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Unformatted text preview: Improper Integrals You are familiar with integrals [Suede where f is bounded and [£1,131 is a bounded interval. Now we allow [(1, b] to be
replaced by an unbounded interval  or f to be unbounded. Integrals over Unbounded Intervals Deﬁnition b r ‘ 9'
Suppose that f is continuous on [0,, 00). We can form fa f dm for all I) > 0..
We deﬁne the improper integral famf(m)da:= lim fﬂmm b—mo if the limit exists and is ﬁnite. Then we say the improper integral f: f do:
converges. Otherwise we say that the improper integral diverges. Remark
If the improper integral converges, its value is the limit above. 00
/ e_wda:
0 converge or diverge? If it converges1 ﬁnd its value.
Solution
The function f = (2—3 is continuous on {0, 00). So we examine Example 1
Does 1:
lim 6.7de b—roo G
. _z m=b
= b11130 [—8 lm=0
= ljm [—e"" + 1]
b—ioo
= 0 + 1 = 1. Thus the improper integral fooo (raids: converges and equals 1. Example 2
Determine for which 13 > 0, converges.
Solution
Here rm:— gap is continuous on [1, 00), and we must check whether 1:
1
lim —dm (2600 1 3p exists and is ﬁnite. Recall that I: Eli(ix has different formulas for p = 1 and
p 75 1, so we consider them separately. “as p = 1
Here
. b 1
/ —dm = lnb
1 m
so
5 1
lim —dx = lim (1n b) = 00.
b—roo 1 m b—too So in this case, the integral diverges, that is the improper integral °° 1
f —d:I:
1 m
diverges. p 79 1
Here 1; 1 b
—d:c / 93—pdx
1 1 mp
[ $—p+1 r25
_p + 1 36:1 b—pil 1
—p+ 1 —p+ 1' How will this behave as b "—4 00? To resolve this, we recall that a positive power
of 5 goes to 00 as b —> 00. Also, a negative power of b goes to 0 as b —> 00. So it
all depends on whether the power —p + 1 is positive or negative. We consider these cases separately:
a) —p 1 > O Weseethat
—p+1>0<=>1>p. 28 Then
lim b‘p‘H z oo b—roo
so b
—p+1
lim i05:1: : lim b —— 1 =
5"“) 1 931’ (HOG —P+l Wp+1
Thus in this case, where p < l, the improper integral diverges.
b) —p + 1 < 0
We see that
—p + 1 < 0 <=> 1 < 33. Then lim b‘p'l'l = 0 b—roo
so b 1 bP+1 1
lim f —dm = lim —
b—roo 1 $33 b—roo —p+1 ep+1
1
z 0 _
"P + 1
_ l
__ p m 1, Thus in this case, Where p > 1, the improper integral converges, and Let us summarize all the above: Summary 11.7.1 (page 566 of SHE)
The improper integral converges for p > 1, and diverges for p S 1. For 32 > 1, the improper integral
1
equals 1:. Remarks _
(a) If C is a real number, then [lef(m)dm=C/amf(m)dm' provided the improper integral on the right converges. Thus you can pull a
constant out of a convergent improper integral. In fact if C is non—zero, and the
integral on the right diverges, so does that on the left. Thus the convergence of
either integral is equivalent to the convergence of the other. 29 There is also the familiar "the improper integral of the sum is the sum
of the improper integralsz" [:0Old?)+Q'(33))darELWf($)dm+Loog(m)dm provided BOTH improper integrals on the right converge.
(b) Note that if f is continuous,
/ new
1 00 f (a?) div 100
converges. In other words, the lower limit does not affect the convergence (only
the value of the integral). So we can often replace lower limits by some suitable large number.
((1) Quite often, we cannot explicitly calculate an improper integral, but just
need to know if it converges. The comparison test can be used for this. will converge if and only if Theorem 11.7 .2 The Comparison Test (Page 567 of SHE, 10th edn.)
Suppose that f and g are continuous, and  O S f(m) S g(a:) for all r E [a,oo). (I)
f g dm converges => / f da: converges.
a G (H) DO 00
f f d2: diverges => / g d2: diverges. Proof The basiclidea for (I) is that f is smaller than 9. So the area under (parts of)
the graph of f is smaller than that for 9. Hence if eventually the area under the
graph of g from b to 00 is becoming small, the same is true for that of f. I Example 1
Determine whether the improper integral DO
/ (2 + marl/‘4 d3:
1 converges.
