A closed formula for a sum involving hypergeometric function

Question

Let ${ }_1 F_1(a ; c ; z)$ be Kummer's function defined by the function, and all its analytic continuations, represented by the infinite series $\sum_{n=0}^{\infty} \frac{(a)_n}{(c)_n} \frac{z^n}{n !}$. That is, $$ \begin{aligned} { }_1 F_1(a ; c ; z) & =\sum_{n=0}^{\infty} \frac{(a)_n}{(c)_n} \frac{z^n}{n !}, c \neq 0,-1,-2, \ldots \\ & =\frac{\Gamma(c)}{\Gamma(a)} \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(c+n)} \frac{z^n}{n !} . \end{aligned} $$

Other symbols used instead of ${ }_1 F_1(a ; c ; z)$ are $\Phi(a, c ; z)$ and $M(a, c, z)$.

Another function closely related to ${ }_1 F_1(a ; c ; z)$ is the Tricomi function défined by $$U(a, c, z)=\frac{\Gamma(1-c)}{\Gamma(a-c+1)}{ }_1 F_1(a, c, x)+\frac{\Gamma(c-1)}{\Gamma(a)} x^{1-c} {}_1F_1(a-c+1,2-c, x) . $$ My question: How to prove this formula for $0<x\leq y$ $$\sum^\infty_{k=0}\frac{(xy)^k}{k!\Gamma(1+\alpha+k)}\Gamma(\beta+k)U(\beta+k,1+\alpha+2k,x+y)=\frac{\Gamma(\beta)}{\Gamma(1+\alpha)}{ }_1F_1(\beta,\alpha+1, x) U(\beta,\alpha+1, y)$$

Thank you in advance.

Answer 1

The formula is wrong (except some special cases like for $x=0$).

The left hand side (LHS) is symmetric in $x$ and $y$. This would imply a symmetry of the RHS.

$${ }_1F_1(\beta,\alpha+1, x) U(\beta,\alpha+1, y)\\=\frac{\Gamma(1+\alpha)}{\Gamma(\beta)}\sum^\infty_{k=0}\frac{(xy)^k}{k!\Gamma(1+\alpha+k)}\Gamma(\beta+k)U(\beta+k,1+\alpha+2k,x+y)\\=\frac{\Gamma(1+\alpha)}{\Gamma(\beta)}\sum^\infty_{k=0}\frac{(yx)^k}{k!\Gamma(1+\alpha+k)}\Gamma(\beta+k)U(\beta+k,1+\alpha+2k,y+x)\\={ }_1F_1(\beta,\alpha+1, y) U(\beta,\alpha+1, x)$$

Clearly, this cannot hold in general. Let's take the example from my comment. If $\alpha=\beta=1$, we get $$_1F_1(1; 2; x)=\frac{e^x-1}{x}$$ and (see https://functions.wolfram.com/07.33.03.0401.01) $$U(1;2;x)=1/y.$$

Therefore $$\frac{e^x-1}{xy}=\frac{e^y-1}{xy},$$ which does imply $x=y$.

Answer 2

Numerically, the equality does seem to be correct for $y>x$. Here is a plot of the left-hand-side (gold) and the right-hand-side (blue) for $\alpha=\beta=x=1$,as a function of $y$.

Plot of $A(1,1,1,y)$ (gold) and $B(1,1,1,y)$ (blue) as a function of $y$.

$$A(\alpha,\beta,x,y)= \sum^\infty_{k=0}\frac{(xy)^k}{k!\Gamma(1+\alpha+k)}\Gamma(\beta+k)U(\beta+k,1+\alpha+2k,x+y),\\ B(\alpha,\beta,x,y)=\frac{\Gamma(\beta)}{\Gamma(1+\alpha)}{ }_1F_1(\beta,\alpha+1, x) U(\beta,\alpha+1, y).$$

The sum over $k$ converges slowly, for this plot I converted it to a Bessel function integral: $$A(\alpha,\beta,x,y)=\int_0^\infty dt\,t^{\beta-1} (t+1)^{\alpha-\beta} e^{-t (x+y)} \bigl[t (t+1) xy\bigr]^{-\alpha/2} I_\alpha\left(2 \sqrt{t(t+1)xy} \right).$$

The two curves depart when $y$ is still a bit bigger than $x$, this is likely an issue of numerical accuracy.

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Concerning a proof, the paper cited in the OP gives a proof of a somewhat different identity, with the $U$ function on the left-hand-side replaced by the product of generalized Laguerre polynomials. I copy that proof below, for the record (the functions $_1F_1$ and $U$ are denoted $\Phi$ and $\Psi$ respectively). It may be possible to rework the left-hand-side of Eq. (11) into the Bessel function integral given above, something to check.

Ref. 7 points to page 140 of H. Buchholz, The Confluent Hypergeometric Function.