Can anyone help me solve this sum?

 p[x_,a_] := PDF[NormalDistribution[0,a], x];
  a[x_] := Exp[x];
  u[x_] := (a[x]/(a[x] + 1));
  wa[x_, t_] := ((t - 3)*a[x])^(u[x] + .02442);
 Sum[p[x, .5]*wa[x, t], {x, 0, Infinity}, {t, 0, 50}]

Output For me it only gives back a symbolic output. But I would like to have the actual result.

  • 3
    $\begingroup$ You mean, a closed form? Do you have any justification that it even exists? $\endgroup$
    – corey979
    Jun 24, 2018 at 13:09
  • $\begingroup$ @corey979 Justification? what do you mean? $\endgroup$
    – MarV
    Jun 24, 2018 at 13:10
  • 2
    $\begingroup$ Do you have strong evidence that a closed form (e.g., the sum is convergent) exists (you didn't answer my first question, so I assume you want a closed form), or is it just an expression that you arrived at, threw it in MMA and just expect it to do wonders? $\endgroup$
    – corey979
    Jun 24, 2018 at 13:17
  • 1
    $\begingroup$ I see. Well, first, you can't obtain a result other than numeric because you have inexact numbers in the formulas. E.g., Sum[1./k^2, {k, 1, Infinity}] yields 1.64493, but if you change 1. to 1, you'll get $\pi^2/6$. Second, examine Table[Plot[Evaluate@ReIm[p[x, .5]*wa[x, t]], {x, 0, 5}, PlotRange -> All, Frame -> True, PlotLegends -> {"Re", "Im"}, PlotLabel -> "t = " <> ToString@t], {t, Range[0, 10]~Join~{100}~Join~{1000}}] // FlipView. For every t, the expression decays very quickly with x, so summing up to infinity is an overkill (...) $\endgroup$
    – corey979
    Jun 24, 2018 at 13:30
  • 2
    $\begingroup$ (...) Second (as a side note), the bigger the t, the greater the values of the real part of the function are. Finally, NSum[p[x, .5]*wa[x, t], {x, 0, 3}, {t, 0, 50}] gave me (after several seconds) 303.282 + 4.15345 I - is that a result that satisfies you - the actual result you are after? [Changing {x, 0, 3} to {x, 0, 4} doesn't change the result.] $\endgroup$
    – corey979
    Jun 24, 2018 at 13:32


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