# A closed formula for a sum involving hypergeometric function

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 $$0 $$\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)$$

• Why do you think identity is true? How did you get it? Commented Nov 15, 2023 at 2:42
• The formula seems to be wrong. The left hand side is symmetric in x and y, but the right hand side is not. Commented Nov 15, 2023 at 7:16
• @losif Pinelis. I found it in the following paper formula 11: " Resolvent of harmonic oscillator Hamiltonian and its application to Fourier transform for generalized functions". Commented Nov 15, 2023 at 14:09
• @zoranVicovic Please determine a domain for the variables. You are right. The trivial case $x=0$ works, but in general the equation will fail. For example: Let $\alpha=\beta=1$, $x=1$, and $y=2$. The rhs evaluates to $\frac{e-1}{2}$. Exchange x and y ($x=2$, $y=1$) and redo the calculation of the rhs. We get $\frac{e^2-1}{2}$. But looking at the lhs, both values should be the same! Commented Nov 15, 2023 at 15:36
• @zoranVicovic The code is Gamma[1]/Gamma[2] Hypergeometric1F1[1, 2, 1] HypergeometricU[1, 2, 2] and Gamma[1]/Gamma[2] Hypergeometric1F1[1, 2, 2] HypergeometricU[1, 2, 1] Commented Nov 16, 2023 at 12:11

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 $$x=y$$.

• You are right. it is true when $0<x\leq y$ it is written in the following paper formula 11: " Resolvent of harmonic oscillator Hamiltonian and its application to Fourier transform for generalized functions". Thank you Commented Nov 16, 2023 at 12:43

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.

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.

• Thank you a lot@Carlo Beenakker Commented Nov 16, 2023 at 14:37
• I will try to verify the identity thank you for your book Commented Nov 17, 2023 at 17:31