1)Is it possible to simplify
$$J_{k,\alpha}={}_1F_1(2\alpha , 2 ,ik\pi) +{}_1F_1(2\alpha , 2 ,-ik\pi)$$
$k\in \{1,2,\cdots\}$ and $\alpha \in (0,1)$. ${}_1F_1$ is confluent hypergeometric functions of the first kind(Kummer's function).
I know $J_{k,\alpha}$ is real valued.
library(hypergeo)
Jk<-function(k,alpha){
U<-c(2*alpha)
L<-c(2)
z<-1i*k*pi
return(genhypergeo(U,L,z)+genhypergeo(U,L,-z))
}
> Jk(1,.2)
[1] 1.336592+0i
there are some value for $J_{(k,\alpha)}$ \begin{array}{c|ccccccccc} \alpha = &0.1& 0.2& 0.3& 0.4 &0.5& 0.6& 0.7 &0.8 &0.9 \\ \hline k=1 & 1.70 & 1.34 & 0.92 & 0.47 & 0.00 & -0.47 & -0.92 & -1.34 & -1.70 \\ k=2 & 1.41 & 0.85 & 0.40 & 0.10 & 0.00 & 0.10 & 0.40 & 0.85 & 1.41 \\ k=3 & 1.32 & 0.77 & 0.39 & 0.16 & 0.00 & -0.16 & -0.39 & -0.77 & -1.32 \\ k=4 & 1.23 & 0.65 & 0.27 & 0.07 & -0.00 & 0.07 & 0.27 & 0.65 & 1.23 \\ \end{array}
by definition
$${}_1F_1(a , b ,z)=\frac{\Gamma(b)}{\Gamma(b)\Gamma(b-a)}\int_0^1 e^{zu} u^{a-1} (1-u)^{b-a-1} du= \sum_{n=0}^{\infty} \frac{(a)_n}{(b)_n} \frac{z^n}{n!}$$
where $(a)_n=a(a+1)\cdots(a+n-1)$. so $${}_1F_1(2\alpha , 2 ,z)=\frac{1}{\Gamma(2\alpha)\Gamma(2-2\alpha)}\int_0^1 e^{zu} u^{2\alpha-1} (1-u)^{1-2\alpha} du = \sum_{n=0}^{\infty} \frac{(2\alpha)_n}{(n+1)!} \frac{z^n}{n!}$$
2)simplify special cases, $J(k=1,\alpha)$ and $J(k=2,\alpha)$ are very important.
3) finding and proving any relation, like relation $J(k=1,\alpha)$ and $J(k=1,1-\alpha)$ , very appreciated.It seems for even $k$ $$J(k,\alpha)=J(k,1-\alpha)$$
and for odd $k$
$$J(k,\alpha)=-J(k,1-\alpha)$$ .
My try to show $J_{(k,1-\alpha)}=(-1)^k J_{(k,\alpha)}$:
define $M(a,b,z)={}_1F_1(a , b ,z)$
By Kummer's_transformation $$M(a,b,z)=e^z M(b-a,b,-z)$$ and \begin{eqnarray} e^{ik\pi}=\cos(k\pi)+i\sin(k\pi)=\cos(k\pi)=\left\{ \begin{array}{cc} -1 & k\in 1,3,\cdots \\ 1 & k\in 2,4,\cdots \end{array} \right. \end{eqnarray}
\begin{eqnarray} J_{(k,\alpha)}&=&M(2\alpha , 2 ,ik\pi) +M(2\alpha , 2 ,-ik\pi) \\ &=& e^{ik\pi}M(2-2\alpha , 2 ,-ik\pi) +e^{-ik\pi}M(2-2\alpha , 2 ,ik\pi) \\ &=& e^{ik\pi}M(2(1-\alpha) , 2 ,-ik\pi) +e^{-ik\pi}M(2(1-\alpha) , 2 ,ik\pi) \\ &=& \left\{ \begin{array}{cc} -M(2(1-\alpha) , 2 ,-ik\pi) -M(2(1-\alpha) , 2 ,ik\pi) & k\in 1,3,\cdots \\ M(2(1-\alpha) , 2 ,-ik\pi) +M(2(1-\alpha) , 2 ,ik\pi) & k\in 2,4,\cdots \end{array} \right. \\ &=& \left\{ \begin{array}{cc} -J(k,1-\alpha) & k\in 1,3,\cdots \\ J(k,1-\alpha) & k\in 2,4,\cdots \end{array} \right. \end{eqnarray} is this true(valid proof)?
By this result, choosing $\alpha=.5$
$$J(k,.5)=0$$
coment on $\alpha = 1/2$
When $\alpha = 1/2$, note $$ {}_1F_1(2\alpha;2;z) = {}_1F_1(1;2;z) = \sum_{n=0}^\infty \frac{z^n}{(n+1)!} = \frac{e^z-1}{z} $$ so that $$ {}_1F_1(2\alpha;2;i k \pi)+{}_1F_1(2\alpha;2;-i k \pi) =\frac{e^{i k \pi}-1}{i k \pi}+\frac{e^{-i k \pi}-1}{-i k \pi} =\frac{e^{i k \pi}-e^{-i k \pi}}{ik\pi} - \frac{1-1}{ik\pi} =\frac{2\sin(k\pi)}{k\pi} $$ This is true for real $k\ne0$. If $k =1,2,3\cdots$ then we get $J_{k,1/2} = 0$.
Similarly, when $\alpha = 1$ we get $J_{(k,1)} = 2 \cos(k\pi)$.
another note
from $$ {}_1F_1(2\alpha , 2 ,z)=\frac{1}{\Gamma(2\alpha)\Gamma(2-2\alpha)}\int_0^1 e^{zu} u^{2\alpha-1} (1-u)^{1-2\alpha} du $$ we get $$ {}_1F_1(2\alpha , 2 ,i\pi k)+{}_1F_1(2\alpha , 2 ,-i\pi k) =\frac{2}{\Gamma(2\alpha)\Gamma(2-2\alpha)}\int_0^1 \cos({\pi k u}) u^{2\alpha-1} (1-u)^{1-2\alpha} du $$ so this question is very close to $I_k=\int_0^1 \frac{1}{\mathbf{B}(\alpha , \beta )} \cos^k (\pi \theta) \theta^{\alpha -1} (1-\theta)^{\beta -1}d\theta $