I came across one such integral in my calculations, for which there is no analytical solution. But it exists numerical solution, so how can I derive analytical solution of this integral from numerical solution? Below is my attempt, it might take some time to run the calculation.

$Assumptions = {{\[Alpha], \[Beta]} \[Element] Reals};
R[r_] = Exp[-(1/\[Alpha])*((r/\[Beta])^\[Alpha] - 3)];
RR[r_] = Integrate[R[r]*r^2, {r, 0, r}];
  RR[r] /. \[Alpha] -> \[Alpha]1}, {\[Alpha]1, {0.13, 0.14, 0.15, 
   0.16, 0.17, 0.18, 0.19}}]

Thanks a lot for everyone's suggestions and help!


1 Answer 1

  1. The integration limit {r,0,r} in your integral is a syntactic error. You should write {r,0,r1} and then do with r1 what you need.

  2. It is impossible to solve your integral numerically since you did not fix [Beta].

  3. However, after correction of the first error the integral can be solved exactly:

    R[r_] = Exp[-(1/\[Alpha])*((r/\[Beta])^\[Alpha] - 3)];

     ` Integrate[R[r]*r^2, {r, 0, r1}, 
      Assumptions -> {\[Alpha] > 0, \[Beta] > 0}]`

with the effect:

enter image description here

Have fun!

  • $\begingroup$ In fact, you can use the same symbol as the integrating variable and the upper limit, e.g. Integrate[x, {x, 0, x}] gives x^2/2 as "expected" without any syntax error. $\endgroup$
    – Domen
    Jun 7, 2023 at 11:27
  • $\begingroup$ @Alexei Boulbitch Thanks a lot for your help, putting the assumptions in the integral grammar solved the problem. This is the first problem I have encountered. The second is how to achieve the integration of this result again to get the analytical solution? A simple example would be adding the following integral Integrate[RR[r]^2/r, {r, 0, r1}] to the above result. Thanks again for your help. $\endgroup$ Jun 8, 2023 at 5:20
  • $\begingroup$ @Domen Thanks for your help. I tried your hint and using {x,0,x} seems to get the result too, but it seems to benefit from putting assumptions inside the integral grammar, only for this problem of me. I updated my next question in a previous comment. I would appreciate it if you have suggestions as well. $\endgroup$ Jun 8, 2023 at 5:31
  • $\begingroup$ @little star The second integral, Integrate[RR[r]^2/r, {r, 0, r1}], does not converge. C.f. Simplify[ Series[R[r]^2/r, {r, 0, 1}] // Normal, {\[Alpha] > 0, \[Beta] > 0}]. It diverges as 1/r at r->0. $\endgroup$ Jun 8, 2023 at 9:21
  • $\begingroup$ @AlexeiBoulbitch I might have made a mistake, the integration expression should be Integrate[RR[r]/r^2, {r, 0, r1}]. For this integral expression, once $\alpha=0.13$ is fixed, the integral seems to have a corresponding numerical solution (Integrate[RR[r]/r^2 /. {\[Alpha] -> 0.13}, {r, 0, r1}]). So, if $\alpha$ is sufficiently multiple, for example $\alpha=0.1,0.11,0.12,0.13...$, can the exact solution be derived from the numerical solution? $\endgroup$ Jun 8, 2023 at 11:46

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