The problem in the link: can be integrated analytically and the answer is 4, however I\
Well, this is strange, because on the poster's similar previous question I claimed this can't be done, and now after having looked at Guddu's answer I realize its not that complicated. What I wrote before, that a numerical integration results in a number but not a function, is true – but beside the point: One can just define a function that evaluates the integral for every given parameter, and this way effectively one does have a function as a result of a numerical integration.
Anyways, here it goes:
function q = outer
f = @(z) (z .* exp(inner(z)));
q = quad(f, eps, 2);
end
function qs = inner(zs)
% compute \int_0^1 1 / (y + z) dy for given z
qs = nan(size(zs));
for i = 1 : numel(zs)
z = zs(i);
f = @(y) (1 ./ (y + z));
qs(i) = quad(f, 0 , 1);
end
end
I applied the simplification suggested by myself in a comment, eliminating x. The function inner
calculates the value of the inner integral over y as a function of z. Then the function outer computes the outer integral over z. I avoid the pole at z = 0 by letting the integration run from eps
instead of 0. The result is
4.00000013663955
inner
has to be implemented using a for
loop because a function given to quad
needs to be able to return its value simultaneously for several argument values.