Let $b$ and $w$ be real parameters subject to $b\neq 1$. Let $x \in {\mathbb R}$. Define: \begin{equation} {\mathfrak N}(w,b):= \frac{2^{\frac{1}{2 (1-b)}+1} \left(\frac{1}{b-1}\right)^{\frac{1}{2 (b-1)}} w^{\frac{1}{2 (b-1)}+1}}{\frac{1}{2 (b-1)}!} \quad (I) \end{equation} Then the following identity holds: \begin{eqnarray} &&x^{\frac{1-b}{2}} e^{-\frac{i w x^{1-b}}{b-1}} {\mathfrak N}(w,b) \cdot \, _1F_1\left(\frac{3}{2}+\frac{1}{2 (b-1)};2+\frac{1}{b-1};\frac{2 i w x^{1-b}}{b-1}\right)=\\ &&\!\!\!\!\left( \left(2 w x^{1-\frac{b}{2}}+i x^{b/2}\right) J_{\frac{1}{2 (b-1)}}\left(\frac{w x^{1-b}}{b-1}\right)-i w x^{1-\frac{b}{2}} \left(J_{\frac{1}{2 (b-1)}-1}\left(\frac{w x^{1-b}}{b-1}\right)-J_{1+\frac{1}{2 (b-1)}}\left(\frac{w x^{1-b}}{b-1}\right)\right)\right) \end{eqnarray} The code below serves as a verification:
b = RandomReal[{-5, 5}]
w = RandomReal[{1, 20}]; x =.; Clear[NN];
NN[w_, b_] := (1/(-1 + b))^(1/(2 (-1 + b))) 2^(1 + 1/(2 (1 - b)))
w^(1 + 1/(2 (-1 + b)))/(1/(2 (-1 + b)))!;
fp[x_] := (-I w x^(
1 - b/2) (BesselJ[-1 + 1/(2 (-1 + b)), (
w x^(1 - b))/(-1 + b)] -
BesselJ[1 + 1/(2 (-1 + b)), (
w x^(1 - b))/(-1 + b)]) + (2 w x^(1 - b/2) +
I x^(b/2)) BesselJ[1/(2 (-1 + b)), (w x^(1 - b))/(-1 + b)]) /
NN[w, b];
fm[x_] := (-I w x BesselJ[-1 + 1/(2 - 2 b), (w x^(1 - b))/(-1 + b)] +
I w x BesselJ[1 + 1/(2 - 2 b), (
w x^(1 - b))/(-1 + b)] + (2 w x + I x^b) BesselJ[1/(2 - 2 b), (
w x^(1 - b))/(-1 + b)]) x^(-b/2) ;
yp[x_] :=
E^(-((I w x^(1 - b))/(-1 + b))) x^((1 - b)/2)
Hypergeometric1F1[3/2 + 1/(2 (-1 + b)),
2 + 1/(-1 + b), ((2 I w x^(1 - b))/(-1 + b))];
ym[x_] :=
E^(-((I w x^(1 - b))/(-1 + b))) x^((1 - b)/2)
HypergeometricU[3/2 + 1/(2 (-1 + b)),
2 + 1/(-1 + b), ((2 I w x^(1 - b))/(-1 + b))];
b = RandomReal[{-5, -1}];
w = RandomReal[{1, 20}]; x = Range[0, 0.1, 0.0001];
ListPlot[First[{Re[#], Im[#]} & /@ {fp[x]/yp[x]}]]
We have obtained the identity above by analyzing solutions to the following ODE: \begin{equation} \left[\frac{d^2}{d x^2} + \left( \frac{1-b^2}{4 x^2} - \imath (-1+b) w x^{-1-b} + w^2 x^{-2 b}\right)\right] \cdot y(x)=0 \end{equation} The ODE in question is on one hand produced from the Whittaker ODE by rescaling the variables appropriately. On the other hand the ODE in question corresponds to a certain gauge transformation (Gauge transformation of differential equations.) of the Bessel ODE. As a matter of fact both sides of equation $(I)$ are solutions to our ODE above and the nomalization constant ${\mathfrak N}(w,b)$ ensures that the the values of the function and its derivative at zero match.
Note that if we take $b=5/4$ we retrieve the following entry http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F1/03/02/13/0020/ in Wolframs database as a special case.
Now having said all this my question would be how would one go about deriving the identity in question using some alternative approach?
Introducing the notations: \begin{equation} \nu=\frac{1}{2(b-1)}\: ; z= i w \frac{x^{1-b}}{b-1} \end{equation} we have to evaluate \begin{equation} Y= \,_1F_1\left(\nu+ \frac{3}{2};2\nu+2;2z\right) \end{equation} Using the derivative formula for the Kummer confluent hypergeometric function: \begin{align} Y&=\frac{\partial}{\partial z}\,_1F_1\left(\nu+ \frac{1}{2};2\nu+1;z\right)\\ &=\Gamma(\nu+1)\frac{\partial}{\partial z}\left[ 2^\nu e^zz^{-\nu}I_\nu(z)\right]\\ &=2^\nu\Gamma(\nu+1)e^z\left( z^{-\nu}-\nu z^{-\nu-1} \right)I_\nu(z) \\ &\qquad\qquad+2^{\nu-1}\Gamma(\nu+1)e^zz^{-\nu}\left[I_{\nu-1}(z)+I_{\nu+1}(z) \right]\\ &=2^{\nu-1}(\nu)!z^{-\nu}e^z\left\lbrace2\left( 1-\nu z^{-1} \right)I_\nu(z)+ \left[I_{\nu-1}(z)+I_{\nu+1}(z) \right] \right\rbrace \end{align} where the hypergeometric representation for the modified Bessel function $I_\nu$ and an expression of its derivative were used. Introducing the representation \begin{equation} I_\mu(z)=I_\mu\left( i w \frac{x^{1-b}}{b-1} \right)=i^\mu J_\mu\left( w \frac{x^{1-b}}{b-1} \right) \end{equation} in the latter expression should give the proposed identity.