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Can anyone please give me the answer of this multiplication in the terms of another MeijerG function?

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

where a > 0 and b > 0.5.

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  • $\begingroup$ Should anyone want to try it out: I couldn't get MeijerGReduce[] to work on the OP's integral. $\endgroup$ – J. M.'s discontentment Aug 9 '16 at 7:36
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This is not an answer, but a request for more information that is too long for a comment.

When I evaluate your expression in V10.4.1, I get

E^(-b r^(-a/2) x^(a/2) - b r^(-a/2) (-x + z)^(a/2)) (-x + z)^(-1 + (a b)/2)

Why is this not a good result? Why do you insist Mathematica return a result in the form of MeijerG[...]?

Note that the documentation of MeijerG says

In many special cases, MeijerG is automatically converted to other functions.

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  • $\begingroup$ i think the problem is in my version , it's 9...and it doesn't give me the result you got, your result is fine actually! can you give me a link to download 10.4.1 please?? $\endgroup$ – dalia Aug 7 '16 at 10:48
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    $\begingroup$ @dalia. I don't think you can just download it. I think you have to purchase it as an upgrade. $\endgroup$ – m_goldberg Aug 7 '16 at 10:50
  • $\begingroup$ okay, no problem...so i guess the problem is in my version, right?? because it doesn't give me what you got! it returns the same input. $\endgroup$ – dalia Aug 7 '16 at 10:53
  • $\begingroup$ @dalia. I don't have access to V9 to test with it, but it is very likely you are right. Without an upgrade you may be out-of-luck. $\endgroup$ – m_goldberg Aug 7 '16 at 10:57
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    $\begingroup$ @dalia. I copied and pasted your code into a V10.4.1 notebook and evaluated it exactly as you wrote it. I made no modifications whatsoever. $\endgroup$ – m_goldberg Aug 7 '16 at 11:49
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If you are using version 9.0 you may only need to update to version 9.0.1

Using version 9.0.1.0 on a Mac

$Version

(*  "9.0 for Mac OS X x86 (64-bit) (January 24, 2013)"  *)

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

(*  E^(((-b)*x^(a/2))/r^(a/2) - 
        (b*(-x + z)^(a/2))/r^(a/2))*
   (-x + z)^(-1 + (a*b)/2)  *)

expr // Simplify

(*  (-x + z)^(-1 + (a*b)/2)/
   E^((b*(x^(a/2) + (-x + z)^
                (a/2)))/r^(a/2))  *)
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  • $\begingroup$ i have just tried to do it on version 10.0 and it works! $\endgroup$ – dalia Aug 7 '16 at 15:18

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