Solution
The idea is to see how the function ﬂm=o+ﬁ) —1/4 30 behaves for large m, and then compare it to something we know. First of all
note that it is a continu0us function on [1, 00). We see that for large as, $8 is
much larger than 2. Thus we expect f to behave roughly like (5034/4 = 3—2. This should suggest using 11.7.1: 00 0C} 1
f x_2dm = / —2dm converges as the exponent p = 2 > 1.
1 1 m How do we make this rigorous? We apply the comparison test with g(x) = x_2.
We see that
2 + $8 2 m8
=> (2+23)”1/4 5(9:8)_1/4 259—2
=> ft?) S 9013)
Since
OSf(m) Sg(s:),z E [1,00)
and since [mg(m)dm=/loom_2dcc floof(w)dm=/;oo (2+m8)_1/4dw converges, 50 also converges. Example 2
Determine whether the improper integral / (TI' + m3)_1/4 d3
0 converges.
Solution
The idea is to see how the function W) = (H933) behaves for large m, and then compare it to something we know. First of all
note that it is a continuous function on [0, 00). We see that for large m, :53 is
much larger than 11'. Thus we expect f to behave roughly like —1/4 (m3)1/4 : 373/4. 31 This should suggest using 11.7.1: 00 0° 1 3
f1 $‘3/4dm = j; E72653 diverges as the exponent p = Z < 1' To make this rigorous, we see that for m 2 2, m3 2 8 > 7r so
77+m3 5 r3+z3 =2ma,
so for so 2 2,
f = (71' £— adj—U4 2 (251:3)_1/4 : 2wl/4m“3/4. Here by 11.7.1 and our other remarks,
()0 00
/ 2—1/4x"3/4dm : 2—1/4 f m_3/4dx diverges.
2 2 By the comparison test, finedwf: («+m3)_1/4d93 diverges. Then by our remarks, co 3 —1/4 .
(17 + m ) dm diverges also.
0 In much the same way we can deﬁne improper integrals over (—oo,a] and we
can also deﬁne integrals over (—00, oo) : Deﬁnition
Let f be continuous in (—00, a]. We deﬁne the improper integral 1:0f(m)da2=b.14ir_nooj:f(m)dm '6— if the limit exists and is ﬁnite. In this case, we say the integral converges. Oth
erwise it diverges. DeﬁnitiOn
Let f be continuous in (—00, 00). We deﬁne the improper integral I £:f(sr)dm=£:f(x)dm+me(w)dcc provided BOTH integrals on the right converge. In this case, we say that the
improper integral / f d1]: converges.
—oo 32 If either 0
f f d9: diverges or f f d2: diverges,
—oo 0 then we say that
f f d2: diverges.
—oo Example
Determine if 00 en:
La 1 + e2md$ converges or diverges. If it converges, determine its value.
Solution Here w
e
f (m) _ 1 4 62”” is continuous (and positive). So we test both and [omf($)dm. few f dx
We must check if the limit b—boo O 11m f d3
b—mo 0
b a:
2 Km e_2dm
5*“ o 1 + (em)
:5
. e 1
— blip; 80 1 "F U2 du (we used the substitution u = 6””, so do = emdm) b
_ . _1 “=3
= lim [tan—1 eb — tan—1 1]
15—400
71'
= lim ten‘1 eh — —
13—900 4
_ 1r 7r _ 7r
* 2 4 _ 4' 33 (Recall that tan‘1 2 Thus the improper integral j? ﬁgd'l: converges, and equals :2.
fan f (2:) dx
We must check if the limit
b—i—oo b “113100 h f (56) d3
“ lim 0 ex
’H—W b 1 4 (6'7")2
30
_ 1
 ab 118261“
(we used the substitution 1.: = 8“, so (in = emdic)
m . _1 11:1
_. b35100 [tan 15] “Mb
= lim tan"1 1 — talf1 ab]
b—r—oo
= E — lim tan—1 6”
4 bur—co
_ 35 _ —1 = E
— 4 tan 0 4. Recall that tan'1 0 r 0 . Thus the improper integral _ 6:: d2: converges,
co 1+8
and equals in Then 00 ea: 0 8;; 00 ea,
(11: d3: dz;
1°01+e2m fml+ezm +1; 1+6?” 71' 71' ‘11" 4 + 4 2
and the improper integral If; ffgﬁd'z: converges, and equals Integrals of Unbounded Emotions We deﬁne integrals such as
1
l
f new
a 3” where p > 0. rI'he function f(.13) : 1/35” is continuous in (0, 1] but unbounded
there. Deﬁnition
Let f be continuous on (a, b] but unbounded on (a, b]. We deﬁne the improper
integral
b _ b
' [Hedwig] f(x)drc if this limit exists and is ﬁnite. Then we say that the improper integral f dag
converges. Otherwise it diverges. Example
Let p > 0. Investigate whether “d5:
0 3p
converges or diverges. Solution
Now f = 1/93? is continuous in {0,1} but unbounded there. So we check if 1 1
lim —dm
c—+0+ c ﬂip
exists and is ﬁnite. Recall that there are different formulas for f z—lpdm if p = 1
and p gé 1. p=1 .
