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Numerical Integration
!Improper Integrals"Change of variable"Elimination of the singularity"Ignoring the singularity"Truncation of the interval"Formulas of Interpolatory and Gauss type"Numerical evaluation of the Cauchy Principal Value
!Indefinite Integration"Indefinite integration via Differential Equations"Application of Approximation Theory
Marialuce Graziadei
Ref. ‘Methods of Numerical Integration’, Philip J. Davis and Philip Rabinowitz.
6/11/2002 Seminar: Numerical Integration 2
Definitions
Improper integrals Integrals whose integrand is unbounded.
1) • is defined on
• is unbounded in the neighbourhood of
)(xf ( ];,ba
.ax =)(xf
∫∫ =+→
b
rar
b
a
dxxfdxxf )(lim)(
2)• is defined on
• is unbounded in the vicinity of with
The Cauchy Principal Value of the integral is defined by the limit
,x c= .bca <<)(xf [ ] { }, \ ;a b c
)(xf
[ ]∫+∫∫ =+
−
→ +
b
rc
rc
ar
b
a
dxxfdxxfdxxfP )()()(0
lim
6/11/2002 Seminar: Numerical Integration 3
Change of variableSometimes it is possible to find a change of variable that eliminates the singularity.
[ ]1,0)( Cxg ∈Example 1
∫ ≥=−1
0
1
2)( ndxxgxI n
The change of variable transforms the integral intoxnt =
∫= −1
0
2 )( dttgtnI nn
But (example 2)1
0
log( ) ( )I x g x dx= ⋅∫ lo gt x= −
becomes
0
( )t tI te g e d t∞
− −= − ∫ Infinite range of integration
which is proper.
with
6/11/2002 Seminar: Numerical Integration 4
Elimination of the singularity
General ideas: subtract from the singular integrand a function! integral is known in closed form;! is no longer singular.This means that has to mimic the behaviour of closely to its singular point.
)(xf ( ) .g x
)()( xgxf −
)( xg )(xf
Example
.1cos21cos1cos 1
0
1
0
1
0
1
0∫
−+∫ ∫ =−+∫ = dxxxdx
xxdx
xdx
xx
But
near , so the last integrand in now in [ ]1,0C2
1cos2xx −≈−
0=x
( )g x
6/11/2002 Seminar: Numerical Integration 5
Ignoring the singularityIt is also possible to avoid the integrand singularities and apply the standard quadrature rules.We want to compute
where is unbounded in the neighbourhood of
!Then we set (or any other value) and use any sequence of rules.
!Another option: use a sequence of rules that do not involve the value of at
,)(1
0∫ dxxf
)( xf .0=x
.0=x)( xf
0)0( =f
6/11/2002 Seminar: Numerical Integration 6
Ignoring the singularityExample 1
32 × S 1.8427 G2 1.65068
64 × S 1.8887 G3 1.75086
128 × S 1.9213 G4 1.80634
256 × S 1.9444 G10 1.91706
512 × S 1.9606 G16 1.94722
1024 × S 1.9721 G32 1.97321 S = Simpson G = Gauss
1
0
2d xx
=∫
But the method of ‘ignoring the singularity’ may not work if the integrand is oscillatory
1 1
0 0
1 1 1 s ins in ( 2 ) .6 2 4 7 1 32
xd x d xx x x
π= − =∫ ∫
No patterns of convergence is discernible
from this computations
32 × S 2.3123
64 × S 1.6946
128 × S -0.6083
256 × S 1.2181
512 × S 0.7215
1024 × S 0.3178 S = Simpson
Example 2
6/11/2002 Seminar: Numerical Integration 7
Ignoring the singularity
∑==
m
kkk xfwfR
1)()(
However let designate a fixed m-point rule of approximate integration in ]1,0[R
∑ =>≤<<<≤=
m
kkkm wwxxx
121 ,1,0,1...0with
and let designate the compound rule that arises by applying to each of the subintervals
nR R
[ ] [ ] [ ] ,1,/)1(,...,/2,/1,/1,0 nnnnn − then
TheoremIf is a monotonic increasing integrable singular function with a singularity at then
)( xf,0=x
.)()(lim1
0∫=
∞→dxxffR nn
0 1n/1 n/2 nn /)1( −
6/11/2002 Seminar: Numerical Integration 8
Proceeding to the limit
Integral to be evaluated:
! is continuous in (may be unbounded in ).! is a sequence of points that converges to 0 (e.g ).
