Integrals with Singularities

Topics in Integration

3.7 Integrals with Singularities

(4) Sometimes it is possible to subtract from the integrandf(x) a function whose indefinite integral is known, and which has the same singularities as f(x). For the above example, xjX is such a function:

I

.b jX sin x dx

'0

.b . b . b

= I

Jx (sin x - x) dx

+ I

^{xJx dx}

⁼ I

Jx(sin x - x) dx

+

!b5 / 2 •

'0 ' 0 ' 0

The new integrand has a continuous third derivative and is therefore better amenable to standard integration methods. In order to avoid cancellation when calculating the difference sin x - x for small x, it is recommended to evaluate the power series

( 1 1 ) ⁰⁰ (-1)"

sin x - x

=

_x3 , - , x²

+ ... =

_x³

L

,x^{2v •} 3. 5. - v=o (2v

+

3).

(5) For certain types of singularities, as in the case of

.b

1 =

I

^{xj(x) dx, 0< IY. < 1,}

'0

withf(x) sufficiently often differentiable on [0, b], the trapezoidal sum T(h) does not have an asymptotic expansion of the form (3.4.1), but rather of the more general form (3.5.1):

T(h) '" ^To

+

T 1h^Yl

+

T2hY2

+ "',

where

{y;}

=

{I

+

^IY., 2, 2

+

^IY., 4, 4

+

^IY., 6, 6

+

IY., ••• }

[see Bulirsch (1964)]. Suitable step-length sequences for extrapolation methods in this case are discussed in Bulirsch and Stoer (1964).

(6) Often the following scheme works surprisingly well: if the integrand of

b

1

= f

^{f(x) dx}

a

is not, or not sufficiently often, differentiable for x

=

a, put aj:=a

+-.-,

b-a

J j

=

1,2, ... ,

in effect partitioning the half-open interval (a, b] into infinitely many subin- tervals over which to integrate separately:

Ij'=

(j

^{f(x) dx,}

.. aj+ 1

using standard methods. Then

1=/1+12 +/3 + ....

The convergence of this sequence can often be accelerated using, for instance, Aitken's ~2 method (see Section 5.10). Obviously, this scheme can be adapted to calculating improper integrals

r

_a ^{f(x} dx.}

(7) The range of improper integrals can be made finite by suitable var- iable transformations. For x

=

lit we have, for instance,

(' f(x) dx = (

~f(!)

^{dt .}

• 1 '0 t t

If the new integrand is singular at 0, then one of the above approaches may be tried. Note that the Gaussian integration rules based on Laguerre and Hermite polynomials (Section 3.6) apply directly to improper integrals of the forms

.00 +00

.10 f(x} dx,

Loo

^{f(x} dx,}

respectively.

EXERCISES FOR CHAPTER 3

1. Let a ,,;; Xo < XI < ... < X" ,,;; b be an arbitrary fixed partition of the interval [a, b]. Show that there exist unique numbers Yo, Yt. ... , Yn with

" .b

LYiP(Xi) =

I

P(x) dx

i=O "a

for all polynomials P with degree(P) ,,;; n. Hint: P(x) = 1, X, ... , x". Compare the resulting system of linear equations with that representing the polynomial inter- polation problem with support abscissas Xi' i = 0, ... , n.

2. By construction, the nth Newton-Cotes formula yields the exact value of the integral for integrands which are polynomials of degree at most n. Show that for even values of n, polynomials of degree n + 1 are also integrated exactly. Hint:

Consider the integrand ^X" + I in the interval [-k, + k], n = 2k + 1.

3. Iff E C²[a, b] then there exists an X E (a, b) such that the error of the trapezoidal rule is expressed as follows:

I

.b f(x) dx - i(b - a)(f(a) + f(b)) = -b(b - a)3j"(x).

Derive this result from the error formula in (2.1.4.1) by showing

'.

thatj"(~(x)) is continuous in x.

4. Derive the error formula (3.1.6) using Theorem (2.1.5.10). Hint: See Exercise 3.

5. Let f E C^{6 [ -} 1,

+

1], and let P E ITs be the Hermite interpolation polynomial with P(Xi) = f(x;), P'(x;) = !'(Xi), Xi = -1, 0, + 1.

(a) Show that

f

+1 P(t) dt = ~f( -1) + Mf(O) + ~f( + 1) + -fif'( -1) - -fif'( + 1).

'-1

(b) By construction, the above formula represents an integration rule which is exact for all polynomials of degree 5 or less. Show that it need not be exact for polynomials of degree 6.

(c) Use Theorem (2.1.5.10) to derive an error formula for the integration rule in (a).

(d) Given a uniform partition Xi = a + ih, i = 0, ... , 2n, h = (b - a)/2n of the interval [a, b], what composite integration rule can be based on the integra- tion rule in (a)?

6. Consider an arbitrary partition !l:= {a = Xo < ... < x. = b} of a given interval [a, b]. In order to approximate

I

.b ^{f(t) dt} 'Q

using the function valuesf(xi), i = 0; ... , n, spline interpolation (see Section 2.4) may be considered. Derive an integration rule in terms off(Xi) and the moments (2.4.2.1) of the" natural" spline (2.4.1.2a).

7. Determine the Peano kernel for Simpson's rule and n = 2 instead of n = 3 in [-1, + 1]. Does it change sign in the interval of integration?

8. Consider the integration rule of Exercise 5.

(a) Show that its Peano kernel does not change its sign in [-1, + 1].

(b) Use (3.2.8) to derive an error term.

9. Prove

using the Euler-Maclaurin summation formula.

10. Integration over the interval [0, 1] by Romberg's method using Neville's algor- ithm leads to the tableau

h5

=

1 Too = T(ho) Tll

hI

=!

^TlO = T(hl) T22

T21

h~

=-h

T20

=

T(h2) Tn

T31 h~

=-A

T30 = T(h3)

In Section 3.4, it is shown that Tll is Simpson's rule.

(a) Show that T22 is Milne's rule.

T33

(b) Show that T33 is not the Newton-Cotes formula for n = 8.

11. Let ho := 'b - a, hl := ho/3. Show that extrapolating T(ho) and T(hl) linearly to h = 0 gives the 3/8-rule.

12. One wishes to approximate the number e by an extrapolation method.

(a) Show that T(h) = (1 + h)llh, h

+

^0,

^!

^{h! <} 1, has an expansion of the form

<Xl

T(h) = e + LTihi

i= 1

which converges if

!

h! < 1.

(b) Modify T(h) in such a way that extrapolation to h = 0 yields, for a fixed value x, an approximation to eX.

13. Consider integration by a polynomial extrapolation method based on a geome- tric step-size sequence hj = ho bj , j = 0, 1, ... ,0 < b < 1. Show that small errors 11 Tj in the computation of the trapezoidal sums T(hJ,j = 0, 1, ... , m, will cause an error Tmm in the extrapolated value Tmm satisfying

!I1Tmm! ,;.;;; CmW) max !I1Tj!,

O~j!!S;m

where Cm(8) is the constant given in (3.5.5). Note that Cm(8) --> 00 as 8 --> 1, so that the stability of the extrapolation method deteriorates sharply as b approaches 1.

14. Consider a weight function w(x);?; 0 which satisfies (3.6.1a) and (b). Show that (3.6.1c) is equivalent to

I

• b ^{w(x) dx > O.}

Hint: The mean-value theorem of integral calculus applied to suitable subinter-

'.

vals of [a, b].

15. The integral

• + 1

(f, g) ,=

I

f(x)g(x) dx

'-1

defines a scalar product for functions f, 9 E C[ -1, + 1]. Show that iff and 9 are polynomials of degree less than n, if Xi, i = 1, 2, ., ., n, are the roots of the nth Legendre polynomial (3.6.18), and if

+1

I'i ,=

f

^{Li(x) dx}

. -I

with

i = 1,2, ... , n, then

n

(f, g) =

L

I'i f(Xi)g(XJ

i= 1

16. Consider the Legendre polynomials pAx) in (3.6.18).

(a) Show that the leading coefficient of Pj(x) has value 1.

(b) Verify the orthogonality of these polynomials: (Pi, Pj) = 0 if i < j.

Hint: Integration by parts, noting that

and that the polynomial

is divisible by x^{2 -} 1 if I < k.

17. Consider Gaussian integration, [a, b] = [-1, + 1], w(x) == 1.

(a) Show that (ji = 0 for i > 0 in the recursion (3.6.5) for the corresponding orthogonal polynomials pAx) (3.6.18). Hint: pAx) == (-l)j+l pj(-x).

(b) Verify

Hint: Repeated integration by parts of the integrand (x2 -

1Y

== (x + 1Y{x -

1Y.

(c) Calculate (Pj, Pj) using integration by parts (see Exercise 16) and the result (b) of this exercise. Show that

2 I ·2

Yi = (2i + 1)(2i - 1) for i > 0 in the recursion (3.6.5).

18. Consider Gaussian integration in the interval [-1, + 1] with the weight function w{x) == ~ 1-x 1 ^2'

In this case, the orthogonal polynomials pAx) are the classical Chebychev poly- nomials, To(x) == 1, TI (x) == x, T2(x) == 2X2 - 1, T3(X) == 4x3 - 3x, ... , 1j+I(X) == 2x1j{x) - 1j_I(X), up to scalar factors.

(a) Prove that pAx) == (1/2j-1 )1j(x) for j ;;,: 1. What is the form of the tridiagonal matrix (3.6.19) in this case?

(b) For n = 3, determine the equation system (3.6.13). Verify that WI = W2 =

W3 = rt13. (In the Chebychev case, the weights Wi are equal for general n.) 19. Denote by T(f; h) the trapezoidal sum of step length h for the integral

r

I f(x) dx.

"0

For ^(J. > 1, T(x'; h) has the asymptotic expansion

T(x"; h) ~

I

• I ^{x' dx + alhlh} + a2h2 + a4h4 + a6 h6 + ....

'0

Show that, as a consequence, every function f(x) which is analytic on a disk

I

z

I

~ r in the complex plane with r > 1 admits an asymptotic expansion of the form

.1

T(x'f(x); h)

-I

^{x'f(x)dx + b1h 1+. + b1hz+, + b3h3h + ...}

'0

+ c1h z + c4h4 + c6 h6 + ....

Hint: Expand f(x) into a power series and apply T(q> + 1/1; h) = T(q>; h) + T(I/I; h).