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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 257 
"" 0 "" {TEXT -1 32 "Lesson 13a: Improper Integrals  " }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 55 "So far in our study of integration, we ha
ve considered " }{XPPEDIT 18 0 "Int(f(x),x = a .. b);" "6#-%$IntG6$-%
\"fG6#%\"xG/F);%\"aG%\"bG" }{TEXT -1 7 " where " }{XPPEDIT 18 0 "f;" "
6#%\"fG" }{TEXT -1 47 " is a bounded function on the bounded interval \+
" }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 47 ".  We now wa
nt to see what happens when either " }{XPPEDIT 18 0 "f;" "6#%\"fG" }
{TEXT -1 17 " or the interval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"
bG" }{TEXT -1 63 " becomes unbounded.  In either case, we have what is
 called an " }{TEXT 256 17 "improper integral" }{TEXT -1 47 " (the int
egrals we have seen so far are called " }{TEXT 257 7 "proper " }{TEXT 
-1 162 "integrals).  As you will see, an improper integral is not defi
ned directly in terms of partitions and sums, but is instead defined a
s a limit of proper integrals." }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 
26 "Type 1: Infinite Intervals" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "
We start with the case where the interval of integration, [" }
{XPPEDIT 18 0 "a,infinity;" "6$%\"aG%)infinityG" }{TEXT -1 53 "), is u
nbounded on the right.  In this case\nwe define" }}}{EXCHG {PARA 256 "
" 0 "" {TEXT -1 0 "" }{XPPEDIT 18 0 "Int(f(x),x = a .. infinity) := li
mit(Int(f(x),x = a .. b),b = infinity);" "6#>-%$IntG6$-%\"fG6#%\"xG/F*
;%\"aG%)infinityG-%&limitG6$-F%6$-F(6#F*/F*;F-%\"bG/F8F." }{TEXT -1 2 
" ," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "provided the limit exists.
  If " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 92 " is a positive func
tion, this limit may be interpreted as the total area under the graph \+
of " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 17 " to the right of " }
{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 113 "; in general, it may be gi
ven all the same interpretations as a proper integral.\n\nWe can start
 by observing that " }{TEXT 258 5 "Maple" }{TEXT -1 62 " is capable of
 evaluating many improper integrals, for example" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 70 "Int(1/(2*x - 5)^3, x=4..infinity) = int(1/(2*
x - 5)^3, x=4..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6
$*&\"\"\"F(*$),&%\"xG\"\"#\"\"&!\"\"\"\"$F(F//F,;\"\"%%)infinityG#F(\"
#O" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "f := x->1/(2*x - 5)^3
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGR6#%\"xG6\"6$%)operatorG%&a
rrowGF(*&\"\"\"F-*$),&9$\"\"#\"\"&!\"\"\"\"$F-F4F(F(F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "p1 := plot(f(x), x=3.5..8,thicknes
s=2):\np2 := plottools[line]([4,0],[4,f(4)],color=blue,thickness=2):\n
plots[display](p1,p2);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 
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45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
141 "This gives us very little insight, however, so let's evaluate thi
s same integral step-by-step.  First, we need to evaluate the integral
 from " }{XPPEDIT 18 0 "4;" "6#\"\"%" }{TEXT -1 27 " to a variable upp
er limit " }{XPPEDIT 18 0 "b;" "6#%\"bG" }{TEXT -1 130 ".  If we were \+
doing this by hand, we would find an anti-derivative for our function \+
and apply the Fundamental Theorem of Calculus:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 19 "ad := int(f(x), x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#adG,$*&\"\"\"F'*$),&%\"xG\"\"#\"\"&!\"\"F,F'F.#F.\"
\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "inttob := subs(x=b, \+
ad) - subs(x=4, ad);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttobG,&*&
\"\"\"F'*$),&%\"bG\"\"#\"\"&!\"\"F,F'F.#F.\"\"%#F'\"#OF'" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 30 "We can get the same result in " }{TEXT 
259 5 "Maple" }{TEXT -1 45 " by asking it to find the integral from 4 \+
to " }{XPPEDIT 18 0 "b;" "6#%\"bG" }{TEXT -1 91 " directly.  (The answ
er appears in a different form, but we can check that it is the same.)
