Question

I am interested in learning an elegant way to use currying in a functional programming language to numerically evaluate multiple integrals. My language of choice is F#.

If I want to integrate f(x,y,z)=8xyz on the region [0,1]x[0,1]x[0,1] I start by writing down a triple integral of the differential form 8xyz dx dy dz. In some sense, this is a function of three ordered arguments: a (float -> float -> float -> float).

I take the first integral and the problem reduces to the double integral of 4xy dx dy on [0,1]x[0,1]. Conceptually, we have curried the function to become a (float -> float -> float).

After the second integral I am left to take the integral of 2x dx, a (float -> float), on the unit interval.

After three integrals I am left with the result, the number 1.0.

Ignoring optimizations of the numeric integration, how could I succinctly execute this? I would like to write something like:

let diffForm = (fun x y z -> 8 * x * y * z)

let result =
    diffForm
    |> Integrate 0.0 1.0
    |> Integrate 0.0 1.0
    |> Integrate 0.0 1.0

Is this doable, if perhaps impractical? I like the idea of how closely this would capture what is going on mathematically.

Answer 1

I like the idea of how closely this would capture what is going on mathematically.

I’m afraid your premise is false: The pipe operator threads a value through a chain of functions and is closely related to function composition. Integrating over an n-dimensional domain however is analogous to n nested loops, i.e. in your case something like

for x in x_grid_nodes do
    for y in y_grid_nodes do
        for z in z_grid_nodes do
            integral <- integral + ... // details depend on integration scheme

You cannot easily map that to a chain of three independet calls to some Integrate function and thus the composition integrate x1 x2 >> integrate y1 y2 >> integrate z1 z2 is actually not what you do when you integrate f. That is why Tomas’ solution—if I understood it correctly (and I am not sure about that…)—essentially evaluates your function on an implicitly defined 3D grid and passes that to the integration function. I suspect that is as close as you can get to your original question.

You did not ask for it, but if you do want to evaluate a n-dimensional integral in practice, look into Monte Carlo integration, which avoids another problem commonly known as the “curse of dimensionality”, i.e. that fact that the number of required sample points grows exponentially with n with classic integration schemes.

Update

You can implement iterated integration, but not with a single integrate function, because the type of the function to be integrated is different for each step of the integration (i.e. each step turns an n-ary function to an (n – 1)-ary one):

let f = fun x y z -> 8.0 * x * y * z

// numerically integrate f on [x1, x2]
let trapRule f x1 x2 = (x2 - x1) * (f x1 + f x2) / 2.0 

// uniform step size for simplicity
let h = 0.1

// integrate an unary function f on a given discrete grid
let integrate grid f =
    let mutable integral = 0.0
    for x1, x2 in Seq.zip grid (Seq.skip 1 grid) do
        integral <- integral + trapRule f x1 x2
    integral

// integrate a 3-ary function f with respect to its last argument
let integrate3 lower upper f =
    let grid = seq { lower .. h .. upper }
    fun x y -> integrate grid (f x y)

// integrate a 2-ary function f with respect to its last argument
let integrate2 lower upper f =
    let grid = seq { lower .. h .. upper }
    fun x -> integrate grid (f x)

// integrate an unary function f on [lower, upper]
let integrate1 lower upper f =
    integrate (seq { lower .. h .. upper }) f

With your example function f

f |> integrate3 0.0 1.0 |> integrate2 0.0 1.0 |> integrate1 0.0 1.0

yields 1.0.