I have an algorithm for which I need to compute some multidimensional integrals. Sometimes, Integrate works but some others Integrate just fails to compute the integral.

A simple example:

$$ \int^{\infty}_{-\infty}\int^{\infty}_{-\infty} -\frac{2 x^2 \left(e^{2 y+1}+1\right) (\text{erf}(y)+\text{erf}(y+1)) e^{-x^2-(y+1)^2}}{\pi } {\rm d}x {\rm d}y$$

Calling integrate,

Integrate[-((2 E^(-x^2 - (1 + y)^2) (1 + E^(1 + 2 y)) x^2 (Erf[y] + Erf[1 + y]))/\[Pi]), {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]


$$ \int_{-\infty }^{\infty } -\frac{e^{-(y+1)^2} \left(e^{2 y+1}+1\right) (\text{erf}(y)+\text{erf}(y+1))}{\sqrt{\pi }} \, {\rm d}y $$

which is 0 (Interestingly, WolframAlpha indeed handles this integral, while my Mathematica does not). Now, Calling NIntegrate,

NIntegrate[-((2 E^(-x^2 - (1 + y)^2) (1 + E^(1 + 2 y)) x^2 (Erf[y] + 
 Erf[1 + y]))/\[Pi]), {x, -Infinity, Infinity}, {y, -Infinity, Infinity}, AccuracyGoal -> 14]

yields $-5.2\times 10^{-16}$ (After a bunch of warnings). I was wondering I there was any way of further calling NIntegrate whenever Integrate fails. I know that I can call


but I have not figured out how to pass method options to this. I am looking for a generic solution that works for higher dimensional integrals. If it helps, my integrals all involve exponentials and error functions similar to that one.

Thanks in advance.


1 Answer 1


This can be achieved with quite a simple function. Let's define a few integrands to test on

integrand1 = -((2 E^(-x^2 - (1 + y)^2) (1 + 
         E^(1 + 2 y)) x^2 (Erf[y] + Erf[1 + y]))/π);
integrand2 = Exp[-x^2 - y^2];
integrand3 = Sin[Sin[Sin[x]]]/x;

Now define the integrator you requested. It checks whether integration has succeeded by seeing whether the result includes the symbol Integrate

int[u__] := Module[{ans},
  ans = Integrate[u];
  If[FreeQ[ans, Integrate], ans, N[ans]]]

Now test them

int[integrand1, {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]
(* 0 *)

[It looks like Mathematica V12 computes your integral is analytically]

Here's another case that is possible analytically

int[integrand2, {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]
(* π *)

and one that isn't

int[integrand3, {x, -2, 2}]
(* 2.75358 *)

------------------WITH OPTIONS-------------

Here is an alternative that (I hope) supports options for both Integrate and NIntegrate routing them appropriately.

int[u__, v___Rule] := Module[{ans},
  ans = Integrate[u, Sequence @@ FilterRules[{v}, Options[Integrate]]];
  If[FreeQ[ans, Integrate], ans, 
    Evaluate[Sequence @@ FilterRules[{v}, Options[NIntegrate]]]]]]

There are a lot of cases to test to prove that this works fully, so here are a few:

int[integrand1, {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]
(* 0 *)

int[integrand2, {x, -Infinity, Infinity}, {y, -Infinity, Infinity}]
(* π *)

int[integrand3, {x, -2, 2}]
(* 2.75358 *)

int[integrand3, {x, -2, 2}, Method -> MonteCarlo]
(* 2.74324 *)

int[1/x, {x, -3, 3}, PrincipalValue -> True]
(* 0 *)
  • 1
    $\begingroup$ The OP seems to want to be able to pass Method options to the NIntegrate call. Is that possible? $\endgroup$
    – Michael E2
    Dec 20, 2019 at 20:38
  • $\begingroup$ Thanks for this. As @MichaelE2 mentions I do want to pass options. I have not tested it yet, but I think this can either be done by changing N[ans] by a call to NIntegrate[u,opts] with some fixed options or extending the pattern to accept options as a second argument. $\endgroup$ Dec 21, 2019 at 5:29
  • $\begingroup$ @MarioE.Villanueva. I've edited to support options for both Integrate and NIntegrate $\endgroup$
    – mikado
    Dec 21, 2019 at 11:52
  • $\begingroup$ Thanks, I am ready to accept the answer as this already works for me. I was wondering tho, if, for the sake of completeness the following case can be addressed: Sometimes, Integrate will do at least one of the integrals and return a simplified integral. Is there any way to use this simplified integral in NIntegrate? (e.g., the integral in my question). In particular, can we extract the integrand and remaining independent variables/limits. $\endgroup$ Dec 22, 2019 at 9:55
  • $\begingroup$ @MarioE.Villanueva I think you have all the pieces to do what you are looking for, if you think it's worthwhile $\endgroup$
    – mikado
    Dec 23, 2019 at 6:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.