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "int(f(x), x=4..b);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,(\"\"%\"\"\"*$)%\"bG\"\"#F'F'*&\"
\"&F'F*F'!\"\"F'*$),&F*F+F-F.F+F'F.#F'\"\"*" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 37 "simplify(inttob - int(f(x), x=4..b));" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "H
aving found the integral up to " }{XPPEDIT 18 0 "b;" "6#%\"bG" }{TEXT 
-1 50 " one way or another, we can now take the limit as " }{XPPEDIT 
18 0 "proc (b) options operator, arrow; infinity end;" "6#R6#%\"bG7\"6
$%)operatorG%&arrowG6\"%)infinityGF*F*F*" }{TEXT -1 2 " :" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "Int(1/(2*x - 5)^3, x=4..infinity) =
 limit(inttob, b=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$Int
G6$*&\"\"\"F(*$),&%\"xG\"\"#\"\"&!\"\"\"\"$F(F//F,;\"\"%%)infinityG#F(
\"#O" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "in agreement with " }
{TEXT 260 7 "Maple's" }{TEXT -1 17 " original answer." }}{PARA 0 "" 0 
"" {TEXT -1 73 "\nIf the interval is unbounded to the left, we proceed
 similarly, defining" }}}{EXCHG {PARA 256 "" 0 "" {XPPEDIT 18 0 "Int(f
(x),x = -infinity .. b) := limit(Int(f(x),x = a .. b),a = -infinity);
" "6#>-%$IntG6$-%\"fG6#%\"xG/F*;,$%)infinityG!\"\"%\"bG-%&limitG6$-F%6
$-F(6#F*/F*;%\"aGF0/F:,$F.F/" }{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 62 "Int(exp(3*x), x=-infinity..2) = int(exp(3*x), x=
-infinity..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$-%$expG6#,
$%\"xG\"\"$/F+;,$%)infinityG!\"\"\"\"#,$-F(6#\"\"'#\"\"\"F," }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "inttoa := int(exp(3*x), x=a.
.2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttoaG,&-%$expG6#\"\"'#\"
\"\"\"\"$*&#F+F,F+-F'6#,$%\"aGF,F+!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "limit(inttoa, a=-infinity);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$-%$expG6#\"\"'#\"\"\"\"\"$" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 41 "Finally, if the integral is unbounded at " }{TEXT 261 
4 "both" }{TEXT -1 36 " ends, we choose a convenient point " }
{XPPEDIT 18 0 "c;" "6#%\"cG" }{TEXT -1 10 " (usually " }{XPPEDIT 18 0 
"c = 0;" "6#/%\"cG\"\"!" }{TEXT -1 19 " is fine)and define" }}}{EXCHG 
{PARA 256 "" 0 "" {XPPEDIT 18 0 "Int(f(x),x = -infinity .. infinity) :
= Int(f(x),x = -infinity .. c)+Int(f(x),x = c .. infinity);" "6#>-%$In
tG6$-%\"fG6#%\"xG/F*;,$%)infinityG!\"\"F.,&-F%6$-F(6#F*/F*;,$F.F/%\"cG
\"\"\"-F%6$-F(6#F*/F*;F8F.F9" }{TEXT -1 2 " ," }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 0 "" }{TEXT 262 53 "provided both integrals on the right-h
and side exist." }{TEXT -1 64 "  This condition is crucial.  For examp
le, consider the integral" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
39 "f := x->(1 + (1+x^2)*sin(x))/(1 + x^2);" }}{PARA 11 "" 1 "" 
