8
$\begingroup$

I want to represent these elementary functions: $x^{2}\sqrt{x}$, $\sin{4x}$, and $x\ln{x}$ as cases of MeijerG. What arguments should I give to MeijerG to get these elementary functions?

For reference, SymPy is able to represent them. http://docs.sympy.org/0.7.2/modules/integrals/g-functions.html

$\endgroup$
  • 1
    $\begingroup$ Why do you go through the stage of the integration? Why not merely ask (for your first question) how to express $a\ x^{3.5}$ in terms of MeijerG? $\endgroup$ – David G. Stork Jan 10 '16 at 3:09
  • $\begingroup$ Unfortunately, there's no argument that available in "MeijerG" to represent the integrand in MeijerG form. $\endgroup$ – làntèrn Jan 10 '16 at 3:18
  • $\begingroup$ I do not know of a direct Mathematica conversion routine for this. Might get some help from e.g. 1, 2, 3. $\endgroup$ – Daniel Lichtblau Jan 10 '16 at 16:59
4
$\begingroup$

A partial answer using the reference that you provided:

Limit[b^-1 MeijerG[{{0}, {}}, {{0}, {}}, x^(-5/2)/b], b -> 0]

(*  x^(5/2)  *)

Assuming[{x >= 0}, 
 Sqrt[π] MeijerG[{{}, {}}, {{1/2}, {0}}, 4 x^2] // Simplify]

(*  Sin[4 x]  *)
$\endgroup$
7
$\begingroup$

Version 11 now has the function MeijerGReduce[], which returns equivalent MeijerG[] expressions in an inactive state. Applied to the OP's examples:

MeijerGReduce[{x^2, Sqrt[x], Sin[4 x], x Log[x]}, x]
   {x^2, Sqrt[x], 
    Sqrt[π] Inactive[MeijerG][{{}, {}}, {{1/2}, {0}}, 2 x, 1/2], 
    x (Inactive[MeijerG][{{1, 1}, {}}, {{1}, {0}}, x] - 
    Inactive[MeijerG][{{1, 1}, {}}, {{1}, {0}}, x, -1])}

where we see that only the last two functions were touched. The first two fail precisely because their Mellin transforms are not in a form suitable for the conversion:

MellinTransform[{x^2, Sqrt[x]}, x, s]
   {DiracDelta[2 + s], 2 DiracDelta[1 + 2 s]}

Contrast this with

MellinTransform[Sin[4 x], x, s]
   4^-s Gamma[s] Sin[(π s)/2]

In its current form, it does not seem to fit the style of the defining Mellin-Barnes integral for Meijer $G$, but using the reflection formula and the duplication formula for the gamma function, we have the identity

4^-s Gamma[s] Sin[(π s)/2] == 
Sqrt[π]/2 Gamma[1/2 + s/2]/Gamma[1 - s/2] 2^-s // FullSimplify
   True 

which can now be compared with the $G$ expression returned by MeijerGReduce[]: Sqrt[π] MeijerG[{{}, {}}, {{1/2}, {0}}, 2 x, 1/2].

$\endgroup$
4
$\begingroup$

There is an undocumented(!) function that essentially performs a Mellin transform through lookup. In your case, however, only one of your four examples actually has a sensible Mellin transform (and thus, a Meijer $G$ representation):

Integrate[Log[x], {x, 0, 1}]; (* force autoloading of the internal function *)

Integrate`ImproperDump`Mellin[Sin[4 x], x]
   (* Sqrt[Pi] Integrate`ImproperDump`MeijerGfunction[{}, {}, {1/2}, {0}, 4 x^2] *)

Check:

Simplify[% /. Integrate`ImproperDump`MeijerGfunction[a_, b_, c_, d_, z_] :>
         MeijerG[{a, b}, {c, d}, z], x > 0]
   (* Sin[4 x] *)

Try a more elaborate example:

Integrate`ImproperDump`Mellin[Hypergeometric2F1[1/2, -1/3, 1, x], x]
   (* Integrate`ImproperDump`MeijerGfunction[{4/3, 1/2}, {}, {0}, {0}, -x]/(Sqrt[Pi] Gamma[-1/3]) *)

As it is a lookup-based function, it might return unevaluated on functions that nevertheless have a $G$ function representation:

Integrate`ImproperDump`Mellin[Log[1 + x]/x, x]
   (* Integrate`ImproperDump`Mellin[Log[1 + x]/x, x] *)

MeijerG[{{0, 0}, {}}, {{0}, {-1}}, x]
   (* Log[1 + x]/x *)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.