A A Calculation of a Form Factor at large ##q^2##

[MathematicalPhysicist]

TL;DR Summary

A question on a calculation in the book of Mueller, Perturbative QCD, from page 180.

On page 180 they write the following passage:

The form factors for the heavy hadrons are normalized by the constraint that the Coulomb contribution to the form factor equals the total hadronic charge at $q^2=0$. Further, by the correspondence principle, the form factor should agree with the standard non-relativistic calculation at small momentum transfer. All of these constraints are satisfied by the form:
F^M_{(0,0)}(q^2)=e_1\frac{16\gamma^4}{(q^2+\gamma^2)^2}(\frac{M_H^2}{m_2^2})^2 \bigg( 1-\frac{q^2}{4M_H^2}\frac{2m_2}{m_1}\bigg)+1\leftrightarrow 2 .
At large $q^2$ the form factor can also be written as:
F^M_{(0,0)}=e_1 \frac{16\pi\alpha_s f_M^2}{9q^2}(\frac{M_H^2}{m_2^2})+(1\leftrightarrow 2), f_M/(2\sqrt{3})=\int_0^1dx \phi(x,Q)
where $f_M=(6\gamma^3/\pi M_H)^{1/2}$ is the meson decay constant.

My question is how do I get $F^M_{(0,0)}$ from $F^M_{(0,0)}(q^2)$ for large $q^2$.
I tried using Mathematica's free input to get the series $1/(1+x^2)^2=1-2x^2+3x^4+O(x^6)$, in our case $x=\gamma/q$.
But it doesn't seem to fit the result from the book, perhaps there's something I am missing, any help?
A remark, $\gamma$ is Lorentz factor.

Thanks!

 

[MathematicalPhysicist]

No one knows?