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Wolfram Alpha tells me for instance $$ \frac{\Gamma(6-1/4)}{\Gamma(12+5/4)}-\frac{\Gamma(12-1/4)}{\Gamma(9+5/4)}=\frac{133259008 \sqrt{2} \pi}{1020857565\Gamma(1/4)^2}. $$ I am now looking for a general formula for the constant $C_{ij}$ such that $$ \frac{\Gamma(|i-j|-1/4)}{\Gamma(|i-j|+5/4)}-\frac{\Gamma(i+j-1/4)}{\Gamma(i+j+5/4)}=C_{ij}\frac{\sqrt{2} \pi}{\Gamma(1/4)^2} $$ for $i,j\in\mathbb{N}$, i.e. $C_{9,3}=C_{3,9}= 133259008/1020857565$.

Thanks for your help.

user427574
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  • Note that $i-j$ and $i+j$ have the same parity. The second formula is not a generalization of the first one. – Jean-Claude Arbaut Jun 06 '18 at 11:43
  • So $i+j=9$ in your example. But then $|i-j|\in {7,5,3,1}$. How do you get $6$? – Dietrich Burde Jun 06 '18 at 11:44
  • You are absolutely right. I made a terrible typo. Please find the fixed version. – user427574 Jun 06 '18 at 11:46
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    I am not sure there is an easy closed form, but you can simplify this expression using a combination of two formulas: $\Gamma(x+1)=x\Gamma(x)$ and $\Gamma(x)\Gamma(1-x)=\dfrac{\pi}{\sin(\pi x)}$ – Jean-Claude Arbaut Jun 06 '18 at 11:48
  • How would you proceed? – user427574 Jun 06 '18 at 11:55
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    Reduce the arguments of $\Gamma$ using the first formula, to bring them near $1$. In the end, the numerator will be $\Gamma(3/4)$ and the denominator $\Gamma(1/4)$ (with additional factors coming from the reduction). Then use the second formula to get rid of $\Gamma(3/4)$. The $\sqrt{2}$ in the result comes from $\sin(\pi/4)$. You can write the general result using falling factorials. – Jean-Claude Arbaut Jun 06 '18 at 12:17

1 Answers1

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The rising factorial is defined by

$$x^{(n)}=x(x+1)\dots(x+n-1)=\frac{\Gamma(x+n)}{\Gamma(x)}$$

Let $a,b,c,d$ be integers. Then, using the formula $\Gamma(1/4)\Gamma(3/4)=\dfrac{\pi}{\sin(\pi/4)}$:

$$\frac{\Gamma(a-1/4)}{\Gamma(b+1/4)}=\frac{\Gamma(a-1+3/4)}{\Gamma(b+1/4)}=\frac{\Gamma(3/4)\cdot(3/4)^{(a-1)}}{\Gamma(1/4)\cdot(1/4)^{(b)}}=\frac{(3/4)^{(a-1)}}{(1/4)^{(b)}}\cdot\frac{\pi}{\Gamma(1/4)^2\sin(\pi/4)}=\frac{(3/4)^{(a-1)}}{(1/4)^{(b)}}\cdot\frac{\pi\sqrt{2}}{\Gamma(1/4)^2}$$

Likewise for $c$ and $d$. Thus

$$\frac{\Gamma(a-1/4)}{\Gamma(b+1/4)}-\frac{\Gamma(c-1/4)}{\Gamma(d+1/4)}=\left(\frac{(3/4)^{(a-1)}}{(1/4)^{(b)}}-\frac{(3/4)^{(c-1)}}{(1/4)^{(d)}}\right)\cdot\frac{\pi\sqrt{2}}{\Gamma(1/4)^2}$$

It's possible to rewrite this with the falling factorial, using the formula $x^{(n)}=(x+n-1)_n$. The factor before $\frac{\pi\sqrt{2}}{\Gamma(1/4)^2}$ only involves finite products of rational numbers.

Jean-Claude Arbaut
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