∫1
0
)( dxxf
)(xf 10 ≤< x 0=x...1 21 >>> rr n
nr−= 2
...)()()()(2
3
1
21
11
0∫ +∫ +∫ +∫ =r
r
r
rrdxxfdxxfdxxfdxxf
proper integrals
The evaluation is terminated when
ε≤∫+1
)(n
n
r
r
dxxf
6/11/2002 Seminar: Numerical Integration 9
Truncation of the interval1 1
0 0
( ) ( ) ( )r
r
f x d x f x d x f x d x= +∫ ∫ ∫Then, if
we can simply evaluate the proper integral
ε≤∫r
dxxf0
)(
Example∫1
)(r
dxxf
∫+
=1
0 31
21
)( dxxx
xgI 1)( ≤xgwith g bounded in e.g.,0 ,1
But in [ ]1,0
21
31
21
2
1)(
xxx
xg ≤+
∫ =≤∫+
rr
rx
dxdxxx
xg0
21
21
0 31
21
2
)(
And, if we take , we get an accuracy of 610−≤r 310−
6/11/2002 Seminar: Numerical Integration 10
Integration Formulas of Interpolatory TypeConsider the integral
∫1
0
)()( dxxfxw
where is a function with a singularity in the neighbourhood of , but such that
)(xw 0=x
∫1
0
)( dxxxw k exist for .,...,1,0 nk =
1
00
( ) ( ) ( )n
i ii
w x p x d x w p x=
= ∑∫
The, for a given sequence of abscissas , we can determine weights such that
1...0 10 ≤<<< nxxxiw
whenever .np P∈This leads to the approximate integration formula
∑≈∫=
n
iii xfwdxxfxw
0
1
0)()()(
6/11/2002 Seminar: Numerical Integration 11
Integration Formulas of Interpolatory TypeExample
12( )w x x
−= 0 1 2
1 2, , 1.3 3
x x x= = =
1 12
1 2 30
1 11 2 2
30
2,
2 2 ,3 3 3
w w w x dx
w w w x xdx
−
−
+ + = =
+ + = =
∫
∫
.52
94
92
1
0
21
321 =∫=++
−
dxxxwww
This leads to the rule
( )1 1
2
0
14 1 8 2 4( ) 1 .5 3 5 3 5
x f x dx f f f− ≈ − +
∫
6/11/2002 Seminar: Numerical Integration 12
Integration Formulas of Gauss TypeSingularities may be accommodated by means of Gauss-type formulas. The integral is written in the form
,)()(∫=b
a
dxxfxwIwhere is a fixed positive weight function. The moments)( xw
∫b
a
n dxxxw )( exist for ,...1,0=nbut may have one or more singularities in the interval)( xw [ ]ba ,The corresponding orthonormal polynomials are and their zeros are)( xp n
nwww ,...,, 21Then (positive constants) can be found such thatbxxxa n <<<<< ...21
∑=∫=
n
kkk
b
a
xpwdxxpxw1
)()()(
∑≈∫=
n
kkk
b
axfwdxxfxw
1)()()(
6/11/2002 Seminar: Numerical Integration 13
Integration Formulas of Gauss TypeWe want to compute the integral
1
0
( ) ( ) ,I L o g x f x d x= ∫• is regular in [0,1]( )f x
We need[0,1].
1
0
( )( ) .nnI L o g x x d x= −∫
yx e−=•
• integration by parts
( 1 )2
0
1 11 ( 1)
m ymI e d y
m m
∞− += =
+ +∫
To get them, we must solve( )nG x polynomials orthonormal to in( )Log x
By Mme Henri Berthod Zabrowski
6/11/2002 Seminar: Numerical Integration 14
Integration Formulas of Gauss Type
• Points • Weights
1
0
( ) i j i jL o g x G G d x δ− =∫• Polynomials
( )
( )
0
1
2
2
3 2
3
1,
12 1 ;47
5 252 180 17;
12 7
7 258800 310500 92016 4679;
9 10849 647
G
G x
x xG
x x xG
=
= −
− +=
− − −=
…
10.25
0.6022770.112009
0.2814610.718539
0.5134050.3919800.094615
0.0638910.3689970.766880
6/11/2002 Seminar: Numerical Integration 15
Integration Formulas of Gauss Type
• Example1
0
( ) 0 .8 2 2 4 6 7 0 .1 1 2
L o g xI d xx
π= − = =+∫
n=3
7
0 .8224485
185 10com puted
exact com puted
I
I I −
=
− = ⋅
6/11/2002 Seminar: Numerical Integration 16
Numerical Evaluation of the Cauchy Principal Value
Reduction of the CPV to one-sided improper integral is possible.
)( xf cx = .bca <<unbounded in with
Suppose that ∫b
a
dxxfP )( exists.
0( ) lim ( ) ( )
b c r b
ra a c r
P f x f x dx f x dx−
→+
= +
∫ ∫ ∫
Consider
Decompose in its odd and even parts
Odd Even
0c =)( xf
.b a=and
[ ])()(21)( xfxfxg −−= [ ])()(
21)( xfxfxh −+=
6/11/2002 Seminar: Numerical Integration 17
Numerical Evaluation of the Cauchy Principal Value
.2 )(
)()()()(
)()(
∫
=∫ ∫+∫+∫
=∫+∫−
−
−
−
−
−
+a
r
a
r
a
r
r
a
r
a
a
r
r
a
dxx
dxxdxxdxxdxx
dxxfdxxf
h
hhgg
Therefore
0.( ) 2lim ( )
a a
ra r
P f x dx h x dx+→
−
=∫ ∫
6/11/2002 Seminar: Numerical Integration 18
Numerical Evaluation of the Cauchy Principal Value
∫−
1
1 xdxP
Example1
01
1
=∫− x
dxP( ) 01121 =
−=
xxxh
Example2
dxx
ePx
∫−
1
1
( ) ( )xxx
exexh
xx
sinh121 =
−+=
−
xdx
xxdxeP
x
∫=∫−
1
0
1
1
)sinh(2
6/11/2002 Seminar: Numerical Integration 19
Numerical Evaluation of the Cauchy Principal Value
The method of subtracting the singularity may also be used.( )( ) ,
b
a
f tI x P d t a x bt x
= < <−∫
Hilbert transform of
)( xf( ) ( )( ) ( )
( ) ( ) ( ) log .
