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Timeline for Nested NIntegrate

Current License: CC BY-SA 3.0

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Mar 2, 2020 at 15:09 comment added a20 Note that if you have defined a function without the ?NumericQ, you need to clear the function definition (easiest method is to restart Kernel) after adding the ?NumericQ, or it will not work.
Oct 5, 2017 at 4:59 comment added annie marie cœur Related you may help ; ) mathematica.stackexchange.com/questions/157158
Feb 3, 2015 at 22:48 comment added Wizzerad Curiously it works only with ?NumericQ. So, it looks like NIntegrate gives numerical values only if the check ?NumericQ is inserted, which looks strange to me. Does anyone has an explanation why it is like this?
Oct 9, 2012 at 17:14 comment added James Matta Thank you for this answer it has solved my problem perfectly (and saved me the trouble of asking a question to boot :) ) Thanks again.
Sep 13, 2012 at 23:16 comment added rm -rf @b.gatessucks If the inner ones have an analytic solution that also evaluates quickly, then yes, doing so might be worthwhile. But I've come across some integrals that have a nice analytical solution, but take a long time to compute whereas NIntegrate gives it instantly. Also, N@Integrate will give the same result as NIntegrate only if the function is well behaved within the default options in NIntegrae (e.g., not highly oscillatory, requiring increased precisions/recursions, etc). So N@Integrate only applies N to the result, whereas NIntegrate will use different algorithms
Sep 13, 2012 at 6:46 comment added b.gates.you.know.what @R.M. Could one use N[Integrate[...]] in case some of the inner integrals have an analytic solution ? If not, N[Integrate[...]]=NIntegrate[...] anyways.
Sep 12, 2012 at 19:43 comment added jmlopez Took a while but it finished. :)
Sep 12, 2012 at 19:36 vote accept jmlopez
Sep 12, 2012 at 19:36 comment added jmlopez This seems to be working on my actual problem. I'm still waiting on the result. Thank you.
Sep 12, 2012 at 19:26 history answered rm -rf CC BY-SA 3.0