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Does anyone have any ideas on howto verify $$\sum_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)}\cdot \frac{1}{3^{n+m+3x-1}} = \Gamma(x)$$ for $x>0$?

I posted this question also on math.stackexchange.

This is not an exercise from a book, but arises due to my research in the study of a probability density.

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  • $\begingroup$ LHS seems to have a pole at $x=-1/3$, while RHS does not. $\endgroup$
    – Alex Gavrilov
    Commented Jul 25, 2017 at 11:16
  • $\begingroup$ (This is a problem because, contrary to Nemo, the series converges everywhere.) $\endgroup$
    – Alex Gavrilov
    Commented Jul 25, 2017 at 11:25
  • $\begingroup$ This is equivalent to double integration $\int_0^1\int_0^1 \frac{u^{x-1} v^{x-1}}{(u+v+1)^{3 x}} dudv=\frac{\Gamma^3(x)}{3\Gamma(3x)}$. $\endgroup$
    – Nemo
    Commented Jul 25, 2017 at 14:04
  • $\begingroup$ This request is clearly much better suited to fit in math.SE. I feel very sorry, I should practice patience. $\endgroup$
    – maliesen
    Commented Jul 25, 2017 at 19:10

1 Answer 1

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This problem can be reduced at least formally to a compact double integral, which might be easier to solve.

Starting with the integral representation for the Gamma function, we write the double sum as an integral of the square of the confluent hypergeometric function ${}_1F_1$, then apply analogue of Euler's transformation formula: \begin{align} &\sum_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)}\cdot \frac{1}{3^{n+m+3x-1}}\\ &=3\int_0^\infty e^{-3t}t^{3x-1}\sum_{n,m=0}^\infty \frac{t^{n+m}}{\Gamma(n+1+x)\Gamma(m+1+x)}dt\\ &=\frac3{\Gamma^2(x+1)}\int_0^\infty e^{-3t}t^{3x-1}~\left[{}_1F_1(1,x+1;t)\right]^2dt\\ &=\frac3{\Gamma^2(x+1)}\int_0^\infty e^{-t}t^{3x-1}~\left[{}_1F_1(x,x+1;-t)\right]^2dt\\ &=\frac3{\Gamma^2(x+1)}\int_0^\infty e^{-t}t^{3x-1}~\sum_{n,m=0}^\infty\frac{(x)_n(x)_m}{(x+1)_n(x+1)_m}\frac{(-t)^{n+m}}{n!m!}dt\\ &=\frac{3x^2}{\Gamma^2(x+1)}\int_0^\infty e^{-t}t^{3x-1}~\sum_{n,m=0}^\infty\frac{1}{(x+n)(x+m)}\frac{(-t)^{n+m}}{n!m!}dt\\ &=\frac{3}{\Gamma^2(x)}\int_0^\infty e^{-t}t^{3x-1}\int_0^1 du\int_0^1 dv\sum_{n,m=0}^\infty\frac{(-t)^{n+m}u^{n+x-1}v^{m+x-1}}{n!m!} dt\\ &=\frac{3}{\Gamma^2(x)}\int_0^\infty e^{-t}t^{3x-1}\int_0^1 du\int_0^1 dv~ e^{-t(u+v)}(uv)^{x-1}dt\\ &=\frac{3\Gamma(3x)}{\Gamma^2(x)}\int_0^1\int_0^1 \frac{u^{x-1} v^{x-1}}{(u+v+1)^{3 x}} dudv. \end{align}

This means that the initial problem is equivalent to $$\int_0^1\int_0^1 \frac{u^{x-1} v^{x-1}}{(u+v+1)^{3 x}} dudv=\frac{\Gamma^3(x)}{3\Gamma(3x)}. $$

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    $\begingroup$ I didn't check any details of this answer, but for fun I had Mathematica spot-check if the double integral equals $\Gamma^3(x)/(3 \Gamma(3x))$. For $x=1,2,3,4,5,6$, at least, Mathematica can verify both sides agree as rational numbers. A couple other random non-integral choices of $x$ also gives agreement up to the default accuracy. $\endgroup$
    – Matt Young
    Commented Jul 25, 2017 at 15:58
  • $\begingroup$ It is easy to prove that $\int_0^\infty\int_0^\infty \frac{u^{x-1} v^{x-1}}{(u+v+1)^{3 x}} dudv=\frac{\Gamma^3(x)}{\Gamma(3x)}$ $\endgroup$
    – Nemo
    Commented Jul 25, 2017 at 18:10
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    $\begingroup$ There is a much simpler proof, just expand the function $\frac{1}{(3-(1-u)-(1-v))^{3x}}$ into the powers of $(1-u)^n(1-v)^m$ and integrate termwise. $\endgroup$
    – Nemo
    Commented Jul 25, 2017 at 19:38

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