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..
Trang 1Example 3 Let us show that the improper integral
∞
a
sin x
x λ dx is convergent for a >0and λ >0
Set f (x) = sin x and g(x) = x–λand verify conditions (i) and (ii) of Theorem 8 We have
(i)
a A sin x dx
= |cosa – 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+ x6sin2x 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◦ If f (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→0ε
∞
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
∞
a + bx
1+ x
dx
(1+ x)2 =
1
b – a
b
a f (x) dx;
∞
0
f (ax) – f (bx)
f(0) – f (∞)ln b
a if a >0, b >0, f (x) is continuous
on [0,∞), and f(∞) = lim
x→∞ f (x) is a finite quantity;
Trang 2∞
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;
∞
ax – b
x
dx= 1
a
∞
0 f(|x|) dx if a >0, b >0;
∞
2f
ax – x bdx= 1
a3
∞
0 (x
2+ ab)f (| x|) dx if a >0, b >0;
∞
ax – x bdx x2 =
1
b
∞
0 f(|x|) dx if a >0, b >0;
∞
x, 1
x
dx
x =2
1
0 f
x, 1
x
dx
x if f (x, y) = f (y, x);
∞
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
∞
x
a + a
x
ln x
x dx = ln a
∞
x
a + a
x
dx
x if a >0;
∞
x p
a + a
x p
ln x
x dx= ln a
p
∞
x p
a + a
x p
dx
x if a >0, p >0;
∞
0 f (x
a + x–a) ln x
1+ x2 dx=0 (a special case of the integral below);
∞
x, 1
x
ln x
1+ x2 dx=0 if f (x, y) = f (y, x) (the integral is assumed to exist);
∞
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) and f (x + π) = f (x);
∞
0 f (x)
sin x
x dx=
π/2
0 f (x) cos x dx if f (x) = f (–x) and f (x + π) = –f (x);
∞
0
f (sin x)
π/2
0
f (sin x) sin x dx if f (–x) = –f (x);
∞
0
f (sin x)
x2 dx=
π/2
0
f (sin x)
sin2x dx if f (x) = f (–x);
∞
0
f (sin x)
x cos x dx =
π/2
0
f (sin x) sin x cos
2x dx if f (–x) = –f (x);
Trang 3∞
0
f (sin x)
x tan x dx =
π/2
0 f (sin x) dx if f (–x) = f (x);
∞
0
f (sin x)
x2+ a2 dx=
sinh(2a)
2a
π/2
0
f (sin x) dx
cosh2a– cos2x if f (–x) = f (x);
∞
x+ 1
x
arctan x
x dx= π
4
∞
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), i2= –1,
where F (z) is a function analytic in a circle of radius r Then the following formulas hold:
∞
0
f (x, r)
x2+ a2 dx=
π
2a F (re–a);
∞
0
xg (x, r)
x2+ a2 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 (x2+ a2) dx=
π
2a2[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
2
f (p) dp,
∞
n f (x) dx = (–1) n+1
d n
dp n f2(p)∞
0 ,
n=1, 2, , (7.2.8.1) where 2f (p) is the Laplace transform of the function f (x), which is defined as
2
f (p) =
∞
–pxf (x) dx.
Short notation for the Laplace transform: 2f (p) =L5f (x)6
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
Trang 4Example 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): L5sin(ax)6
= a
a2+ p2 Substitute this expression into the first formula
in (7 2 8 1 ) and integrate to obtain
∞
0
sin(ax)
x dx=
∞
0
a dp
a2+ p2 = arctan
p a
∞
0
= π
2.
Example 2 Evaluate Frullani’s integral
∞
0
e–ax – e–bx
x dx , where a >0and 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 L5e–ax6
= 1
p + a Integrating yields
∞
0
e–ax – e–bx
∞
0
1
p + a –
1
p + b
dp= lnp + a
p + b
∞
0
= – lna
b = lnb
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◦ For λ → ∞, 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 J0(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 have2n+1partial derivatives with respect to x for x≥ 0
that monotonically go to zero as x → ∞.
Trang 53◦ For λ → ∞, 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–xtdx is the Laplace transform of the function f (x).
4◦ For λ → ∞, 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–xtdx 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
≤π2 ∞
2(x) dx ∞
2f2(x) dx
holds, provided the integral on the left-hand side exists The constant π2is best in the sense
that there exist functions f (x)0for 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) be defined and continuous for a≤x < b, but lim
x→b–0f (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
Trang 6If no finite limit exists, the integral is called divergent.
If lim
x→a+0f (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].
THEOREM1 (CAUCHY’S CONVERGENCE CRITERION) For the integral (7.2.10.1) to be
convergent is it necessary and sufficient that for any ε >0there exists a number δ >0such
that for any δ1and δ2satisfying0< δ1 < δand0< δ2< δthe following inequality holds:
b–δ2
b–δ1
f (x) dx
< ε.
THEOREM2 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) dxis 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
ϕ2(x) + √
1– x
is convergent and does not exceed 2, since 1
ϕ2(x) + √
1– x <
1
√
1– x, while the integral
1
0
dx
√
1– x is
convergent and is equal to 2.
THEOREM3 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
Trang 7THEOREM4 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)≠ 0holds
Then: (i) if λ <1and ϕ(x)≤c<∞, then the integral b
a f (x) dxis convergent; (ii) if
λ≥ 1and 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
a
f (x) dx = (b – a)
∞
1
f
a – b
z + bdz
z2, 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 x1[a, b] and
x2 [a, b] we have
|f (x2) – f (x1)|< A| x2– x1|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 If λ >1, then by condition (7.2.11.1) the derivative f
x (x) vanishes everywhere, and f (x) must be constant Therefore, we assume that0< λ≤ 1 For
λ=1, the H¨older condition is often called the Lipschitz condition Sometimes the H¨older
condition is called the Lipschitz condition of order λ.
If x1 and x2 are sufficiently close to each other and if the H¨older condition holds for
some exponent λ1, then this condition certainly holds for each exponent λ < λ1 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 f1(x) and f2(x) satisfy the H¨older condition with exponents λ1and λ2, 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(λ1, λ2)
If f (x) is differentiable and has a bounded derivative, then f (x) satisfies the Lipschitz
condition In general, the converse assertion fails