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Fp\"[!pN%F1$!3NKG6nSg5SF-7$$\"3]nmmmxGpVF1$!3Ge%z'*eB([GF-7$$\"3'f7GQd
!\\#Q%F1$!3%)G6(f/?Ac\"F-7$$\"3S%e*)4Q$p&R%F1$!3K)oJXsMV[#F^r7$$\"3&G/
^\")='*)3WF1$\"312p6xwwp5F-7$$\"3I,DJ&**)4AWF1$\"33kWK8\"Q%pBF-7$$\"3w
fRZ-=INWF1$\"3o^3;*4bzi$F-7$$\"3A=aj4Y][WF1$\"3ariqBATB[F-7$$\"3mwoz;u
qhWF1$\"3V@iO22+NfF-7$$\"3SM$eRA5\\Z%F1$\"3j0Ci,;PVpF-7$$\"3f]7GQeJ,XF
1$\"3%)*)=t\\MN#e)F-7$$\"3]nTg_9sFXF1$\"3ag9k-rtE'*F-7$$\"3&fil(fU#4a%
F1$\"3]ZD*)G-c,**F-7$$\"3S%3Fp1FTb%F1$\"3ED'Hu.7/+\"Ffn7$$\"3'Ga)3u)Ht
c%F1$\"3#\\us](=jK**F-7$$\"3K,+D\"oK0e%F1$\"3kr7t!eQ$)o*F-7$$\"3],vVBi
!eg%F1$\"3+aXYjwgc()F-7$$\"3q,]il(z5j%F1$\"3sq))e)*y\")osF-7$$\"3!=v=n
`;Pk%F1$\"3;t8^a-rWjF-7$$\"3!>]7yI`jl%F1$\"3N[[+#4%\\>`F-7$$\"3+_i!*y+
**oYF1$\"3m;M2/(=&4UF-7$$\"3Q,++]oi\"o%F1$\"3hVVy*3&[KIF-7$$\"3y]P4@OE
%p%F1$\"3Jhr\\mO;2=F-7$$\"3(3](=#R+pq%F1$\"3?(*y\"[F\\4`&F^r7$$\"3(4D
\"Gjr`>ZF1$!33UA%H\"=A(4(F^r7$$\"32,]PMR<KZF1$!3)e,!e&>\\6'>F-7$$\"3<^
(oaq5[u%F1$!3i.6B#HH7=$F-7$$\"3G,DcwuWdZF1$!3Cej'=b00N%F-7$$\"3Q^ilZU3
qZF1$!3EX&z#\\7L]aF-7$$\"3Y,+v=5s#y%F1$!3_U1BBs;jkF-7$$\"3k^iS\"*3))4[
F1$!3mBS\"3LWAF)F-7$$\"35,D1k2/P[F1$!3)>`X\"4k`u%*F-7$$\"3[D1R+2i][F1$
!3bD'3@%\\l=)*F-7$$\"3c](=nj+U'[F1$!3'z8FTu->)**F-7$$\"358G)[g!*4([F1$
!3)>2i0))GY***F-7$$\"3mvo/t0yx[F1$!3?Ar9[VFh**F-7$$\"3?Q4@T0d%)[F1$!3?
lXq=H*>))*F-7$$\"3u+]P40O\"*[F1$!3)[IKB-]rv*F-7$$\"3\"4DJ?Q?&=\\F1$!39
.%)[0\"fn\"))F-7$$\"3Q+voa-oX\\F1$!3&zl\"fhnnHsF-7$$\"3uCc,\">g#f\\F1$
!3k\"zkH4[$GiF-7$$\"3%)\\PMF,%G(\\F1$!3aFY6N4E7^F-7$$\"3#\\(=nj+U')\\F
1$!3`%y99Pm>!RF-7$$\"#]F*$!3,W&*RO,v>EF--%'COLOURG6&%$RGBG$\"#5!\"\"$F
*F*Fg_x-%+AXESLABELSG6$Q\"x6\"Q!6\"-%*THICKNESSG6#\"\"#-%%VIEWG6$;F(F
\\_x%(DEFAULTG" 1 2 0 1 10 2 2 9 1 4 2 1.000000 45.000000 45.000000 0 
0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "The plot indicate
s an integral over the whole real line might not exist.  If we were ca
reless, we might try to evaluate this as" }{XPPEDIT 18 0 "Limit(Int((1
+(1+x^2)*sin(x))/(1+x^2),x = -a .. a),a = infinity);" "6#-%&LimitG6$-%
$IntG6$*&,&\"\"\"F+*&,&F+F+*$%\"xG\"\"#F+F+-%$sinG6#F/F+F+F+,&F+F+*$F/
F0F+!\"\"/F/;,$%\"aGF6F:/F:%)infinityG" }{TEXT -1 1 " " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "finiteint := int((1 + (1+x^2)*sin(x
))/(1+x^2),x = -a .. a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*finitei
ntG,$-%'arctanG6#%\"aG\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
66 "Int(f(x) ,x = -infinity..infinity) = limit(finiteint, a=infinity);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&,&\"\"\"F)*&,&F)F)*$)%
\"xG\"\"#F)F)F)-%$sinG6#F.F)F)F)F+!\"\"/F.;,$%)infinityGF3F7%#PiG" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 263 14 "THIS IS WRONG!" }
{TEXT -1 76 "  To compute the integral correctly, we must let the two \+
endpoints go to +/-" }{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }
