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I would like to compute the following function:

Integrate[Gamma[k,0,Exp[-2 e x^2]],x]

k is an integer and e a parameter. I would like to be able to have an closed form expression for it to fit a curve (actually to fit convolutions of this function, but it's the same), or at least a function to compute numerical approximate values, given the values of k and e. I have tried Integrate and NIntegrate, but they do not work, and do not return a value or an expression.

Is there a way I can do this with Mathematica? Or maybe another software that could do it?

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  • $\begingroup$ Do you know the integration limits? $\endgroup$
    – Ivan
    Commented May 1, 2015 at 4:47
  • $\begingroup$ Ideally, it would be from 0 to Infinity. $\endgroup$
    – Bill
    Commented May 1, 2015 at 6:15
  • $\begingroup$ Even the simple $\int \exp(-\exp(-x^2))\mathrm dx$ does not seem to have a closed form, so I'm even more doubtful that your integral is tractable. You may have to content yourself with numerics. $\endgroup$ Commented May 2, 2015 at 6:46

1 Answer 1

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Having a closed form for integral of an incomplete gamma function should be a big deal(!), but having a numerical approximation is simple. Define:

f[k_,e_,limit_]:=NIntegrate[Gamma[k, 0, Exp[-2 e x^2]], {x, -limit, limit}];

For example:

f[2,0.2,3]
(* 0.591563 *)

You can also see how f changes with k and e:

Plot[{Legended[f[k, 0.2, 3.], "e=0.2"], 
  Legended[f[k, 2., 3.], "e=2."], Legended[f[k, 5., 3.], "e=5."]}, {k,
   0, 10}, AxesLabel -> {"k", "f"}]

enter image description here

and

Plot[{Legended[f[1, e, 3.], "k=1"], Legended[f[2, e, 3.], "k=2"], 
  Legended[f[3, e, 3.], "k=3"]}, {e, 0, 10}, AxesLabel -> {"e", "f"}]

enter image description here

You can easily change the limit of integral to $0$ to $\infty$:

f[k_, e_] := NIntegrate[Gamma[k, 0, Exp[-2 e x^2]], {x, 0, Infinity}];
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  • $\begingroup$ This is very helpful and detailed thanks. Do you know if I have any chance of computing this approximate value of the integral with the integration limits being 0 and +Infinity? I guess I can then use f as a function of k, to compute a convolution of values of f? $\endgroup$
    – Bill
    Commented May 1, 2015 at 6:14

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