﻿FunctionMath.Integrate Method (Func(Double, Double), Double, Double, IntegrationSettings)

Evaluates a definite integral with the given evaluation settings.

Namespace:  Meta.Numerics.Analysis
Assembly:  Meta.Numerics (in Meta.Numerics.dll) Version: 4.1.4
Syntax
```public static IntegrationResult Integrate(
Func<double, double> integrand,
double start,
double end,
IntegrationSettings settings
)```

#### Parameters

integrand
Type: SystemFuncDouble, Double
The function to be integrated.
start
Type: SystemDouble
The left integration endpoint.
end
Type: SystemDouble
The right integration endpoint.
settings
Type: Meta.Numerics.AnalysisIntegrationSettings
The settings which control the evaluation of the integral.

#### Return Value

Type: IntegrationResult
The result of the integral.
Exceptions
ExceptionCondition
ArgumentNullExceptionThe integrand is .
NonconvergenceExceptionThe maximum number of function evaluations was exceeded before the integral could be determined to the required precision.
Remarks

To do integrals over infinite regions, simply set start or end to NegativeInfinity or PositiveInfinity.

Our integrator handles smooth functions extremely efficiently. It handles integrands with discontinuities or kinks at the price of slightly more evaluations of the integrand. It can handle oscillatory functions, as long as cancelation between positive and negative regions is not too severe. It can integrate logarithmic and mild power-law singularities.

Strong power-law singularities will cause the algorithm to fail with a NonconvergenceException. This is unavoidable for essentially any double-precision numerical integrator. Consider, for example, the integrable singularity x-1/2. Since ε = ∫0δ x-1/2 dx = 2 δ1/2, points within δ ∼ 10-16 of the end-points, which as a close as you can get to a point in double precision without being on top of it, contribute at the ε ∼ 10-8 level to our integral, well beyond limit that nearly-full double precision requires. Said differently, to know the value of the integral to ε ∼ 10-16 precision, we would need to evaluate the contributions of points within δ ∼ 10-32 of the endpoints, which is far closer than we can get.

If you need to evaluate an integral with such a strong singularity, try to make an analytic change of variable to absorb the singularity before attempting numerical integration. For example, to evaluate I = ∫0b f(x) x-1/2 dx, substitute y = x1/2 to obtain I = 2 ∫0√b f(y2) dy.