{TEXT -1 23 " separately.  Choosing " }{XPPEDIT 18 0 "c = 0;" "6#/%\"c
G\"\"!" }{TEXT -1 30 ", the improper integral up to " }{XPPEDIT 18 0 "
infinity;" "6#%)infinityG" }{TEXT -1 15 " is computed as" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "inttob := int(f(x), x=0..b);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttobG,(-%'arctanG6#%\"bG\"\"\"-%$
cosGF(!\"\"F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "Int(f(x)
, x=0..infinity) = limit(inttob, b=infinity);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%$IntG6$*&,&\"\"\"F)*&,&F)F)*$)%\"xG\"\"#F)F)F)-%$sin
G6#F.F)F)F)F+!\"\"/F.;\"\"!%)infinityG;,$%#PiG#F)F/,&F/F)*&F;F)F:F)F)
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "The strange form of the answ
er shows that something has gone wrong.  Indeed, you should be able to
 see that the expression " }{TEXT 264 6 "inttob" }{TEXT -1 25 " doesn'
t have a limit as " }{XPPEDIT 18 0 "proc (b) options operator, arrow; \+
infinity end;" "6#R6#%\"bG7\"6$%)operatorG%&arrowG6\"%)infinityGF*F*F*
" }{TEXT -1 103 " .  There is no need to go further: one of the semi-i
nfinite improper integrals does not exist, and so " }{XPPEDIT 18 0 "In
t((1+(1+x^2)*sin(x))/(1+x^2),x = -infinity .. infinity);" "6#-%$IntG6$
*&,&\"\"\"F(*&,&F(F(*$%\"xG\"\"#F(F(-%$sinG6#F,F(F(F(,&F(F(*$F,F-F(!\"
\"/F,;,$%)infinityGF3F7" }{TEXT -1 16 " does not exist." }}}{SECT 0 
{PARA 4 "" 0 "" {TEXT -1 10 "Question 1" }}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 88 "Find the following improper integrals, or show that they \+
do not exist.  (Don't just ask " }{TEXT 265 5 "Maple" }{TEXT -1 75 " t
o evaluate them in one step: compute them as limits of proper integral
s.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "Int((2*x + 1)/(x^4 +
 1), x=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&,&%
\"xG\"\"#\"\"\"F*F*,&*$)F(\"\"%F*F*F*F*!\"\"/F(;\"\"!%)infinityG" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "inttob := int((2*x + 1)/(x^4
 + 1), x=0..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttobG,**&-%%sq
rtG6#\"\"#\"\"\"-%#lnG6#,$*&,(*$)%\"bGF*F+F+*&F4F+F'F+F+F+F+F+,(F2!\"
\"F5F+F+F7F7F7F+#F+\"\")*(#F+\"\"%F+F'F+-%'arctanG6#,&F5F+F+F+F+F+*(F;
F+F'F+-F>6#,&F5F+F+F7F+F+-F>6#F2F+" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "limit(inttob, b=infinity);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,&*&-%%sqrtG6#\"\"#\"\"\"%#PiGF)#F)\"\"%*&#F)F(F)F*F)F)
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "Int(1/(x*ln(x)), x=2..i
nfinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&\"\"\"F'*&%\"x
GF'-%#lnG6#F)F'!\"\"/F);\"\"#%)infinityG" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "inttob := int(1/(x*ln(x)), x=2..b);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%'inttobG,&-%#lnG6#-F'6#%\"bG\"\"\"-F'6#-F'6#\"\"#!
