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I'm trying to solve this integral including the MeijerG function using Mathematica :

Integrate[
 x^((a*b/2) - 1)*
  MeijerG[{{}, {}}, {{0}, {}}, (b *x^(l/k))/r^(
   a/2)]*(z - x)^((a*b/2) - 1)*
  MeijerG[{{}, {}}, {{0}, {}}, (b *(z - x)^(l/k))/r^(a/2)], {x, 0, z},]

But, it keeps giving me the same integral I write as follows:

Integrate[
 E^(-b r^(-a/2) x^(l/k) - b r^(-a/2) (-x + z)^(l/k))
   x^(-1 + (a b)/2) (-x + z)^(-1 + (a b)/2), {x, 0, z}]

What can i do to get the output of this integration??

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10
  • 1
    $\begingroup$ Are you even sure that there's supposed to be a closed form entirely in terms of Meijer $G$? $\endgroup$ Commented Aug 6, 2016 at 11:48
  • $\begingroup$ With the explicit settings {a = 1, b = 1, k = 1, l = 2, r = 1}, it still returns the Integrate[], which indicates that Mathematica does not know how to solve it (at least in that form). $\endgroup$
    – Michael E2
    Commented Aug 6, 2016 at 12:54
  • $\begingroup$ Rubi (apmaths.uwo.ca/~arich) cannot do it, either. $\endgroup$
    – Michael E2
    Commented Aug 6, 2016 at 12:57
  • $\begingroup$ does it mean that the integral has no answer?? $\endgroup$
    – dalia
    Commented Aug 6, 2016 at 13:08
  • $\begingroup$ All it means is that there might indeed be one, but Mathematica is not sufficiently capable to find it. $\endgroup$ Commented Aug 6, 2016 at 13:11

1 Answer 1

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With

$Version

(* "10.4.1 for Microsoft Windows (64-bit) (April 11, 2016)" *)

the integral does have a symbolic solution for k == l (and the spurious comma deleted from just before the final bracket).

Integrate[(x^((a*b/2) - 1)*MeijerG[{{}, {}}, {{0}, {}}, (b *x^(l/k))/r^(a/2)]*
    (z - x)^((a*b/2) - 1)*MeijerG[{{}, {}}, {{0}, {}}, (b *(z - x)^(l/k))/r^(a/2)]) /. 
    k -> l, {x, 0, z}]

(* ConditionalExpression[(E^(-b r^(-a/2) z) z^(-1 + a b) Gamma[(a b)/2]^2)/Gamma[a b], 
   Re[a b] > 0 && Re[z] > 0 && Im[z] == 0] *)

This solution is not as general as that requested but may still be of use.

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14
  • $\begingroup$ I guess it should be noted at this juncture that OP's integral is an autoconvolution integral… $\endgroup$ Commented Aug 6, 2016 at 19:06
  • $\begingroup$ i copied and pasted your integration but i didn't get your answer, i got the same integration again!!! $\endgroup$
    – dalia
    Commented Aug 7, 2016 at 9:10
  • $\begingroup$ why it keeps giving me the same integration?!! $\endgroup$
    – dalia
    Commented Aug 7, 2016 at 9:21
  • $\begingroup$ What version and OS are you on, @dalia? $\endgroup$ Commented Aug 7, 2016 at 10:22
  • $\begingroup$ i have Mathematica version 9 $\endgroup$
    – dalia
    Commented Aug 7, 2016 at 10:26

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