0

The Gamma function has certain properties that allow it being evaluated for certain complex arguments. Is there any known expression simplifying

$\Gamma(-n + i x) \Gamma(-n - i x)$

for integers $n \ge 0$ and real variables $x$? I checked Abramovitz and Stegun, but didn't find anything. Mathematica can simplify this for $n=0$ and $n=1$, but doesn't simplify for larger $n$.

Thanks in advance!

Sah
  • 1
  • After some numerical analysis, I think the solution has the form $\frac{\pi \operatorname{cosech}{(\pi x)}}{\sum_{j=0}^{n} a_j x^{2j+1}}$, but I am unable to work out the $a_j$. – Sah May 08 '19 at 12:58

1 Answers1

3

By the recursive relation $\Gamma(x+1)=x\Gamma(x)$ one has

$$\Gamma(z-n)=\Gamma(z)\prod_{k=1}^n\frac1{z-k}$$

and hence,

$$\Gamma(ix-n)\Gamma(-ix-n)=\Gamma(ix)\Gamma(-ix)\prod_{k=1}^n\frac1{k^2-x^2}$$

but by the reflection formula, $\Gamma(z)\Gamma(-z)=-\frac\pi{z\sin(\pi z)}$, giving us

$$\Gamma(ix-n)\Gamma(-ix-n)=-\frac\pi{ix\sin(\pi ix)}\prod_{k=1}^n\frac1{k^2-x^2}$$

or, without imaginary numbers,

$$\Gamma(ix-n)\Gamma(-ix-n)=\frac\pi{x\sinh(\pi x)}\prod_{k=1}^n\frac1{k^2-x^2}$$

where the product multiplies out into the form of

$$\prod_{k=1}^n(k^2-x^2)=\sum_{k=0}^na_kx^{2k}$$

where

$$a_k=(-1)^k\sum_{1\le i_1<i_2<\cdots<i_{n-k}\le n}(i_1i_2\cdots i_{n-k})^2$$

are given by Vieta's formulas.

Simply Beautiful Art
  • 74,685
  • Thanks for this! I think you have an error in the product part though (you have $i x \times -i x$ so it should go to $+x^2$). It should then be $k^2+x^2$. (I also checked this numerically) – Sah May 13 '19 at 09:57