Hereif0<c<l,
1
f ldc.~:=1111—1nc:—lnc :12 so 1
lim 3oh: 2 lim (wlnc) = 00.
c—*O+ c :1: c—r0+ So in this case, the integral diverges, that is the improper integral 1
f 1do:
0 3
diverges. p as 1
Here 1 1
ids: / 2:"pdm
C [ m_:p_i_1 ]m=1
WP + 1 9::c H How will this behave as c —> 0+? To resolve this we recall that a positive power
of 6 goes to 0 as c —> 0+. Also, a negative power of 6 goes to 00 as c —+ 0+. SO
it all depends on whether the power —p + 1 is positive or negative. We consider
these cases separately: a) —p+ l > 0
We see that
—p+ 1 > 0 <2} 1 > p.
Then
lim cwp+1=0
040+
so 1 +1
1 —P _
lim —dx=lim 1 —C 2 1 =;.
<:—»0+c m? c—>0+ —p+1 —p+l ep+1 1—}? Thus in this case, where p < 1, the improper integral converges and 1
i03:1: = 0 3” 1 —P
b) —p + l < 0
We see that
—p+ 1 < 0 <=> 1 < 10.
Then
lim 6"?“ = 00
c—>O+
so
1 _
1 1 p+1
lirn ——d$ = lim — C
c——>0+ ,3 mi" c—+ﬂ+ —p+1 up+1
= 00. Thus in this case, where p > 1, the improper integral diverges (to 00). Let us
summarize all the above: Summary
The improper integral 1
1
/ —dm
0 33” diverges for p _>_ 1, and converges for p < 1. For p < 1, the improper integral equals 1—}5. Remark ,
In exactly the same way, we can deﬁne the improper integral ofa function f that is continuous in [(1, b) but unbounded as we approach I) : 36 Deﬁnition
Let f be continuous on [9,, b) but unbounded on [(1.6). We deﬁne the improper integral b
farengggfnmm if this limit exists and is ﬁnite. Then we say that the improper integral f: f (1:) d2:
converges. Otherwise it diverges. So far we have dealt with the case where f becomes unbounded at an endpoint
(a. or b) of the interval of integration. What happens if it becomes unbounded
inside? Then we split at the point where it is unbounded, and deﬁne the im
proper integral as a sum of two improper intregrals: Deﬁnition
' Let f be continuous on [0,,b] except at some point e in (11,13). We deﬁne the
improper integral fussmmfﬁemmfnwm provided BOTH improper integrals on the right converge. Then we say that the
improper integral 1: f 03$ converges. Otherwise it diverges. Example
Does the improper integral 1
1
(1’1:
[—1 [3511/2 '
converge or diverge? Solution “‘1
Here f = 121—1172 is continuous in [—1, 1] except at .1: = 0. So We split 1 1
1 l 1
(its: [0 da:+ f d1}:
[1 :1:1/2 H1 Inf/2 0 1241/2 and must check if both improper integrals on the right converge. Firstly 1
fa Wax
Since x :1: for :1: > O, l l
1 1
(LT: ——‘—d.'1:
f0 {sf/2 [0 $112 and by the summary above, this converges (p = i < 1) and equals 0
L1 de Since _: ——$ for m > O,
0  0
1 f 1 .
aim = d3:
.[qmu2 _m—mU2
—‘ 1' c 1 d
— ail6L _1 (_m)1/2 a:
1
, 1
a salt}?— _C (substitution t = —a:) 1
35%
H
H
: H
N)
E:

UN by the above. Theﬁ 1 1
d3: converges
V/m1 ﬁll/2 1 I 0 r 1
1 f 1 f 1
d2: dm + dm
11m“? _mmV2 0 m“?
2+2=¢ and I! 38 ...
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This note was uploaded on 09/07/2010 for the course CALC 1502 taught by Professor Morley during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Morley

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