufficient that for any ε > 0 there exists a number δ > 0 such that for any δ 1 and δ 2 satisfying 0 < δ 1 < δ and 0 < δ 2 < δ the following inequality holds:      b–δ 2 b–δ 1 f(x) dx     < ε. T HEOREM 2. If 0 ≤ f(x) ≤ g(x) for a ≤ x < b , then the convergence of the inte- gral  b a g(x) dx implies the convergence of the integral  b a f(x) dx , with  b a f(x) dx ≤  b a g(x) dx . If the integral  b a f(x) dx is divergent, then the integral  b a g(x) dx is also divergent. Example. For any continuous function ϕ(x) such that ϕ(1)=0, the improper integral  1 0 dx ϕ 2 (x)+ √ 1 – x is convergent and does not exceed 2, since 1 ϕ 2 (x)+ √ 1 – x < 1 √ 1 – x , while the integral  1 0 dx √ 1 – x is convergent and is equal to 2. THEOREM 3. Let f(x) and g(x) be continuous functions on [a, b) and let the following limit exist: lim x→b f(x) g(x) = K (0 < K < ∞). Then both integrals  b a f(x) dx,  b a g(x) dx are either convergent or divergent simultaneously. 310 INTEGRALS THEOREM 4. Let a function f(x) be representable in the form f(x)= ϕ(x) (b – x) λ (λ > 0), where ϕ(x) is continuous on [a, b] and the condition ϕ(b) ≠ 0 holds. Then: (i) if λ < 1 and ϕ(x) ≤ c < ∞ , then the integral  b a f(x) dx is convergent; (ii) if λ ≥ 1 and ϕ(x) ≥ c > 0 , this integral is divergent. Remark. The issue of convergence of the integral of an unbounded function (7.2.10.1) at x = b can be reduced by a simple change of variable to the issue of convergence of an improper integral with an infinite limit:  b a f(x) dx =(b – a)  ∞ 1 f  a – b z + b  dz z 2 , z = b – a b – x . 7.2.10-3. Calculation of integrals using infinite sums of special form. Let a function f (x) be continuous and monotonic on the interval (0, 1), whose endpoints can be singular. If the integral (proper or improper)  1 0 f(x) dx exists, then the following limiting relations hold:  1 0 f(x) dx = lim n→∞ 1 n n–1  k=1 f  k n  = lim n→∞ 1 n n  k=1 f  2k – 1 2n  . 7.2.11. Cauchy-Type Singular Integrals 7.2.11-1. H ¨ older and Lipschitz conditions. We say that f(x) satisfies the H ¨ older condition on [a, b] if for any two points x 1 [a, b]and x 2 [a, b]wehave |f(x 2 )–f(x 1 )| < A|x 2 – x 1 | λ ,(7.2.11.1) where A and λ are positive constants. The number A is called the H ¨ older constant and λ is called the H ¨ older exponent.Ifλ > 1, then by condition (7.2.11.1) the derivative f  x (x) vanishes everywhere, and f(x) must be constant. Therefore, we assume that 0 < λ ≤ 1.For λ = 1,theH ¨ older condition is often called the Lipschitz condition. Sometimes the H ¨ older condition is called the Lipschitz condition of order λ. If x 1 and x 2 are sufficiently close to each other and if the H ¨ older condition holds for some exponent λ 1 , then this condition certainly holds for each exponent λ < λ 1 . In general, the converse assertion fails. The smaller λ, the broader is the class of H ¨ older continuous functions. The narrowest class is that of functions satisfying the Lipschitz condition. It follows from the last property that if functions f 1 (x)andf 2 (x) satisfy the H ¨ older condition with exponents λ 1 and λ 2 , respectively, then their sum and the product, as well as their ratio provided that the denominator is nonzero, satisfy the H ¨ older condition with exponent λ =min(λ 1 , λ 2 ). If f(x) is differentiable and has a bounded derivative, then f(x) satisfies the Lipschitz condition. In general, the converse assertion fails. . notation for the Laplace transform:  f(p)=L  f(x)  . Section 11.2 presents properties and methods for determining the Laplace transform, and Section T3.1 gives tables of the Laplace transforms of. method for the calculation of improper integrals using the theory of functions of a complex variable. 7.2.8-5. Calculation of improper integrals using the Laplace transform. The following classes of. Asymptotic Formulas for the Calculation of Improper Integrals Below are some general formulas, involving arbitrary functions and parameters, that may be useful for determining the asymptotic behavior of