Obtain the conversion of Gamma function?

Questions
1

In

eq1 = F1[1][T] == 
  E^-T B1[1] (-((2 E^(-(Rx^2/(4 T))) (Rx^2/T)^(-Rv/2))/Rv) - 
     2^-Rv Gamma[-(Rv/2), R0^2/(4 T)] + 
     2^-Rv Gamma[-(Rv/2), Rx^2/(4 T)]),

I want to realize this conversion:

(-2^-Rv Gamma[-(Rv/2), R0^2/(4 T)] + 
   2^-Rv Gamma[-(Rv/2), Rx^2/(4 T)]) -> 
 2^-Rv Gamma[-(Rv/2), Rx^2/(4 T), R0^2/(4 T)]

Would you like to tell me how to realize it?
Thanks.

2

I would think that you want to have a conversion that doesn’t require you to know the specific Gamma coefficients. Does the following do what you want?

eq1 = F1[1][T] == E^-T B1[1] (-((2 E^(-(Rx^2/(4 T))) (Rx^2/T)^(-Rv/2))/Rv) - 
     2^-Rv Gamma[-(Rv/2), R0^2/(4 T)] + 2^-Rv Gamma[-(Rv/2), Rx^2/(4 T)])

FullSimplify[eq1] /. Gamma[a_, b_] - Gamma[a_, c_] -> Gamma[a, b, c]

(* F1[1][T] == E^-T B1[1] (-((2 E^(-(Rx^2/(4 T))) (Rx^2/T)^(-Rv/2))/Rv) + 
    2^-Rv Gamma[-(Rv/2), Rx^2/(4 T), R0^2/(4 T)]) *)
3

Are your expressions in general always going to have only two instances of the incomplete gamma function with opposite signs? If you have four of them, then you already have two ways of pairing them up, and it will only get combinatorically worse from there.

4

Yes, Your codes are what I wanted.
Actually it also works with the coefficient.

In[88]:= eq1 = 
 F1[1][T] == 
  E^-T B1[1] (-((2 E^(-(Rx^2/(4 T))) (Rx^2/T)^(-Rv/2))/Rv) - 
     2^-Rv Gamma[-(Rv/2), R0^2/(4 T)] + 
     2^-Rv Gamma[-(Rv/2), Rx^2/(4 T)])

Out[88]= F1[1][T] == 
 E^-T B1[1] (-((2 E^(-(Rx^2/(4 T))) (Rx^2/T)^(-Rv/2))/Rv) - 
    2^-Rv Gamma[-(Rv/2), R0^2/(4 T)] + 
    2^-Rv Gamma[-(Rv/2), Rx^2/(4 T)])

In[89]:= eq2 = 
 eq1 /. {2^-Rv  Gamma[a_, b_] - 2^-Rv  Gamma[a_, c_] -> 
    2^-Rv Gamma[a, b, c]}

Out[89]= F1[1][T] == 
 E^-T B1[1] (-((2 E^(-(Rx^2/(4 T))) (Rx^2/T)^(-Rv/2))/Rv) + 
    2^-Rv Gamma[-(Rv/2), Rx^2/(4 T), R0^2/(4 T)])