How to speedup integration of interpolated function with logarithmized data?

Question

Consider some dataset corresponding to the grid x1, x2, function[x1,x2]:

function[x1_, x2_] = Exp[-x2/25]*Cos[20*x1]^2*Sin[x1];
dataset = 
  Flatten[Table[{x1, x2, function[x1, x2]}, {x1, 
     Table[10^
      x, {x, -5., Log10[1/20. Pi/2], (Log10[1/20. Pi/2] + 5.)/
       20}]}, {x2, 
     Table[10^
      x, {x, Log10[5.3], Log10[200.], (Log10[200.] - Log10[5.3])/
       20}]}], {1, 2}];
{x1min, x1max} = MinMax[dataset[[All, 1]]]
{x2min, x2max} = MinMax[dataset[[All, 2]]]

I interpolate it in two ways:

int1[x1_, x2_] = 10^(Interpolation[{Log10[#[[1]]], Log10[#[[2]]], Log10[#[[3]] + 10^-90]}&/@dataset, InterpolationOrder -> 1][Log10[x1], Log10[x2]]);
int2[x1_, x2_] = Interpolation[{#[[1]], #[[2]], #[[3]]}&/@dataset, InterpolationOrder -> 1][x1, x2];

The integration of the int1 is much slower than int2:

testfunction[x1_, x2_] = Sqrt[(1 - Cos[x1])]*Exp[5/x2];
NIntegrate[int1[x1, x2], {x1, x1min, x1max}, {x2, x2min, x2max}, 
  Method -> "InterpolationPointsSubdivision"] // AbsoluteTiming
NIntegrate[int2[x1, x2], {x1, x1min, x1max}, {x2, x2min, x2max}, 
  Method -> "InterpolationPointsSubdivision"] // AbsoluteTiming
NIntegrate[
  int1[x1, x2]*testfunction[x1, x2], {x1, x1min, x1max}, {x2, x2min, 
   x2max}, Method -> 
   "InterpolationPointsSubdivision"] // AbsoluteTiming
NIntegrate[
  int2[x1, x2]*testfunction[x1, x2], {x1, x1min, x1max}, {x2, x2min, 
   x2max}, Method -> 
   "InterpolationPointsSubdivision"] // AbsoluteTiming

{0.960714, 0.0146339}

{0.0011156, 0.0189933}

{1.01974, 0.000414336}

{0.0718938, 0.000659965}

On the other hand, int1 has better behavior in case if function changes fast between the neighboring values of x1, x2 in the grid.

Switching to "AdaptiveMonteCarlo" integration method somehow improves the situation, but non-negligible difference in timing (up to a few times) remains (it's funny that it slows down much the integration of int2 without testfunction).

Could you please tell me how to speed up the integration with int1?

Answer 1

I suggest that you try to use the substitution formula:

$$ \int_a^b \int_c^d f( \varphi(x),\psi(y)) \, \varphi'(x) \, \psi'(y) \, \mathrm{d}x \, \mathrm{d}y = \int_{\varphi(a)}^{\varphi(a)} \int_{\psi(c)}^{\psi(d)} f(u,v) \, \mathrm{d}u \, \mathrm{d}v. $$

So you could apply this to $\varphi(t) = \psi(t) = \log_{10}(t)$ or $\varphi(t) = \psi(t) = 10^{t}$. (I don't know at the moment which direction would be suitable for you.)

If $\varphi'(x) \neq 0$ and $\psi'(y) \neq 0$ for all $x \in [a,b]$ and $y \in [c,d]$, then you may define

$$ g(u,v) := f(u,v) \, \varphi'(\varphi^{-1}(u)) \, \psi'(\psi^{-1}(v)), $$ which leads to the following identity: $$ \int_a^b \int_c^d g( \varphi(x),\psi(y)) \, \mathrm{d}x \, \mathrm{d}y = \int_{\varphi(a)}^{\varphi(a)} \int_{\psi(c)}^{\psi(d)} g(u,v) \, \frac{1}{\varphi'(\varphi^{-1}(u))} \, \frac{1}{\psi'(\psi^{-1}(v)))} \, \mathrm{d}u \, \mathrm{d}v. $$

The point is to apply this to the following function $g$:

ip = Interpolation[Transpose[{Log10[x], Log10[y], Log10[z + 10^-90]}],
InterpolationOrder -> 1];
g[x1_, x2_] := 10^ip[x1, x2];

Method -> "InterpolationPointsSubdivision" should allow for fast integration of the integral on the right. And if I understand correctly, it is the integral on the left that you want to compute in the end.