I am trying to find the value of following expression as $x$ tends to $\pm\infty$
$$\exp[-i\omega_+x-i\omega_-\ln(2\cosh x)]\times\;_2F_1(1+i\omega_-,i\omega_-;1-i\omega_{in};\frac{1}{2}(1+\tanh x))$$ where $\omega_{in}$ and $\omega_{out}$ are some arbitrary constant and $\omega_{\pm}=\frac{1}{2}(\omega_{out}\pm\omega_{in})$
So far what I have been able to do is this:
as $x\rightarrow \pm\infty$ then $\cosh x \approx\frac{e^{\pm x}}{2}$ therefore the first factor in the expression becomes $$\exp[-i\omega_+x-i\omega_-ln(2\frac{e^{\pm x}}{2})]$$ $$=\exp[-i\omega_+x\mp i\omega_- x)]$$ $$=\exp[-i(\omega_+\pm \omega_- )x)]$$ $$=e^{-\omega_{out}x},e^{-\omega_{in}x} $$
As for the second term term it can be approximated as following for $x\rightarrow \pm\infty$, $\tanh x\approx \pm 1$ so the second term becomes $$_2F_1(1+i\omega_-,i\omega_-;1-i\omega_{in};0)\hspace{2mm}\text{or}\hspace{2mm}_2F_1(1+i\omega_-,i\omega_-;1-i\omega_{in};1)$$
as last term is zero for second factor when $x\rightarrow -\infty$ it should be $0$ and for case of $+\infty$ I am trying to use the identity $$_2F_1(a,b;c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$ but I am getting nowhere, basically I stops at following expression $$\frac{\Gamma(1-i\omega_{in})\Gamma(-i\omega_{in}-i\omega_--i\omega_-)}{\Gamma(-i\omega_{in}-i\omega_-)\Gamma(1-i\omega_{in}-i\omega_-)}$$ Can the above expression be further simplified?
I'll assume that $\omega_{\text {in}}$ and $\omega_{\text {out}}$ are real and $\omega_{\text {out}}$ is not zero. Since ${_2 F_1}(a, b; c; 0) = 1$, $$F(x) = e^{-i \omega_{\text {in}} x} + o(1), \quad x \to -\infty.$$ The leading terms in the expansion of ${_2 F_1}(a, b; c; z)$ around $z = 1$ are $$\frac {\Gamma(c) \Gamma(a + b - c)} {\Gamma(a) \Gamma(b)} (1 - z)^{c - a - b} + \frac {\Gamma(c) \Gamma(c - a - b)} {\Gamma(c - a) \Gamma(c - b)}.$$ Since we have $\operatorname {Re} (c - a - b) = 0$, neither term is negligible (Gauss's theorem requires $\operatorname {Re} (c - a - b) > 0$). Therefore $$F(x) = \frac {\Gamma(1 - i \omega_{\text {in}}) \Gamma(i \omega_{\text {out}})} {\Gamma(i \omega_-) \Gamma(1 + i \omega_-)} e^{i \omega_{\text {out}} x} + \frac {\Gamma(1 - i \omega_{\text {in}}) \Gamma(-i \omega_{\text {out}})} {\Gamma(-i \omega_+) \Gamma(1 - i \omega_+)} e^{-i \omega_{\text {out}} x} + o(1), \quad x \to \infty.$$