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I've come across a certain hypergeometric series and have tried to express it in a different way. So far I've got the following sum:

$$\sum_{k=0}^{\infty}4^{-k}\left | \Gamma(-k+\frac{i}{2}) \right |^2$$

Which just screams Parseval's identity, but it runs from $0$ to $\infty$ instead of from $-\infty$ to $\infty$, so I am a bit stuck and don't really know where to go from here.

To clarify: I know it can be expressed using a Hypergeometric function, but I am looking for a different way of evaluating it.

Tian Vlašić
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Noa Arvidsson
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  • I think summation $0$ to $\infty$ need not be a problem. But do you know some function where the Fourier series involves $\Gamma(-k+i/2)$ ?? That is the reason I doubt Parseval will help . – GEdgar Aug 08 '23 at 16:48

1 Answers1

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The integral representation (which might have something to do with Bessel functions)

$$ \sum_{k=0}^\infty 4^{-k}\left|\Gamma\left(-k+\frac i2\right)\right|^2 =\color{blue}{\frac\pi{\sinh^2(\pi/2)}\int_{-\pi/2}^{\pi/2}e^{\pm x}\cosh\cos x\,dx} $$

is easy to obtain from the known integral (for $a,b\in\mathbb{C}$ with $\Re a>-1$): $$ \int_{-\pi/2}^{\pi/2}e^{ibx}\cos^a x\,dx=\frac\pi{2^a}\frac{\Gamma(1+a)}{\Gamma\left(1+\frac{a+b}2\right)\Gamma\left(1+\frac{a-b}2\right)}; $$ using the reflection formula for $\Gamma$, at $a=2k$ and $b=\pm i$ we get \begin{align*} 4^{-k}\left|\Gamma\left(-k+\frac i2\right)\right|^2 &=\frac{\pi^2}{\sinh^2(\pi/2)}\frac1{2^{2k}\Gamma\left(1+k+\frac i2\right)\Gamma\left(1+k-\frac i2\right)} \\&=\frac\pi{\sinh^2(\pi/2)}\frac1{(2k)!}\int_{-\pi/2}^{\pi/2}e^{\pm x}\cos^{2k}x\,dx, \end{align*} giving the announced result after summation over $k$.

metamorphy
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