\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "limit(inttob, b=inf
inity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%)infinityG" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 43 "(So this improper integral does not exist
.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Int((2*x + 1)/(x^4 + \+
1), x=-infinity..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG
6$*&,&%\"xG\"\"#\"\"\"F*F*,&*$)F(\"\"%F*F*F*F*!\"\"/F(;,$%)infinityGF/
F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "part1 := int((2*x + 1
)/(x^4 + 1), x=-a..0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&part1G,**
&-%%sqrtG6#\"\"#\"\"\"-%#lnG6#,$*&,(*$)%\"aGF*F+!\"\"*&F4F+F'F+F+F+F5F
+,(F2F+F6F+F+F+F5F5F+#F5\"\")*&#F+\"\"%F+*&F'F+-%'arctanG6#,&F6F5F+F+F
+F+F5*&#F+F<F+*&F'F+-F?6#,&F6F5F+F5F+F+F5-F?6#F2F5" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 35 "limit1 := limit(part1, a=infinity);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit1G,&*&-%%sqrtG6#\"\"#\"\"\"%#P
iGF+#F+\"\"%*&#F+F*F+F,F+!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 42 "part2 := int((2*x + 1)/(x^4 + 1), x=0..b);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%&part2G,**&-%%sqrtG6#\"\"#\"\"\"-%#lnG6#,$*&,(*$)%
\"bGF*F+F+*&F4F+F'F+F+F+F+F+,(F2!\"\"F5F+F+F7F7F7F+#F+\"\")*(#F+\"\"%F
+F'F+-%'arctanG6#,&F5F+F+F+F+F+*(F;F+F'F+-F>6#,&F5F+F+F7F+F+-F>6#F2F+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit2 := limit(part2, \+
b=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit2G,&*&-%%sqrtG
6#\"\"#\"\"\"%#PiGF+#F+\"\"%*&#F+F*F+F,F+F+" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 116 "Since both pieces of the improper integral separately \+
have limits, we can add them together to get the final answer." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "Int((2*x + 1)/(x^4 + 1), x=-
infinity..infinity) = limit1 +    limit2;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%$IntG6$*&,&%\"xG\"\"#\"\"\"F+F+,&*$)F)\"\"%F+F+F+F+!
\"\"/F);,$%)infinityGF0F4,$*&-%%sqrtG6#F*F+%#PiGF+#F+F*" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 45 "Int(sin(x)/(x^2 + 1), x=-infinity..infinity);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&-%$sinG6#%\"xG\"\"\",&F+F+*
$)F*\"\"#F+F+!\"\"/F*;,$%)infinityGF0F4" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "part1 := int(sin(x)/(x^2 + 1), x=-a..0);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%&part1G,0*(^#\"\"\"F(-%#SiG6#F'F(-%%coshG6#F(
F(F(*&-%#CiGF+F(-%%sinhGF.F(F(*(^##!\"\"\"\"#F(F2F(%#PiGF(F(*(F5F(-F*6
#,&%\"aGF(F'F(F(F,F(F(*&#F(F8F(*&-F16#,&F>F7^#F7F(F(F2F(F(F7*(^##F(F8F
(-F*6#,&F>F(FEF(F(F,F(F(*&#F(F8F(*&-F16#,&F>F7F'F(F(F2F(F(F7" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit1 := limit(part1, a=inf
inity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit1G*&^##\"\"\"\"\"#F
(,(*&-%#SiG6#^#F(F(-%%coshG6#F(F(F)*(^#!\"#F(-%#CiGF.F(-%%sinhGF2F(F(*
&F8F(%#PiGF(!\"\"F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "part
2 := int(sin(x)/(x^2 + 1), x=0..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%&part2G,0*(^##!\"\"\"\"#\"\"\"-%#SiG6#,&%\"bGF+^#F)F+F+-%%coshG6#F+
F+F+*(#F+F*F+-%#CiGF.F+-%%sinhGF4F+F+*(^#F6F+-F-6#,&F0F+^#F+F+F+F2F+F+
*(F6F+-F8F>F+F9F+F+*(F1F+-F-6#F@F+F2F+F+*&-F8FEF+F9F+F)*(F<F+F9F+%#PiG
F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit2 := limit(par
t2, b=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit2G*&^##\"
\"\"\"\"#F(,(*&-%#SiG6#^#F(F(-%%coshG6#F(F(!\"#*(^#F)F(-%#CiGF.F(-%%si
nhGF2F(F(*&F8F(%#PiGF(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "Th
e limits involve some unfamiliar functions, but they do (both) exist, \+
and so we can add them together." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 63 "Int(sin(x)/(x^2 + 1), x=-infinity..infinity) = limit1
 + limit2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&-%$sinG6#%\"
