Page 333 - Schaum's Outline of Theory and Problems of Advanced Calculus

324 IMPROPER INTEGRALS [CHAP. 12

u 1
ð ð
Let ð Þ¼ f ðx; Þ dx þ Rðu; Þ; where Rðu; Þ¼ f ðx; Þ dx:
a u
ð u
Then ð þ hÞ¼ f ðx; þ hÞ dx þ Rðu; þ hÞ and so
a
u
ð
ð þ hÞ ð Þ¼ f f ðx; þ hÞ f ðx; Þg dx þ Rðu; þ hÞ Rðu; Þ
a
Thus
ð u
j ð þ hÞ ð Þj @ j f ðx; þ hÞ f ðx; Þjdx þjRðu; þ hÞj þ jRðu; Þj ð1Þ
a
Since the integral is uniformly convergent in 1 @ @ 2 ,we can, for each > 0, find N independent
of such that for u > N,
jRðu; þ hÞj < =3; jRðu; Þj < =3 ð2Þ
Since f ðx; Þ is continuous, we can find > 0corresponding to each > 0suchthat
u
ð
j f ðx; þ hÞ f ðx; Þj dx < =3 for jhj < ð3Þ
a
Using (2) and (3)in (1), we see that j ð þ hÞ ð Þj < for jhj < ,sothat ð Þ is continuous.
Note that in this proof we assume that and þ h are both in the interval 1 @ @ 2 . Thus, if
¼ 1 ,for example, h > 0 and right-hand continuity is assumed.
Also note the analogy of this proof with that for infinite series.
Other properties of uniformly convergent integrals can be proved similarly.

ð ð
1 1
x x
12.21. (a) Show that lim e dx 6¼ lim e dx: ðbÞ Explain the result in (a).
!0þ 0 0 !0þ
ð
1
lim e x dx ¼ lim ¼ 1by Problem 12.19ðaÞ:
ðaÞ
!0þ 0 !0þ
ð ð
1 x 1
lim e dx ¼ 0 dx ¼ 0. Thus the required result follows.
0 !0þ 0
ð
1 ax
e dx is not uniformly convergent for A 0 (see Problem 12.19), there is no
(b)Since ð Þ¼
0
guarantee that ð Þ will be continuous for A 0. Thus lim ð Þ may not be equal to ð0Þ.
!0þ
ð
1
x
12.22. (a)Prove that e cos rx dx ¼ 2 2 for > 0 and any real value of r.
0 þ r
(b)Prove that the integral in (a) converges uniformly and absolutely for a @ @ b, where
0 < a < b and any r.
(a)From integration formula 34, Page 96, we have
ð M e x M
lim e x cos rx dx ¼ lim ðr sin rx cos rxÞ ¼ 2 2
2
2
M!1 0 M!1 þ r 0 þ r
(b) This follows at once from the Weierstrass M test for integrals, by noting that je x cos rxj @ e x and
ð
1 x
e dx converges.
0