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Is it possible to find some of Gamma's non-integer values by using some formulas such as: $$\Gamma(x)\Gamma(1-x)={\pi \over sin(\pi x)}$$

I know that the only known value that When $x=1-x$ and hence we can determine $\Gamma (\frac12)=\sqrt{\pi}$ because of the special case $\Gamma(\frac12)=\Gamma(1-\frac12)$ and by this value we can find all half-integers values by using Gamma property $\Gamma(x+1)=x\Gamma(x)$.

But we can note that there are infinite number of real solutions for$\Gamma(x)=\Gamma(1-x)$ which could help to find some real values.

For example let $\alpha \approx0.026042634$

So if we let $\Gamma(\alpha)=\Gamma(1-\alpha)$ it will not make difference because $\Gamma(\alpha) \neq \Gamma(1-\alpha)$

but actually, I can graphically and numerically check that $\Gamma(\alpha+4)=\Gamma(1-(\alpha+4))\approx 6.20011657$

and by using the reflection formula I have,

$$\Gamma(\alpha+4)^2={\pi \over sin(\pi \alpha)}$$ or $$\Gamma(\alpha)^2={\pi \over [\alpha(\alpha+1)(\alpha+2)(\alpha+3)]^2sin(\pi \alpha)}$$

Now the problem is that I should use an other formula or useful relation to find the value of $\alpha$.

I tried to use the product of sin $Sin(\pi x)=x\prod_{n=1}^{\infty} (1-\frac{x^2}{n^2})$ but it seems that this formula will not help.

So is there a hope to find values for Gamma's non-integer numbers?

I heard that $\Gamma(\frac13)$ and $\Gamma(\frac14)$ has been proven to be transcendental which made me feel that finding such values is just wasting of time for the mathematicians who are interested in it.

Pentapolis
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  • Is your question, for which values $x$ one gets for $\Gamma(x)=\Gamma(1-x)$ ?! Maybe it helps a bit, if you use the product or the integral for $\Gamma$. – user90369 Oct 06 '16 at 12:01
  • @user90369 does this product can give these values exactly I mean without approximation? – Pentapolis Oct 06 '16 at 12:09
  • No, only approximation. – user90369 Oct 06 '16 at 12:32
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    Addition: One can use the functional equation of the Riemann zeta function. Then one gets e.g. the condition $\enspace \displaystyle (2\pi)^{2x-1}\tan\frac{\pi x}{2}=(\frac{\zeta(x)}{\zeta(1-x)})^2$ where the zeta function is analytically extended. – user90369 Oct 06 '16 at 12:46

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