Page 332 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 12] IMPROPER INTEGRALS 323

Write the integral as
ð 1 ð 1
e
x n 1 x dx þ x n 1 x dx ð1Þ
e
0 1
(a)If n A 1, the first integral in (1) converges since the integrand is continuous in 0 @ x @ 1.
If 0 < n < 1, the first integral in (1)is animproper integral of the second kind at x ¼ 0. Since
lim x 1 n x n 1 x ¼ 1, the integral converges by Theorem 3(i), Page 312, with p ¼ 1 n and a ¼ 0.
e
x!0þ
Thus, the first integral converges for n > 0.
If n > 0, the second integral in (1)is animproper integral of the first kind. Since
2
e
lim x x n 1 x ¼ 0(by L’Hospital’s rule or otherwise), this integral converges by Theorem 1ðiÞ,
x!1
Page 309, with p ¼ 2.
Thus, the second integral also converges for n > 0, and so the given integral converges for n > 0.
(b)If n @ 0, the first integral of (1)diverges since lim x x n 1 x ¼1 [Theorem 3(ii), Page 312].
e
x!0þ
e
If n @ 0, the second integral of (1)converges since lim x x n 1 x ¼ 0[Theorem 1(i), Page 309].
x!1
Since the first 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
Chapter 15.
UNIFORM CONVERGENCE OF IMPROPER INTEGRALS
ð
1
e x dx for > 0.
12.19. (a) Evaluate ð Þ¼
0
(b) Prove that the integral in (a) converges uniformly to 1 for A 1 > 0.
(c) Explain why the integral does not converge uniformly to 1 for > 0.
ð b b
ðaÞ ð Þ¼ lim e x dx ¼ lim e x ¼ lim 1 e b ¼ 1 if > 0

x¼0 b!1
b!1 a
b!1
.
Thus, the integral converges to 1 for all > 0.
(b) Method 1,using definition:
The integral converges uniformly to 1 in A 1 > 0iffor each > 0wecan find N,depending on
ð u
x

e dx < for all u > N.
but not on ,such that 1

0

ð u
x u u 1 u 1 1
e dx ¼j1 ð1 e Þj ¼ e @ e < for u > ln ¼ N,the result fol-

Since 1

lows. 0 1
Method 2,using the Weierstrass M test: 1
2
Since lim x e x ¼ 0for A 1 > 0, we can choose j e x j < 2 for sufficiently large x,say
1 1 dx
x!1 ð x
x A x 0 . Taking MðxÞ¼ and noting that converges, it follows that the given integral is
x 2 x 0 x 2
uniformly convergent to 1 for A 1 > 0.
(c) As 1 ! 0, the number N in the first method of (b)increases without limit, so that the integral cannot
be uniformly convergent for > 0.
ð
1
f ðx; Þ dx is uniformly convergent for 1 @ @ 2 , prove that ð Þ is continuous in
12.20. If ð Þ¼
0
this interval.