- #1

- 103

- 14

##p(x) = (e^{x}-1)^{2i}\left({}_{2}F_{1}(a,b;c;e^{x}) + {}_{2}F_{1}(a+1,b+1;c+1;e^{x})\right)## where ##a,b,c## are all complex valued and we have ##\Re(c-a-b)>0## and we have ##\Re((c+1)-(a+1)-(b+1))=0## but ##(a+1)+(b+1)-(c+1) = 2i##. My question is, how do I take the proper limit of ##p(x)## as ##x\rightarrow 0##. I believe the relevant identities are (15.4.20) and (15.4.22) from https://dlmf.nist.gov/15.4 but there is still the limit of figuring out what ##0^{i}## means.

Let me know if you guys have any ideas. Thanks!