b b
a a
b
a
f t f x dtI x dt f x Pt x t xf t f x b xdt f xt x x a
−= + =− −− −+− −
∫ ∫
∫Consider the function
( ) ( )( , ) ,
( , ) '( ) .
f t f xt x t xt x
x x f x t x
φ
φ
−= ≠−
= =band solve∫a
dtxt ),(φ∗ Interpolatory-type and Gauss-type formulas have been developed for Cauchy Principal Value integrals.
6/11/2002 Seminar: Numerical Integration 20
Numerical Evaluation of the Cauchy Principal Value
It may be useful to consider
( ) ( )( , ) .x h h
x h h
f t x f xt x dt dtt
φ+
− −
+ −=∫ ∫If can be expanded in a Taylor series at then we have( )f x ,t x=
2
3
''( ) '''( )( , ) '( ) ...2! 3!
'''( )2 '( ) ...9
x h h
x h h
t f x t f xt x dt f x dt
h f xhf x
φ+
− −
= + + +
= + +
∫ ∫
6/11/2002 Seminar: Numerical Integration 21
Indefinite IntegrationWe want to compute
bxadttfxFx
a≤≤= ∫ )()(
or the more complicated
bxadttxfxFx
a≤≤= ∫ ),()(
Two choices
•Regard the as a definite integral over a variable range;
•Regard as the solution of the differential equation
)( xF
)( xF
( ), ( ) 0.dF f x F adx = =
The simplest approach:
•divide the interval of integration into a set of subintervals;
•apply a rule of approximate integration to each subinterval.
Simpson’s rule is widely used
bxa ≤≤
6/11/2002 Seminar: Numerical Integration 22
Indefinite Integration via Differential EquationsWe can use familiar rules. Consider, for example, the classical Runge-Kutta method for the solution of
.)(),( 00 yxyyxgdxdy ==
The relevant formulas are
1 1 2 3 4
11 2
23 2 3
( 2 2 ),6
( , ), ( , ),2 2
( , ), ( , ).2 2
m m
m m m m
m m m m
hy y k k k k
hkhk g x y k g x y
hkhk g x y k g x h y hk
+ = + + + +
= = + +
= + + = + +
A general multistep method for indefinite integration would consist in computing the value of the integral at the next step, in terms of the values of the integral previously computed, and in terms of the values of the integrand,
1+ny,...,, 1−nn yy
),...(),(),( 11 −+ nnn xfxfxf
6/11/2002 Seminar: Numerical Integration 23
Indefinite Integration-Approximation Theory
∞<<<∞−≤≤= ∫ babxadttfxFx
a)()(
Suppose we can approximate with )( xf
,)()(...)()()( 10 bxaxxxxxf n ≤≤++++= εφφφ
( ) ,x a x bε ε≤ ≤ ≤
( ) ( )x
i ia
x t dtψ φ= ∫Then
and that
is simple to calculate.
),()(...)()()()( 10 xxxxdttfxF n
x
aηψψψ ++++∫ ==
with .)()()( εεη abdttxx
a
−≤∫=
6/11/2002 Seminar: Numerical Integration 24
Indefinite Integration-Approximation TheoryChebyshev Polynomials
11...,,1,0
...)1(2
)coscos()( 22
≤≤−=
+−
+=⋅= −
xn
xxn
xxarnxT nnn
or
0 1
1 1
( ) 1, ( ) ,( ) 2 ( ) ( ), 2,3,...,n n n
T x T x xT x xT x T x n+ −
= == − =
If the function satisfies a Lipschitz condition in it can be expanded in an uniformely convergent series of Chebishev polynomials.
)( xf [ ]1,1 ,−
...)()(21)( 22110 +++= xTaxTaaxf
6/11/2002 Seminar: Numerical Integration 25
Indefinite Integration-Approximation Theory
The coefficients of the series are given by
Orthogonality
1
21
, 0,( ) ( ) , 0,
210, .
m n
m nT x T x dx m n
xm n
ππ
−
= == = ≠
− ≠
∫
∫∫−
=−
=π
ϑϑϑππ 0
1
12
cos)(cos21
)()(2 drfdxx
xTxfa rr
For many functions the sequence decreases to zero rapidly. So,..., 10 aa
0 1 1 11 2
2
( ) ( )( ) ( ) ( )2 4 2 1 1
r r r
r
a a a T x T xf t dt T x T x consr r+ −
=
= + + − + + − ∑∫
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