xG\"\"\",&F,F,*$)F+\"\"#F,F,!\"\"/F+;,$%)infinityGF1F5,&*&^##F,F0F,,(*
&-%#SiG6#^#F,F,-%%coshG6#F,F,F0*(^#!\"#F,-%#CiGF>F,-%%sinhGFBF,F,*&FHF
,%#PiGF,F1F,F,*&F8F,,(F;FE*(^#F0F,FFF,FHF,F,FJF,F,F," }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 147 "The final answer is less surprising if you rea
lise that the function being integrated is odd, and the limits on the \+
integral are symmetric about 0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 47 "Int(x*sin(x)/(x^2 + 1), x=-infinity..infinity);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#-%$IntG6$*&*&%\"xG\"\"\"-%$sinG6#F(F)F),&F)F)*$)
F(\"\"#F)F)!\"\"/F(;,$%)infinityGF1F5" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "part1 := int(x*sin(x)/(x^2 + 1), x=-a..0);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%&part1G,,*&-%%sinhG6#\"\"\"F*%#PiGF*#F*\"
\"#*(F,F*-%#SiG6#,&%\"aGF*^#F*F*F*-%%coshGF)F*F**(^##!\"\"F-F*-%#CiG6#
,&F3F:^#F:F*F*F'F*F**(F,F*-F06#,&F3F*F?F*F*F5F*F**(^#F,F*-F<6#,&F3F:F4
F*F*F'F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit1 := lim
it(part1, a=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit1G,&
*&-%%sinhG6#\"\"\"F*%#PiGF*#!\"\"\"\"#*(#F*F.F*F+F*-%%coshGF)F*F*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "part2 := int(x*sin(x)/(x^2 +
 1), x=0..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&part2G,,*&-%#SiG6#
,&%\"bG\"\"\"^#!\"\"F,F,-%%coshG6#F,F,#F,\"\"#*(^#F2F,-%#CiGF)F,-%%sin
hGF1F,F,*(F2F,-F(6#,&F+F,^#F,F,F,F/F,F,*(^##F.F3F,-F7F<F,F8F,F,*&#F,F3
F,*&F8F,%#PiGF,F,F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limi
t2 := limit(part2, b=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'
limit2G,$*&%#PiG\"\"\",&-%%coshG6#F(F(-%%sinhGF,!\"\"F(#F(\"\"#" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "Int(x*sin(x)/(x^2 + 1), x=-i
nfinity..infinity) = limit1 +    limit2;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%$IntG6$*&*&%\"xG\"\"\"-%$sinG6#F)F*F*,&F*F**$)F)\"\"#F*F*!\"
\"/F);,$%)infinityGF2F6,(*&-%%sinhG6#F*F*%#PiGF*#F2F1*(#F*F1F*F<F*-%%c
oshGF;F*F**(F?F*F<F*,&F@F*F9F2F*F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "Int(2*x/(x^2 + 5), x=-infinity..infinity);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*&%\"xG\"\"\",&*$)F(\"\"#F)F)\"\"
&F)!\"\"F-/F(;,$%)infinityGF/F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 37 "part1 := int(2*x/(x^2 + 5), x=-a..0);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&part1G,&-%#lnG6#\"\"&\"\"\"-F'6#,&*$)%\"aG\"\"#F*F*F
)F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit1 := limit
(part1, a=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'limit1G,$%)
infinityG!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "Since one of th
e two parts of the improper integral doesn't exist, we can stop: " }
{XPPEDIT 18 0 "Int(2*x/(x^2+5),x = -infinity .. infinity);" "6#-%$IntG
6$*(\"\"#\"\"\"%\"xGF(,&*$F)F'F(\"\"&F(!\"\"/F);,$%)infinityGF-F1" }
{TEXT -1 17 "  does not exist." }}}}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 27 "Type 2: Unbo
unded Functions" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 96 "The second type
 of improper integral is that in which the function is unbounded on th
e interval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 251 "
 .  A common example of this is when the function becomes unbounded at
 one or other endpoint of the interval, and we will look at this case.
  (As explained in class, other cases can be reduced to this one anywa
y.)  Suppose first that, for every small " }{XPPEDIT 18 0 "0 < epsilon
;" "6#2\"\"!%(epsilonG" }{TEXT -1 3 ",  " }{XPPEDIT 18 0 "f;" "6#%\"fG
" }{TEXT -1 34 " is integrable on every interval (" }{XPPEDIT 18 0 "a+
epsilon,b;" "6$,&%\"aG\"\"\"%(epsilonGF%%\"bG" }{TEXT -1 45 "] .  Then
 the improper integral is defined as" }}}{EXCHG {PARA 256 "" 0 "" 
{XPPEDIT 18 0 "Int(f(x),x = a .. b) := limit(Int(f(x),x = a+epsilon ..
