Norm is a way to estimate the sum of the residuals and Quiet stops the beeping that I was getting from warnings. Here is an example:

dat = {d1, d2, d3};
model = {m1, m2, m3};
resid = model - dat;
Norm@resid // InputForm (* \
Sqrt[Abs[-d1+m1]^2+Abs[-d2+m2]^2+Abs[-d3+m3]^2] *)

Thanks a lot for helping me!

I fix the code, but the fit is not good.

1. I read the data here my x

xmdatapar = 
  ReadList["/home/roberto/F?ica/Pesquisa/<X^m>/kaon/lib/pares/xm_par.\
dat", {Number}];
xmpardata = Table[xmdatapar[[i, 1]], {i, 1, Length[xmdatapar]}]
{0.243098, 0.127303, 0.0834994, 0.0609518, 0.0473757, 0.0384113, \
0.0320716, 0.027407, 0.0237647, 0.020894, 0.0185901, 0.0167308, \
0.0151393, 0.013839, 0.0127182, 0.0117066, 0.0108577, 0.0101237, \
0.00948312, 0.00887854, 0.00834716, 0.00788257, 0.00746653, \
0.00705183, 0.00669624}

2. i need to fit this variables *Npar, (\[Alpha]RLpar , \[Alpha]2*

\[Phi]par[x_] := 
 Npar*(x (1 - x))^(\[Alpha]RLpar - 
    1/2 ) (1 + \[Alpha]2*GegenbauerC[2, \[Alpha]RLpar, (2 x - 1)]);

3.I did a model to fit

Table[{m, Integrate[\[Phi]par[x] ( 2*x - 1)^m, {x, 0, 1}]}, {m, 2, 6, 
   2}];

fitxm = Integrate[\[Phi]par[x] ( 2*x - 1)^m, {x, 0, 1}];

1. where m is numbers even 2,4,6...50 and x my moments, [Alpha]2 need to be a negative low value like around -0.30~-0.50

     FindFit[xmpardata, fitxm, {\[Alpha]RLpar, \[Alpha]2, Npar}, m]
   
       FindFit::nrlnum: The function value {-0.243098+0. I,-0.000149731-1.55718*10^-17 I,-0.0834994+0. I,0.00219318 -1.54661*10^-17 I,-0.0473757+0. I,0.000892465 -1.444*10^-17 I,-0.0320716+0. I,<<11>>,-0.00948312+0. I,-0.000897592-9.77385*10^-18 I,-0.00834716+0. I,-0.00090295-3.41989*10^-17 I,-0.00746655+0. I,-0.000880321-9.0694*10^-18 I,-0.00669639+0. I} is not a list of real numbers with dimensions {25} at {\[Alpha]RLpar,\[Alpha]2,Nc} = {1.00001,-0.0133867,1.31278}.
   
       {\[Alpha]RLpar -> 1.00001, \[Alpha]2 -> -0.0133867, Npar -> 1.31278}
       ListPlot[xmpardata]
       In[8]:= Table[
        fitxm /. {\[Alpha]RLpar -> 
           1.00001204228849, \[Alpha]2 -> -0.0133867063850715, 
          Npar -> 1.3127808940632832}, {m, 2, 50, 2}]
   
       Out[8]= {0.127153, 0.063145, 0.0393038, 0.0274371, 0.0205373, \
       0.0161126, 0.0130764, 0.010887, 0.00924693, 0.00798095, 0.00697962, \
       0.00617151, 0.00550909, 0.00495871, 0.00444164, 0.00412147, \
       -0.000131836, -0.036299, 0.1193, -0.170002, -4.57313, -135.358, \
       1499.52, 4139.2, 44729.1}
   
       In[10]:= ListPlot[
        Table[fitxm /. {\[Alpha]RLpar -> 
            1.00001204228849, \[Alpha]2 -> -0.0133867063850715, 
           Npar -> 1.3127808940632832}, {m, 2, 50, 2}]]
   

If you look closely at you expression for $\phi par$, then you will see that you have 4 parameters versus the 3 specified in FindFit. One of them is most likely a typo. You can check the unique symbols in your expression as shown below:

\[Phi]par := 
  Npar*(x (1 - x))^
    aRL\[Alpha]par (1 + \[Alpha]2RLpar*
      GegenbauerC[2, \[Alpha]RLpar, (2 x - 1)]);
DeleteDuplicates@Cases[\[Phi]par, _Symbol, Infinity]
(* {Npar, x, aRL\[Alpha]par, \[Alpha]2RLpar, \[Alpha]RLpar} *)

If your end goal is to perform the numerical integration, then you are probably making it much more difficult than it needs to be. If you plot the data with ListLogLogPlot, it is very linear suggesting LinearModelFit is in order.

data = {{2, 0.243098}, {4, 0.127303}, {6, 0.0834994}, {8, 
    0.0609518}, {10, 0.0473757}, {12, 0.0384113}, {14, 
    0.0320716}, {16, 0.027407}, {18, 0.0237647}, {20, 0.020894}, {22, 
    0.0185901}, {24, 0.0167308}, {26, 0.0151393}, {28, 0.013839}, {30,
     0.0127182}, {32, 0.0117066}, {34, 0.0108577}, {36, 
    0.0101237}, {38, 0.00948312}, {40, 0.00887854}, {42, 
    0.00834716}, {44, 0.00788257}, {46, 0.00746653}, {48, 
    0.00705183}, {50, 0.00669624}};
fit = LinearModelFit[Log@data, x, x]
phipar[x_] := Exp@Normal@fit;
fit["ParameterTable"]
Plot[Normal@fit, {x, 0.5, 4}, Axes -> False, Frame -> True, 
 Epilog -> {PointSize@.02, Point@Log@data}]
f[m_] := NIntegrate[(2*x - 1)^m (phipar[x]), {x, 0, 1}]
f[2] (* 0.130834 *)

With your current formulation, $aRL\alpha par$ must be even integers to avoid complex numbers since $x\geq 2$ making the argument negative.

$$\phi_{par}=N_{par}\left ( x\left ( 1-x \right ) \right )^{aRL\alpha par}....$$