I have this MeijerG function

$$f(n,z)=\frac{z^2 \Gamma \left(n+\frac{3}{2}\right)}{2 \sqrt{\pi } K_2(z)}G_{1,3}^{3,0}\left(\frac{z^2}{4}| \begin{array}{c} n \\ -\frac{3}{2},0,\frac{1}{2} \\ \end{array} \right)$$

f[n_,z_] :=((z^2 Gamma[(3/2) + n])/(2 Sqrt[\[Pi]]BesselK[2,z])) MeijerG[{{}, {n}}, {{-(3/2), 0, 1/2}, {}}, z^2/4]

For n=1 I get

f[1,z]=(3 MeijerG[{{}, {2}}, {{-(1/2), 1, 3/2}, {}}, z^2/4])/(2 BesselK[2, z])

$$f(1,z)=\dfrac{3}{2 K_2(z)}G_{1,3}^{3,0}\left(\frac{z^2}{4}| \begin{array}{c} 2 \\ -\frac{1}{2},1,\frac{3}{2} \\ \end{array} \right)$$

The result depends on the two functions, MeijerG and the modified Bessel function $K_2(z)$.

I want to express all in terms of $K_2(z)$. According to the Wolfram function site link there is a relationship between the two functions but for my MeijerG function I couldn't find this relation. For example

$$G_{1,3}^{3,0}\left(z\left| \begin{array}{c} a_1 \\ b_1,b_2,b_3 \\ \end{array} \right.\right)=\pi ^2 \left(\frac{z^{b_1} \csc \left(\pi \left(b_2-b_1\right)\right) \csc \left(\pi \left(b_3-b_1\right)\right) \, _1\tilde{F}_2\left(-a_1+b_1+1;b_1-b_2+1,b_1-b_3+1;z\right)}{\Gamma \left(a_1-b_1\right)}+\frac{z^{b_2} \csc \left(\pi \left(b_1-b_2\right)\right) \csc \left(\pi \left(b_3-b_2\right)\right) \, _1\tilde{F}_2\left(-a_1+b_2+1;-b_1+b_2+1,b_2-b_3+1;z\right)}{\Gamma \left(a_1-b_2\right)}+\frac{z^{b_3} \csc \left(\pi \left(b_1-b_3\right)\right) \csc \left(\pi \left(b_2-b_3\right)\right) \, _1\tilde{F}_2\left(-a_1+b_3+1;-b_1+b_3+1,-b_2+b_3+1;z\right)}{\Gamma \left(a_1-b_3\right)}\right)\text{/;}\neg b_2-b_1\in \mathbb{Z}\land \neg b_3-b_1\in \mathbb{Z}\land \neg b_3-b_2\in \mathbb{Z}$$


$$G_{1,3}^{3,0}\left(z\left| \begin{array}{c} a \\ a-1,b,b+\frac{1}{2} \\ \end{array} \right.\right)=\sqrt{\pi } 2^{2 a-2 b-1} z^{a-1} \Gamma \left(-2 a+2 b+2,2 \sqrt{z}\right)$$


$$ G_{1,3}^{3,0}\left(z\left| \begin{array}{c} a \\ a-1,b,b+\frac{1}{2} \\ \end{array} \right.\right)=2 \sqrt{\pi } z^b E_{2 a-2 b-1}\left(2 \sqrt{z}\right) $$

$$G_{1,3}^{3,0}\left(z\left| \begin{array}{c} a \\ a-\frac{1}{2},2 a-b-1,b \\ \end{array} \right.\right)=\frac{2 z^{a-\frac{1}{2}} K_{-a+b+\frac{1}{2}}\left(\sqrt{z}\right){}^2}{\sqrt{\pi }}$$

$$G_{1,3}^{3,0}\left(z\left| \begin{array}{c} a \\ a-\frac{1}{2},2 a-b,b \\ \end{array} \right.\right)=\frac{2 z^a K_{-a+b-\frac{1}{2}}\left(\sqrt{z}\right) K_{-a+b+\frac{1}{2}}\left(\sqrt{z}\right)}{\sqrt{\pi }}$$

Can MATHEMATICA transform this MeijerG function into a $K_2(z)$?

  • $\begingroup$ FunctionExpand would be your best bet, but in this case it doesn't simplify your MeijerG expression. Are you aware of a transformation rule or relationship that could be applied by hand? $\endgroup$
    – MarcoB
    Jun 22 at 17:18
  • $\begingroup$ @MarcoB According to this link there is a relationship between the two functions but for my MeijerG function I couldn't do it. $\endgroup$
    – Gallagher
    Jun 22 at 17:29
  • $\begingroup$ Which relationship specifically did you find in that page that you think applies to your MeijerG expression? $\endgroup$
    – MarcoB
    Jun 22 at 17:43
  • $\begingroup$ @MarcoB the second or third relationship, but the term b+(1/2) does not verify mine which is equal to 3/2. $\endgroup$
    – Gallagher
    Jun 22 at 17:52
  • 1
    $\begingroup$ The $G$-function you have there corresponds to a logarithmic case (i.e. if you expand it at $z=0$, it will have a term containing $\log z$), so the large formula you have there that expands it in terms of ${}_1 F_2$ will not directly apply. It's not clear to me why you believe it's supposed to also be expressible in terms of $K$ (which is logarithmically singular at the origin as well, but the similarity seems to end there). $\endgroup$ Jun 22 at 19:48


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