 b),epsilon = 0,right);" "6#>-%$IntG6$-%\"fG6#%\"xG/F*;%\"aG%\"bG-%&li
mitG6%-F%6$-F(6#F*/F*;,&F-\"\"\"%(epsilonGF9F./F:\"\"!%&rightG" }
{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Here is an exam
ple.  We can start by seeing that " }{TEXT 266 5 "Maple" }{TEXT -1 53 
" can evaluate this type of improper integral as well." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Int(1/sqrt(x), x=0..2) = int(1/sqrt
(x), x=0..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*
$-%%sqrtG6#%\"xGF(!\"\"/F-;\"\"!\"\"#,$*$-F+6#F2F(F2" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 45 "To check this, we evaluate the integral from " 
}{XPPEDIT 18 0 "epsilon;" "6#%(epsilonG" }{TEXT -1 14 " to 2 and let \+
" }{XPPEDIT 18 0 "epsilon;" "6#%(epsilonG" }{TEXT -1 59 " approach 0 f
rom the right.  (Note the use of the argument " }{TEXT 267 5 "right" }
{TEXT -1 23 " in the limit command.)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "ad := int(1/sqrt(x), x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#adG,$*$-%%sqrtG6#%\"xG\"\"\"\"\"#" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 49 "properint := subs(x=2, ad) - subs(x=epsil
on, ad);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*properintG,&*$-%%sqrtG6
#\"\"#\"\"\"F**&F*F+-F(6#%(epsilonGF+!\"\"" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 59 "Int(1/sqrt(x), x=0..2) = limit(properint, epsilon=0
,right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*$-%%s
qrtG6#%\"xGF(!\"\"/F-;\"\"!\"\"#,$*$-F+6#F2F(F2" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 44 "By the way, notice that the two-sided limit " }
{XPPEDIT 18 0 "limit(sqrt(epsilon),epsilon = 0);" "6#-%&limitG6$-%%sqr
tG6#%(epsilonG/F)\"\"!" }{TEXT -1 44 " does not exist but, at least in
 this case, " }{TEXT 268 5 "Maple" }{TEXT -1 20 " doesn't tell us so:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "limit(properint, epsilo
n=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$-%%sqrtG6#\"\"#\"\"\"F(" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "\n\nIn case " }{XPPEDIT 18 0 "f
;" "6#%\"fG" }{TEXT -1 56 " becomes unbounded at the right-hand endpoi
nt, we define" }}}{EXCHG {PARA 256 "" 0 "" {XPPEDIT 18 0 "Int(f(x),x =
 a .. b) := limit(Int(f(x),x = a .. b-epsilon),epsilon = 0,right);" "6
#>-%$IntG6$-%\"fG6#%\"xG/F*;%\"aG%\"bG-%&limitG6%-F%6$-F(6#F*/F*;F-,&F
.\"\"\"%(epsilonG!\"\"/F:\"\"!%&rightG" }{TEXT -1 2 " ." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Int(1/(x-2)^2, x=0..2) = int(1/(x-2
)^2, x=0..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*
$),&%\"xGF(\"\"#!\"\"F-F(F./F,;\"\"!F-%)infinityG" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 0 "" }{TEXT 269 7 "Maple's" }{TEXT -1 78 " answer sugg
ests that this improper integral doesn't exist.  Let's check this:" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ad := int(1/(x-2)^2, x);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#adG,$*&\"\"\"F',&%\"xGF'\"\"#!\"\"
F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "properint := subs(x
=2-epsilon, ad) - subs(x=0, ad);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
*properintG,&*&\"\"\"F'%(epsilonG!\"\"F'#F'\"\"#F)" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 30 "You can see that the limit as " }{XPPEDIT 18 0 "p
roc (epsilon) options operator, arrow; 0 end;" "6#R6#%(epsilonG7\"6$%)
operatorG%&arrowG6\"\"\"!F*F*F*" }{TEXT -1 24 " doesn't exist.  So can
 " }{TEXT 270 6 "Maple:" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "limit(properint, epsilon=0, right);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "
\n\nFinally, if " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 80 " is unbo
unded at both endpoints, we split the interval at some convenient poin
t " }{XPPEDIT 18 0 "c;" "6#%\"cG" }{TEXT -1 12 ", and define" }}}
{EXCHG {PARA 256 "" 0 "" {XPPEDIT 18 0 "Int(f(x),x = a .. b) := Int(f(
x),x = a .. c)+Int(f(x),x = c .. b);" "6#>-%$IntG6$-%\"fG6#%\"xG/F*;%
\"aG%\"bG,&-F%6$-F(6#F*/F*;F-%\"cG\"\"\"-F%6$-F(6#F*/F*;F6F.F7" }
{TEXT -1 2 " ," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "provided both i
mproper integrals on the right-hand side exist." }}}{SECT 0 {PARA 4 "
" 0 "" {TEXT -1 10 "Question 2" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "
Evaluate the following improper integrals, or show that they do not ex
ist." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Int(1/sqrt(1-x), x=
0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&\"\"\"F'*$-%%sqrt
G6#,&F'F'%\"xG!\"\"F'F./F-;\"\"!F'" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "inttoe := int(1/sqrt(1-x), x=0..1-epsilon);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%'inttoeG,&*$-%%sqrtG6#%(epsilonG\"\"\"!\"#
\"\"#F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "Int(1/sqrt(1-x),
 x=0..1) = limit(inttoe, epsilon=0, right);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*$-%%sqrtG6#,&F(F(%\"xG!\"\"F(F//F.
