I want to obtain a good numerical approximation (up to 10 decimal place would be ok for me) to an integral:

$$ \int^{\infty}_{0} f(r)r^2dr $$

I am using the function $f(r)$, which is related to the function

$$g(r)=-\frac{\sqrt[3]{3} \sqrt[3]{e^{-2 r}}}{\pi ^{2/3}}-\frac{\sqrt[3]{2 \pi }}{5 \sqrt[3]{e^{-2 r}} \left(\frac{3 \sqrt[3]{\pi } \sinh ^{-1}\left(\frac{2 \sqrt[3]{2 \pi }} {\sqrt[3]{e^{-2 r}}}\right)}{5\ 2^{2/3} \sqrt[3]{e^{-2 r}}}+1\right)}$$

by the relation

$$ f(r)=-\frac{1}{4\pi}\nabla^2_{r,\theta,\phi} g(r) $$

Obviously, explicit integration is impossible. The product $f(r)r^2$ is well-behaved and integrable for sure. The function $f(r)$ decays faster than $\frac{1}{r^2}$.

When I try to increase WorkingPrecision in NIntegrate, Mathematica says the expression I am integrating itself is not specified so precisely. How can I overcome this? Any tips/ hints?

I am asking for a general strategy to obtain a precise value of the integral:

NIntegrate[f[r]*4*π*r^2, {r, 0, y}, WorkingPrecision -> x]   

where y and x are some numbers.

P.S., I've been using Mathematica for only two days.

  • 3
    $\begingroup$ Can you update your post with your definition of f and the NIntegrate command you use? $\endgroup$
    – ssch
    Dec 19, 2012 at 0:02
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the FAQs! 3) When you see good Q&A, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. ALSO, remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign` $\endgroup$ Dec 19, 2012 at 7:24

2 Answers 2


Next time it would be really nice of you to give actual Mathematica code for your function definitions. Well, for now, thanks to @ssch and ToExpression we have the following definition of g[r]:

g[r_]=-((3^(1/3) (E^(-2 r))^(1/3))/\[Pi]^(2/3))-
(2 \[Pi])^(1/3)/(5 (E^(-2 r))^(1/3) (1+(3 \[Pi]^(1/3) 
ArcSinh[(2 (2 \[Pi])^(1/3))/(E^(-2 r))^(1/3)])/(5 2^(2/3) (E^(-2 r))^(1/3))));


enter image description here

Find function f[r]:

f[r_] = Simplify[PowerExpand[-(1/(4 Pi)) D[r^2 g'[r], r]/r^2]];

Note, you need the simplifications above so NIntegrate won't choke on too many terms. Now your function has a cool asymptotic behavior:

Limit[r^4 f[r], r -> Infinity]

enter image description here

So indeed r^2 f[r] will decay nicely. Let's plot the monster you're trying to integrate:

Plot[4 Pi f[r] r^2, {r, 0, 25}, PlotRange -> All, Filling -> Axis]

enter image description here

Define the integrating function:

h[s_] := NIntegrate[f[r] 4 Pi r^2, {r, 0, s}, MaxRecursion -> 12]

It works fine:

h /@ {.1, 1, 10, 100, Infinity}

{-0.00396818, -0.2323, -0.647235, -0.942626, -0.999967}

You can even plot it:

ListPlot[Table[{s, h[s]}, {s, 0, 10, .1}], 
 PlotStyle -> Directive[Red, PointSize[.01]], Filling -> 0, AspectRatio -> 1/3]

enter image description here

  • $\begingroup$ thnx a lot! i am a newbie, so please sorry if my question may seem to be easy and straightforward $\endgroup$
    – molkee
    Dec 19, 2012 at 14:49
  • $\begingroup$ using PowerExpand indeed speeds up the calculation! $\endgroup$
    – molkee
    Dec 19, 2012 at 14:59
  • $\begingroup$ Nice use of PowerExpand, I tried all kinds of simplifications to no avail :) $\endgroup$
    – ssch
    Dec 19, 2012 at 17:35

This is not an answer, just too long for a comment.

To start of with your NIntegrate call looks peculiar, the f(r) part is not a function call but is interpreted as f*r a function call looks like f[r]

(* Obtained by doing ToExpression[...I pasted latex stuff here..., TeXForm]
 and replaced e with E ( /.e->E) *)
g[r_] := -((3^(1/3) (E^(-2 r))^(1/3))/\[Pi]^(2/3)) - (2 \[Pi])^(1/3)/(
   5 (E^(-2 r))^(
    1/3) (1 + (
      3 \[Pi]^(1/3) ArcSinh[(2 (2 \[Pi])^(1/3))/(E^(-2 r))^(1/3)])/(
      5 2^(2/3) (E^(-2 r))^(1/3))));
(* I presume the laplacian simplifies to 1/r^2 D[r^2 D[g[r], r], r] in this case *)
f[r_] := Evaluate[Simplify[ -1/(4 Pi) 1/r^2 D[r^2 D[g[r], r], r], r \[Element] Reals]]
(* Sanity check, gives 0, so that's good *)
Limit[f[r] r^2, r -> Infinity]

This is basically how far I got, this spits out errors about under/overflow and $MaxExtraPrecision being reached

Block[{$MaxExtraPrecision = 1000000.}, 
  N[f[10^20] (10^20)^2, {Infinity, 10}]]
  • $\begingroup$ As a general comment: the laplacian does not simplifies to D[g[r],{r,2} in this case. $\endgroup$
    – molkee
    Dec 19, 2012 at 1:07
  • $\begingroup$ @molkee Sorry about that, changed to 1/r^2 D[r^2 D[g[r], r], r] $\endgroup$
    – ssch
    Dec 19, 2012 at 1:14

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