Integration of norminv function

Is there any way to integrate norminv function for big number of dimensions?

For example in case of 3-dimensional integration integration:

fun3=@(x,y,z) exp(-norminv(x,0,1)).*(abs(sin(norminv(y,0,1)))+abs(sin(norminv(z,0,1)))+abs(sin(norminv(x,0,1))));
q3 = integral3(fun3,0,1,0,1,0,1);
vpa(q3)

Matlab returns the following warnings:

Warning: Reached the maximum number of function evaluations (10000). The result passes the global error test. 
> In integral2Calc>integral2t (line 136)
  In integral2Calc (line 9)
  In integral3/innerintegral (line 146)
  In integralCalc/iterateScalarValued (line 314)
  In integralCalc/vadapt (line 132)
  In integralCalc (line 75)
  In integral3 (line 121)
  In ICDFd (line 20)

Can somebody please give me an advise?

Answer 1

[Andrew Newell]

The problem is that norminv(x,0,1) goes to -Inf as x goes to zero and Inf as x goes to 1 (and ditto for y and z), so it's hard to integrate accurately. If you can manage with a larger tolerance,

q3 = integral3(fun3,0,1,0,1,0,1,'AbsTol',1e-3);

does not return any warnings (the default tolerance is 1e-10). If you need higher precision, you'll just need to be patient, as the warnings are not fatal. However, I don't know if the precision is actually met under such circumstances.

[Mariia Vasileva]

Thank you a lot)) Do you know may be how to solve the similar problem for multidimensional integration??? I am using this code:

syms a b c 
y =  exp(-norminv(a,0,1)).*(abs(sin(norminv(a,0,1)))+abs(sin(norminv(b,0,1)))+abs(sin(norminv(c,0,1))));
y1=int(y, c, 0, 1);
y2=int(y1, b, 0, 1);
y3=int(y2, a, 0, 1); 
vpa(y3)

- But always have multiple errors:

Error using symengine
Unable to prove 'a < 0 | 1 < a' literally. Use 'isAlways' to test the statement mathematically.

Error in sym/subsindex (line 792)
                X = find(mupadmex('symobj::logical',A.s,9)) - 1;

Error in sym/privsubsasgn (line 1067)
                L_tilde2 = builtin('subsasgn',L_tilde,struct('type','()','subs',{varargin}),R_tilde);

Error in sym/subsasgn (line 904)
            C = privsubsasgn(L,R,inds{:});

Error in norminv (line 57)
p(p < 0 | 1 < p) = NaN;

Error in untitled (line 2)
y =  exp(-norminv(a,0,1)).*(abs(sin(norminv(a,0,1)))+abs(sin(norminv(b,0,1)))+abs(sin(norminv(c,0,1))));

[Andrew Newell]

The source of the error is your attempt to use norminv with symbolic variables. This gives the same errors:

syms a
norminv(a,0,1)

Trying to come up with your own substitute for integral3 is tricky, and probably not worth the effort as integral3 will almost certainly do a better job.

[Mariia Vasileva]

Yes, but I have problems with multidimensional (greater than 3) integrals...