;\"\"!F(\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Int(1/sqrt
(4-x^2), x=0..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&\"\"\"
F'*$-%%sqrtG6#,&\"\"%F'*$)%\"xG\"\"#F'!\"\"F'F2/F0;\"\"!F1" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "inttoe := int(1/sqrt(4-x^2), x=0..2
-epsilon);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttoeG,$-%'arcsinG6#
,&!\"\"\"\"\"*&#F+\"\"#F+%(epsilonGF+F+F*" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 61 "Int(1/sqrt(4-x^2), x=0..2) = limit(inttoe, epsilon=
0, right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&\"\"\"F(*$-%
%sqrtG6#,&\"\"%F(*$)%\"xG\"\"#F(!\"\"F(F3/F1;\"\"!F2,$%#PiG#F(F2" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "Int(1/(x^2 - 1), x=1..2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%$IntG6$*&\"\"\"F',&*$)%\"xG\"\"#F'F'F'!\"\"F-/F+;F'F,
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "inttoe := int(1/(x^2 - \+
1), x=1+epsilon..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttoeG,(*&
^##\"\"\"\"\"#F)%#PiGF)F)*&#F)F*F)-%#lnG6#\"\"$F)!\"\"-%(arctanhG6#,&F
)F)%(epsilonGF)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "limit(
inttoe, epsilon=0, right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%)infini
tyG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "This improper integral doe
s not exist." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "Int(exp(sqr
t(x))/sqrt(x), x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&
-%$expG6#*$-%%sqrtG6#%\"xG\"\"\"F/*$-F,6#F.F/!\"\"/F.;\"\"!F/" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "inttoe := int(exp(sqrt(x))/s
qrt(x), x=epsilon..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'inttoeG,&
-%$expG6#\"\"\"\"\"#*&F*F)-F'6#*$-%%sqrtG6#%(epsilonGF)F)!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "Int(exp(sqrt(x))/sqrt(x), x=
0..1) = limit(inttoe, epsilon=0,   right);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%$IntG6$*&-%$expG6#*$-%%sqrtG6#%\"xG\"\"\"F0*$-F-6#F/
F0!\"\"/F/;\"\"!F0,&-F)6#F0\"\"#F;F4" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "Int(1/(x^2 - 5*x + 6), x=1..4);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%$IntG6$*&\"\"\"F',(*$)%\"xG\"\"#F'F'*&\"\"&F'F+F'!\"
\"\"\"'F'F//F+;F'\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Note th
at the denominator of this function is 0 at two places: " }{XPPEDIT 
18 0 "x = 2;" "6#/%\"xG\"\"#" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x = \+
3;" "6#/%\"xG\"\"$" }{TEXT -1 179 " , so we have to split the integral
 into (at least) 3 parts, evaluate them separately, and add them toget
her.  First we deal with the integral from 1 to 2, which is improper a
t 2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "inttoe := int(1/(x^
2 - 5*x + 6), x=1..2-epsilon);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'i
nttoeG,(-%#lnG6#,$%(epsilonG!\"\"F+-F'6#,&F+\"\"\"F*F+F/-F'6#\"\"#F+" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "part1 := limit(inttoe, ep
silon=0, right);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&part1G%)infinit
yG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 99 "We can stop: the first part
 of the integral does not exist, so the complete integral cannot eithe
r." }}}}}}{MARK "2 6 1 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }