NAG Toolbox: nag_quad_1d_indef (d01ar)

Purpose

Syntax

Description

Definite Integrals

Indefinite Integrals

References

Parameters

Compulsory Input Parameters

1:     $\mathrm{a}$ – double scalar

$a$, the lower limit of integration.

2:     $\mathrm{b}$ – double scalar

$b$, the upper limit of integration. It is not necessary that $a<b$.

3:     $\mathrm{fun}$ – function handle or string containing name of m-file

fun must return the value of the integrand $f$ at a specified point.

[result] = fun(x)

Input Parameters

1:     $\mathrm{x}$ – double scalar

The point in $\left[a,b\right]$ at which the integrand $f$ must be evaluated.

Output Parameters

1:     $\mathrm{result}$ – double scalar

The value of $f\left(x\right)$ evaluated at x.

If $\mathbf{iparm}=2$, fun is not called.

4:     $\mathrm{relacc}$ – double scalar

The relative accuracy required. If convergence according to absolute accuracy is required, relacc should be set to zero (but see also Accuracy). If $\mathbf{relacc}<0.0$, its absolute value is used.

If $\mathbf{iparm}=2$, relacc is not used.

5:     $\mathrm{absacc}$ – double scalar

The absolute accuracy required. If convergence according to relative accuracy is required, absacc should be set to zero (but see also Accuracy). If $\mathbf{absacc}<0.0$, its absolute value is used.

If $\mathbf{iparm}=2$, absacc is not used.

6:     $\mathrm{maxrul}$ – int64int32nag_int scalar

The maximum number of successive rules that may be used.

Constraint: $1\le \mathbf{maxrul}\le 9$. If maxrul is outside these limits, the value $9$ is assumed.

If $\mathbf{iparm}=2$, maxrul is not used.

7:     $\mathrm{iparm}$ – int64int32nag_int scalar

Indicates the task to be performed by the function.

$\mathbf{iparm}=0$

Only the definite integral over $\left[a,b\right]$ is evaluated.

$\mathbf{iparm}=1$

As well as the definite integral, the expansion of the integrand in Legendre polynomials over $\left[a,b\right]$ is calculated, using the same values of the integrand as used to compute the integral. The expansion coefficients, and some other quantities, are returned in alpha for later use in computing indefinite integrals.

$\mathbf{iparm}=2$

$f\left(t\right)$ is integrated analytically over $\left[a,b\right]$ using the previously computed expansion, stored in alpha. No further evaluations of the integrand are required. The function must previously have been called with $\mathbf{iparm}=1$ and the interval $\left[a,b\right]$ must lie within that specified for the previous call. In this case only the arguments a, b, iparm, ans, alpha and ifail are used.

Constraint: $\mathbf{iparm}=0$, $1$ or $2$.

8:     $\mathrm{alpha}\left(390\right)$ – double array

If $\mathbf{iparm}=2$, alpha must contain the coefficients of the Legendre expansions of the integrand, as returned by a previous call of nag_quad_1d_indef (d01ar) with $\mathbf{iparm}=1$ and a range containing the present range.

If $\mathbf{iparm}=0$ or $1$, alpha need not be set on entry.

Optional Input Parameters

None.

Output Parameters

1:     $\mathrm{acc}$ – double scalar

If $\mathbf{iparm}=0$ or $1$, acc contains the absolute value of the difference between the last two successive estimates of the integral. This may be used as a measure of the accuracy actually achieved.

If $\mathbf{iparm}=2$, acc is not used.

2:     $\mathrm{ans}$ – double scalar

The estimated value of the integral.

3:     $\mathrm{n}$ – int64int32nag_int scalar

When $\mathbf{iparm}=0$ or $1$, n contains the number of integrand evaluations used in the calculation of the integral.

If $\mathbf{iparm}=2$, n is not used.

4:     $\mathrm{alpha}\left(390\right)$ – double array

If $\mathbf{iparm}=1$, the first $m$ elements of alpha hold the coefficients of the Legendre expansion of the integrand, and the value of $m$ is stored in $\mathbf{alpha}\left(390\right)$. alpha must not be changed between a call with $\mathbf{iparm}=1$ and subsequent calls with $\mathbf{iparm}=2$.

If $\mathbf{iparm}=2$, the first $m$ elements of alpha are unchanged on exit.

5:     $\mathrm{ifail}$ – int64int32nag_int scalar

$\mathbf{ifail}=\mathbf{0}$ unless the function detects an error (see Error Indicators and Warnings).

Error Indicators and Warnings

Cases prefixed with W are classified as warnings and do not generate an error of type NAG:error_n. See nag_issue_warnings.

W  $\mathbf{ifail}=1$

If $\mathbf{iparm}=0$ or $1$, this indicates that all maxrul rules have been used and the integral has not converged to the accuracy requested. In this case ans contains the last approximation to the integral, and acc contains the difference between the last two approximations. To check this estimate of the integral, nag_quad_1d_indef (d01ar) could be called again to evaluate

$$\underset{a}{\overset{b}{\int }}f\left(t\right)dt\text{\hspace{1em} as \hspace{1em}}\underset{a}{\overset{c}{\int }}f\left(t\right)dt+\underset{c}{\overset{b}{\int }}f\left(t\right)dt\text{\hspace{1em} for some }a<c<b\text{.}$$

If $\mathbf{iparm}=2$, this indicates failure of convergence during the run with $\mathbf{iparm}=1$ in which the Legendre expansion was created.

   $\mathbf{ifail}=2$

On entry, $\mathbf{iparm}\ne 0$, $1$ or $2$

   $\mathbf{ifail}=3$

The function is called with $\mathbf{iparm}=2$ but a previous call with $\mathbf{iparm}=1$ has been omitted or was invoked with an integration interval of length zero.

   $\mathbf{ifail}=4$

On entry, with $\mathbf{iparm}=2$, the interval for indefinite integration is not contained within the interval specified when nag_quad_1d_indef (d01ar) was previously called with $\mathbf{iparm}=1$.

   $\mathbf{ifail}=-99$

An unexpected error has been triggered by this routine. Please contact NAG.

   $\mathbf{ifail}=-399$

Your licence key may have expired or may not have been installed correctly.

   $\mathbf{ifail}=-999$

Dynamic memory allocation failed.

Accuracy

Further Comments

Example

Open in the MATLAB editor: d01ar_example

function d01ar_example


fprintf('d01ar example results\n\n');

absacc = 1e-05;
relacc = 0;
maxrul = int64(0);
alpha  = zeros(390,1);

% Integral 1
iparm = int64(0);
a = 0;
b = 1;

[acc1, ans1, n1, alpha, ifail] = ...
    d01ar(...
           a, b, @f1, relacc, absacc, maxrul, iparm, alpha);

fprintf('Definite integral of 4/(1+x*x) over (0,1)\n');
fprintf('Estimated value of the integral = %9.5f\n', ans1);
fprintf('Estimated absolute error        = %10.1e\n', acc1);
fprintf('Number of points used           = %4d\n\n', n1);

% Integral 2
iparm = int64(1);
a = 1;
b = 2;

[acc2, ans2, n2, alpha, ifail] = ...
    d01ar(...
           a, b, @f2, relacc, absacc, maxrul, iparm, alpha);

fprintf('Definite integral of x^(1/8) over (1,2)\n');
fprintf('Estimated value of the integral = %9.5f\n', ans2);
fprintf('Estimated absolute error        = %10.1e\n', acc2);
fprintf('Number of points used           = %4d\n\n', n2);

% Integral 3
iparm = int64(2);
a = 1.2;
b = 1.8;

[acc3, ans3, n3, alpha, ifail] = ...
    d01ar(...
           a, b, @f2, relacc, absacc, maxrul, iparm, alpha);

fprintf('Indefinite integral of x^(1/8) over (1.2,1.8)\n');
fprintf('Estimated value of the integral = %9.5f\n', ans3);



function f = f1(x)
  f = 4/(1+x^2);

function f = f2(x)
  f = x^(1/8);

d01ar example results

Definite integral of 4/(1+x*x) over (0,1)
Estimated value of the integral =   3.14159
Estimated absolute error        =    1.8e-08
Number of points used           =   15

Definite integral of x^(1/8) over (1,2)
Estimated value of the integral =   1.04979
Estimated absolute error        =    5.9e-07
Number of points used           =    7

Indefinite integral of x^(1/8) over (1.2,1.8)
Estimated value of the integral =   0.63073