Double Exponential Integration

While the AdaptiveIntegrator works well in most cases, it is not perfect. It sometimes has problems with some singularities on the boundaries of the integration interval. In such cases, the DoubleExponentialIntegrator may be a better option.

This integrator works by implicitly transforming the integration interval using tanh and sinh functions. Convergence is very fast in most cases. For smooth integrands, more traditional methods often win out.

An example of a difficult integral that is computed successfully with double exponential integration is:

$$\int_0^\inf \sqrt{frac{x}{1-x^2}}\,dx$$

The code below shows the difference in effectiveness. The standard adaptive integrator fails completely while the double exponential integrator requires just 51 function evaluations to get a result within the desired tolerance.

Func<double, double> fn = x => Math.Sqrt(x / (1 - x * x));
const double knownResult = 1.198140234735592;
var adaptive = new AdaptiveIntegrator();
var exponential = new DoubleExponentialIntegrator();
var result1 = adaptive.Integrate(fn, 0, 1);
Console.WriteLine($"Adaptive: {adaptive.Status}");
var result2 = exponential.Integrate(fn, 0, 1);
Console.WriteLine($"Double exponential: {exponential.Status}");
Console.WriteLine($"  Result: {result2}");
Console.WriteLine($"  Estimated error: {exponential.EstimatedError}");
Console.WriteLine($"  Actual error: {result2 - knownResult}");
Console.WriteLine($"  Function evaluations: {exponential.EvaluationsNeeded}");

Visual Basic

Dim fn = function(x) Math.Sqrt(x / (1 - x * x))
Const knownResult = 1.198140234735592
Dim adaptive = new AdaptiveIntegrator()
Dim exponential = new DoubleExponentialIntegrator()
Dim result1 = adaptive.Integrate(fn, 0, 1)
Console.WriteLine($"Adaptive: {adaptive.Status}")
Dim result2 = exponential.Integrate(fn, 0, 1)
Console.WriteLine($"Double exponential: {exponential.Status}")
Console.WriteLine($"  Result: {result2}")
Console.WriteLine($"  Estimated error: {exponential.EstimatedError}")
Console.WriteLine($"  Actual error: {result2 - knownResult}")
Console.WriteLine($"  Function evaluations: {exponential.EvaluationsNeeded}")

Visual Basic

No code example is currently available or this language may not be supported.

Visual Basic

let fn = fun x -> sqrt(x / (1.0 - x * x))
let knownResult = 1.198140234735592
let adaptive = new AdaptiveIntegrator()
let exponential = new DoubleExponentialIntegrator()
let result1 = adaptive.Integrate(fn, 0, 1)
printf $"Adaptive: {adaptive.Status}"
let result2 = exponential.Integrate(fn, 0, 1)
printf $"Double exponential: {exponential.Status}"
printf $"  Result: {result2}"
printf $"  Estimated error: {exponential.EstimatedError}"
printf $"  Actual error: {result2 - knownResult}"
printf $"  Function evaluations: {exponential.EvaluationsNeeded}"

References

Hidetosi Takahasi and Masatake Mori, Double Exponential Formulas for Numerical Integration, Publications of the Research Institute for Mathematical Sciences, 9 (3): 721–741