﻿FunctionMath.Integrate Method (Func(Double, Double), Interval, EvaluationSettings)
 FunctionMathIntegrate Method (FuncDouble, Double, Interval, EvaluationSettings)
Evaluates a definite integral with the given evaluation settings.

Namespace:  Meta.Numerics.Analysis
Assembly:  Meta.Numerics (in Meta.Numerics.dll) Version: 3.1.0.0 (3.1.0.0)
Syntax
```public static IntegrationResult Integrate(
Func<double, double> integrand,
Interval range,
EvaluationSettings settings
)```

Parameters

integrand
Type: SystemFuncDouble, Double
The function to be integrated.
range
Type: Meta.NumericsInterval
The range of integration.
settings
Type: Meta.Numerics.AnalysisEvaluationSettings
The settings which control the evaulation of the integal.

Return Value

Type: IntegrationResult
The result of the integral, which includes an estimated value and an estimated uncertainty of that value.
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 the lower bound of the range to NegativeInfinity or the upper bound to PositiveInfinity.

Our numerical integrator uses a Gauss-Kronrod rule that can integrate efficiently, combined with an adaptive strategy that limits function evaluations to those regions required to achieve the desired accuracy.

Our integrator handles smooth functions extremely efficiently. It handles integrands with discontinuities, or discontinuities of derivatives, at the price of slightly more evaluations of the integrand. It can handle oscilatory functions, as long as not too many periods contribute significantly to the integral. It can integrate logarithmic and mild power-law singularities.

Strong power-law singularities will cause the alrorighm to fail with a NonconvergenceException. This is unavoidable for essentially any double-precision numerical integrator. Consider, for example, the integrable singularity 1/√x. 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 prescision, 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, 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.