12.14. Determine whether value sense. in Visual Studio .NET

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5 12.14. Determine whether value sense.
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(a) By de nition,
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dx converges (a) in the usual sense, (b) in the Cauchy principal 3 1 x 1
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1 1 5 dx dx dx lim lim 3 1 !0 1 x 1 3 2 !0 1  x 1 3 1 x 1 2     1 1 1 1 2 lim lim 1 !0 8 2 !0 22 32 21 2
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and since the limits do not exist, the integral does not converge in the usual sense. (b) Since & 1 
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dx x 1 3
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' & ' dx 1 1 1 1 3 lim 2 2 3 !0 8 32 32 2 2 1  x 1
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the integral exists in the Cauchy principal value sense. The principal value is 3/32.
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12.15. Investigate the convergence of: 3 5 dx dx p (a) (c) 2=3 2 3 5 x x 1 2 x x 8 1  1 sin 1 x sin x 2 dx (b dx (d) x3 0 1 1 x
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(a)  2=3 1 1 1 1 lim 2 2 p . 2=3 x!2 x!2 x x 2x 4 x2 x3 8 8 3 18 Theorem 3(i), Page 312. lim x 2 2=3
=2 (e)
dx ; n > 1: cos x 1=n
Hence, the integral converges by
lim x2 sin x 1. x3
IMPROPER INTEGRALS
[CHAP. 12
x!0
Hence, the integral diverges by Theorem 3(ii) on Page 312. 3 dx p 5 x x 1 1 5 dx p : 5 x x 1 3
c Write the integral as
1 1 Since lim x 1 1=2 p , the rst integral converges. x!1 5 x x 1 2 Since 1 1 lim 5 x 1=2 p , the second integral converges. 5 x x 1 2
x!5
Thus, the given integral converges. (d) lim 1 x 2sin x 2=2 . Hence, the integral diverges. 1 x 1
x!1
Another method: 2sin x 2 =2 A , and 1 x 1 x e lim =2 x 1=n
dx diverges. 1 x
Hence, the given integral diverges.
x!1=2
  1 =2 x 1=n lim 1: 1=n x!1=2 cos x cos x
Hence the integral converges.
1 12.16. If m and n are real numbers, prove that
xm 1 1 x n 1 dx (a) converges if m > 0 and n > 0
simultaneously and
(b) diverges otherwise.
(a) For m A 1 and n A 1 simultaneously, the integral converges, since the integrand is continuous in 0 @ x @ 1. Write the integral as 1=2 1 xm 1 1 x n 1 dx xm 1 1 x n 1 dx 1
0 1=2
If 0 < m < 1 and 0 < n < 1, the rst integral converges, since lim x1 m xm 1 1 x n 1 1, using x!0 Theorem 3(i), Page 312, with p 1 m and a 0. 1 n xm 1 1 x n 1 1, using Theorem Similarly, the second integral converges since lim 1 x x!1 4(i), Page 312, with p 1 n and b 1. Thus, the given integral converges if m > 0 and n > 0 simultaneously. (b) If m @ 0, lim x xm 1 1 x n 1 1. Hence, the rst integral in (1) diverges, regardless of the value
x!0
of n, by Theorem 3(ii), Page 312, with p 1 and a 0. Similarly, the second integral diverges if n @ 0 regardless of the value of m, and the required result follows. Some interesting properties of the given integral, called the beta integral or beta function, are considered in 15.
 12.17. Prove that
1 1 sin dx converges conditionally. x x
Letting x 1=y, the integral becomes
sin y dy and the required result follows from Problem 12.12. y
IMPROPER INTEGRALS OF THE THIRD KIND 1 xn 1 e x dx (a) converges if n > 0 and (b) diverges if n @ 0. 12.18. If n is a real number, prove that
CHAP. 12]
IMPROPER INTEGRALS
Write the integral as 1
xn 1 e x dx
xn 1 e x dx
(a) If n A 1, the rst integral in (1) converges since the integrand is continuous in 0 @ x @ 1. If 0 < n < 1, the rst integral in (1) is an improper integral of the second kind at x 0. Since lim x1 n xn 1 e x 1, the integral converges by Theorem 3(i), Page 312, with p 1 n and a 0.
x!0
Thus, the rst integral converges for n > 0. If n > 0, the second integral in (1) is an improper integral of the rst kind. Since lim x2 xn 1 e x 0 (by L Hospital s rule or otherwise), this integral converges by Theorem 1 i ,
Page 309, with p 2. Thus, the second integral also converges for n > 0, and so the given integral converges for n > 0. (b) If n @ 0, the rst integral of (1) diverges since lim x xn 1 e x 1 [Theorem 3(ii), Page 312]. If n @ 0, the second integral of (1) converges since lim x xn 1 e x 0 [Theorem 1(i), Page 309].
x!1 x!0
Since the rst integral in (1) diverges while the second integral converges, their sum also diverges, i.e., the given integral diverges if n @ 0. Some interesting properties of the given integral, called the gamma function, are considered in 15.
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