17 evaluation of improper integrals
TRANSCRIPT
Chapter 7: Evaluation of Improper Integrals
• Advice 1:
• Page No. 257: Q. Nos.: 1 - 5
exists.
RHSon limit theprovided
)(lim)(
then 0, x
allfor continuous is f(x)Let )1(
0 0
R
Rdxxfdxxf
-
)(then
x.allfor continuous is f(x)Let 2
dxxf
exist. RHSon limits both the provided
,)(lim)(lim0
1
2
021
R
R
RRdxxfdxxf
exist. RHS.on limit theprovided
,)( limdxf(x) P.V.
number theis (2) integral theof(P.V.) valueprincipalCauchy
R
R--R
dxxf
dxxf
dxxf
)(P.V.
of existence theimplies )(
integralimproper of Existence (1):Remark
-
not true. is converse But
02
xlim
xdxlimdx)x(f P.V.
x.Then f(x)Let .Ex
R
R
2
R
R
R- R
2lim
2lim
limlim
)(
22
2
21
1
2
02
0
11
RR
xdxxdx
dxxfBut
RR
R
RR
R
.existtofailsdx)x(f
integralimproper The
exist to fails RHS onLimit
If the function f(x) ( ) is an even function i.e. f(-x) =f(x) for all x, then the symmetry of the graph of y = f(x) with respect to y axis leads to
x
RR
R
dxxfdxxf0
)(2)(
When f(x) is an even function and the Cauchy pricncipal value exists, then
0
)(2)()(.. dxxfdxxfdxxfVP
To evaluate improper integral of Even Rational Functions
f(x)=p(x)/q(x)
• p(x) and q(x) are polynomials with real coefficients and no factors in common
• q(z) has no real zeros but has at least one zero above the real axis.
Method
• Identify all distinct zeros of the polynomial q(z) that lie above the real axis
• They will be finite in number
• May be labeled as z1, z2,…..zn where n is less than or equal to the degree of q(z)
• Now, integrate the quotient
f(z)=p(z)/q(z)
around the positively oriented boundary of the semicircular region.
The simple closed contour consists of• The segment of the real axis from
z = -R to z =R and
• The top half of the circle Rz described counterclockwise and denoted by CR.
Remark:
•The positive number R is large enough that the points z1, z2,…zn all lie inside closed path.
.
-R R
CR
O
ziz1
zfidzzfdxxfn
kRC
R
R
1 kzz
Res2)()(
From Cauchy Residue theorem
zfidxxfVPn
k
1 kzz
Res2)(..
followsit then
,0)(lim RC
RdzzfIf
If f(x) is even, then
zfidxxf
and
zfidxxf
n
k
n
k
1 kzz0
1 kzz
Res)(
Res2)(
022
2
4x1x
dxxIEvaluate4.Q
-R R
cR
RCURRCzz
,&41
z f(z)Let
22
2
R
R RC
skzz zfidzzf
xx
dxx
then
)(2)(41
Re22
2
C.region theoutside liezi- -i,zbut f(z) of 1 oforder of poles are2,
iizclearly
632
4)(Re
2
2
2
i
i
iziz
zzfs iz
iz
343
4
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i
i
izz
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iz
3/
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41
1
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xx
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R RC
4141)(
22
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22
2
zz
z
zz
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.
• …
i
22
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2
22
2
Rezon
)4)(1(R
R
4141)(
R
zz
z
zz
zzf
Ras
RRR
RR
RRdzzf
Hence
04
11
1
R L as )inequality L M(
41
.)(
22
22
2
CR
341
yieldswhich
22
2 xx
dxx
0
22
2
341
2 xx
dxx
0
22
2
641
xx
dxx
Sec 73 & 74
Advice 2:
Page No. 265, Q. Nos.: 1 and 6
0y plane halfupper in the bounded is
.e
modulus thefact that ith thetogether w(iaz
ayiaziaxayiyxia eeeee
R
R
R
R
iaxR
Rdxaxxfidxaxxfdxexf sin)(cos)()(
dxaxxf cos)(
Evaluation of improper integral of form and
dxaxxf sin)(
0,
cos1.
2222ba
bxax
dxxIQ
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cR
RCRRC
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iz
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yields
ab
22
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cos
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0y plane halfupper the in z points
allat analytic is f(z) function a (i)
that Suppose
:Lemma sJordan'
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.0lim
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42
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sin4
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SEC 78
• Advice 3:
• Page No. 280: Q. Nos. : 1 - 6
:cosines and sines involving integrals Definite
dF2
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integral heConsider t
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