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I am using Integrate function to do the following double integral

Integrate[3.1 x^2.1 Exp[-y^2/2]/(2 π)^0.5, {y, -1, 3}, {x, 0, y + 35}]

The result is 1204.63.

When I check with the following integral

N[Integrate[Integrate[3.1 x^2.1, {x, 0, y + 35}, Assumptions :> y ∈ Reals]*
   Exp[-y^2/2]/(2 π)^0.5, {y, -1, 3}]]

The result is 52774.4, which also agrees with what I got in Matlab. I thought the first one is a nice and clean way to do double integral than the second one, but it seems that the result is not what I want if the function w.r.t. x (x^2.1 in this case) does not have integer power.

I would be very thankful if anyone can answer my question.

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  • $\begingroup$ NIntegrate[3.1 x^2.1 Exp[-y^2/2]/(2 \[Pi])^0.5, {y, -1, 3}, {x, 0, y + 35}] is perhaps a nice compromise. Integrate is primarily for exact computation. It tries with inexact input, but in this case fails. Note that, with exact input, N@Integrate[31/10 x^(21/10) Exp[-y^2/2]/(2 \[Pi])^(1/2), {y, -1, 3}, {x, 0, y + 35}] also gives the right answer. $\endgroup$
    – Michael E2
    Feb 14, 2015 at 18:22
  • $\begingroup$ BTW, you don't seem to be asking a question. No "?" for example. Just observations about the behavior of Integrate. It's not clear whether you want an explanation or other workarounds or what. $\endgroup$
    – Michael E2
    Feb 14, 2015 at 18:24
  • $\begingroup$ Switch Integrate to NIntegrate. It seems that the machine precision values in your expression make it impossible to get the correct answer symbolically when posed in this way. $\endgroup$
    – Oleksandr R.
    Feb 14, 2015 at 18:29
  • $\begingroup$ Thanks Michael And Oleksandr very much for your comment, it resolves my concern definitely! Sorry for being unable to make the question clear. I was using the first expression a lot to get numerical experiments without noticing the existence problem of inconsistency like this. I thought the second one is correct because it got the same result with Matlab, I was just stuck in the confusion and unable to figure out the problem with the first expression. BTW, when the power of x is integer instead of 2.1 in this case, the results of both expressions always agree. Thanks again :) $\endgroup$
    – Kai He
    Feb 14, 2015 at 18:42
  • $\begingroup$ Related (perhaps a duplicate?): mathematica.stackexchange.com/questions/51809/… $\endgroup$
    – Michael E2
    Feb 14, 2015 at 19:27

1 Answer 1

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This does not answer the question why but I post for interest. The region of integration:

ir = ImplicitRegion[0 < x < y + 35 && -1 < y < 3, {x, y}];
RegionPlot[ir]

enter image description here

The integral must be < Area[ir]f[38,0]:369981. (not a helpful bound). $0<x<38 \land -1<y<3$ would be a closer. 

SetAttributes[dis, HoldFirst];
dis[u_] := {Style[#, Bold], ReleaseHold[#]} & @ HoldForm @ u;
f[x_, y_] := 3.1 x^2.1 Exp[-y^2/2]/(2 π)^0.5;
r1 = dis[NIntegrate[f[x, y], {y, -1, 3}, {x, 0, 35 + y}]];
r2 = dis[NIntegrate[f[x, y], {x, 0, 38}, {y, -1, 3}]];
r3 = dis[NIntegrate[f[x, y], {x, y} ∈ ir]];
r4 = dis[Integrate[f[x, y], {x, y} ∈ ir]]
Grid[{r1, r2, r3,r4}, Alignment -> Left, Frame -> All]

As has been observed by MichaelE2 and OleksandrR numerical integration yields correct result. Interestingly, Integrate works with region (presumably switching to numerical integration). The results are summarized below:

enter image description here

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  • $\begingroup$ Yes, the last two entries are effectively equivalent. Integrate over a region will call NIntegrate with WorkingPrecision set to the minimum precision of the integrand and region, when the precision is less than infinity. $\endgroup$
    – Michael E2
    Feb 15, 2015 at 13:38

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