UNIFORM CONVERGENCE OF IMPROPER INTEGRALS 1 e x dx for > 0. 12.19. (a) Evaluate  in VS .NET

Encoding QR Code 2d barcode in VS .NET UNIFORM CONVERGENCE OF IMPROPER INTEGRALS 1 e x dx for > 0. 12.19. (a) Evaluate 

UNIFORM CONVERGENCE OF IMPROPER INTEGRALS 1 e x dx for > 0. 12.19. (a) Evaluate 
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(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.
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a  lim b b  e x dx lim e x  
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. Thus, the integral converges to 1 for all > 0. (b) Method 1, using de nition: The integral converges uniformly to 1in A 1 > 0 if for each  > 0 we can nd N, depending on  u    but not on , such that 1 e x dx <  for all u > N.   0   u   1 1 Since 1 e x dx j1 1 e u j e u @ e 1 u <  for u > ln N, the result fol  1  0 lows. Method 2, using the Weierstrass M test: 1 Since lim x2 e x 0 for A 1 > 0, we can choose j e x j < 2 for su ciently large x, say 1 x!1 x 1 dx x A x0 . Taking M x 2 and noting that converges, it follows that the given integral is 2 x x0 x uniformly convergent to 1 for A 1 > 0. (c) As 1 ! 0, the number N in the rst method of (b) increases without limit, so that the integral cannot be uniformly convergent for > 0.
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f x; dx is uniformly convergent for 1 @ @ 2 , prove that  is continuous in
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IMPROPER INTEGRALS 1 f x; dx R u; ; where R u;
[CHAP. 12

f x; dx:
Then
 h
f x; h dx R u; h and so u
 h  Thus j h  j @
f f x; h f x; g dx R u; h R u;
j f x; h f x; jdx jR u; h j jR u; j
Since the integral is uniformly convergent in 1 @ @ 2 , we can, for each  > 0, nd N independent of such that for u > N, jR u; h j < =3; jR u; j < =3 2
Since f x; is continuous, we can nd  > 0 corresponding to each  > 0 such that u j f x; h f x; j dx < =3 for jhj < 
Using (2) and (3) in (1), we see that j h  j <  for jhj < , so that  is continuous. Note that in this proof we assume that and h are both in the interval 1 @ @ 2 . Thus, if 1 , for example, h > 0 and right-hand continuity is assumed. Also note the analogy of this proof with that for in nite series. Other properties of uniformly convergent integrals can be proved similarly.
1 12.21. (a) Show that lim
1 a
!0 0
!0 0
e x dx 6
 lim e x dx:
b Explain the result in (a).
e x dx lim 1 by Problem 12.19 a :
 1 lim e x dx 0 dx 0. Thus the required result follows.
1 (b) Since 
e ax dx is not uniformly convergent for A 0 (see Problem 12.19), there is no Thus lim  may not be equal to  0 .
guarantee that  will be continuous for A 0.
1 12.22. (a) Prove that
e x cos rx dx
for > 0 and any real value of r. 2 r 2
(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  M e x r sin rx cos rx M  lim e x cos rx dx lim  M!1 0 M!1 2 r2 2 r2 0 (b) This follows at once from the Weierstrass M test for integrals, by noting that je x cos rxj @ e x and 1 e x dx converges.
CHAP. 12]
IMPROPER INTEGRALS
EVALUATION OF DEFINITE INTEGRALS =2  ln sin x dx ln 2. 12.23. Prove that 2 0
The given integral converges [Problem 12.42( f )]. Letting x =2 y, =2 =2 =2 ln sin x dx ln cos y dy ln cos x dx I
0 0 0
Then 2I   sin 2x ln dx 2 0 0 =2 =2 =2  ln sin 2x dx ln 2 dx ln sin 2x dx ln 2 2 0 0 0 =2 ln sin x ln cos x dx =2
Letting 2x v, & '  1  1 =2 ln sin v dv ln sin v dv ln sin v dv 2 0 2 0 0 =2 1 I I I (letting v  u in the last integral) 2   Hence, (1) becomes 2I I ln 2 or I ln 2. 2 2 =2 ln sin 2x dx
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