Is there any known $x\in (0,1)\setminus \{1/2\}$? such that a simple closed form for $\Gamma(x)$ exists?

Question

Motivation: The idea for this question came from an situation I encountered recently. A friend of mine was working on a math problem and tried to compute $ \Gamma\left(\frac{1}{4}\right) $, thinking it might be required to find an exact closed form. I quickly told him that it wasn’t worth the effort, as no 'nice' closed form exists for $ \Gamma\left(\frac{1}{4}\right) $. I also added that, the only $ x \in (0,1) $ with a simple closed form for $ \Gamma(x) $ is $ \frac{1}{2} $ and it is impossible to find a simple closed form for the rest, so he won't waste his time trying to find $\Gamma(x) $ since it isn't required or even possible. But this got me wondering "really? no $x \in (0,1)\setminus\{1/2\}$ has a nice closed form for $\Gamma(x)$?!"

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It is well-known that $\Gamma\left(\frac 12\right)=\sqrt \pi$ This made me curious: are there any other values $x\in (0,1)\setminus\{1/2\}$ for which $\Gamma(x)$ has a simple closed form?.

Any parameter $x$ involved should also be expressible in a simple closed form.

I searched on Wikipedia but I couldn’t find any such $x$ , Additionally, the lack of simple closed forms for $\Gamma\left(\frac 14\right)$ and $ \Gamma\left(\frac 13\right)$ seems to suggest that no such $x$ exists. However, I wonder if there could be some very specific, unusual value of $a$ for which we have a simple closed form for $\Gamma(x)$

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Simple or nice closed form: is a finite combination of basic elementary functions, expressed in terms of integer arguments..

Basic elementary functions are $x^n, e^x, \sin(x)$ for all $n \in\mathbb{Z}$ and their inverses.

A combination of two functions $f$, $g$ formed through operations such as addition $(f+g)$, multiplication ($f\cdot g$), or composition ( $f\circ g$).

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Update: I asked this question on MO here.

Answer 1

If you can accept $\pi$ as a closed-form, then that may be a start as there are actually generalizations of $\pi$ using closed curves other than a circle.

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I. Integrals

First define the $\pi_k$ as,

\begin{align} \pi_2 &= 2\int_0^1\frac1{(1-x^2)^{1/2}}\,dx = 3.1415926535\dots\\[4pt] \pi_3 &= 3\int_0^1\frac1{(1-x^3)^{2/3}}\,dx = 5.2999162508\dots\\[4pt] \pi_4 &= 4\int_0^1\frac1{(1-x^4)^{3/4}}\,dx = 7.4162987092\dots\\ \end{align}

Alternatively,

\begin{align} \pi_2 &= \;\Gamma\big(\tfrac12\big)\,\Gamma\big(\tfrac12\big)\,=\,\pi\\[4pt] \pi_3 &= \frac{\;\Gamma\big(\tfrac13\big)\,\Gamma\big(\tfrac13\big)}{\;\Gamma\big(\tfrac23\big)\,\Gamma\big(\tfrac33\big)} = \frac{\sqrt3\,\Gamma^3\big(\tfrac13\big)}{2\pi}\\[4pt] \pi_4 &= \frac{\sqrt2\,\Gamma\big(\tfrac14\big)\,\Gamma\big(\tfrac24\big)}{\,\Gamma\big(\tfrac34\big)\,\Gamma\big(\tfrac44\big)}=\frac{\Gamma^2\big(\tfrac14\big)}{\sqrt\pi}\\ \end{align}

Incidentally,

• The half-period $\dfrac{\pi_3}{2\sqrt3} = \omega_2 \approx 1.529954$ is also called the omega_2 constant.
• The half-period $\dfrac{\pi_4}{2\sqrt2} = \varpi = L \approx 2.622057$ is also called the lemniscate constant.

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II. As arc lengths

Note that,

• $\pi_2 \approx 3.14159$ is arc length for the unit circle.
• $\pi_3\sqrt3 \approx 9.179724$ is arc length for the unit trefoil.
• $\pi_4/\sqrt2 \approx 5.244115$ is arc length for the unit lemniscate.

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III. Role in $x^k+y^k=1$

Note that,

• $\pi_2$ for the trigonometric functions and $x^2+y^2=1$, with $\color{blue}{2\sin^{-1}(1)=\pi_2}$.
• $\pi_3$ for the Dixon elliptic functions and $x^3+y^3=1$, with $\color{blue}{3\operatorname{sm}^{-1}(1)=\pi_3}$.
• $\pi_4$ for the lemniscate elliptic functions and $x^4+y^4=1$, with $\color{blue}{4\sqrt2\,\operatorname{sl}^{-1}(1)=\pi_4}$.

$$\sin^2(z)+\cos^2(z)=1$$ $$\operatorname{sm}^3(z)+\operatorname{cm}^3(z)=1$$ $$\big(1+\operatorname{sl}^2(z)\big)\big(1+\operatorname{cl}^2(z)\big)=2$$

with the latter two being the Dixon and lemniscate functions, respectively. The $x^4+y^4=1$ is obeyed by the hyperbolic lemniscate functions.

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IV. Conclusion

Therefore, if we allow the $\pi_k$ as closed-forms and fundamental constants in $x^k+y^k=1$, then $\Gamma\big(\tfrac12\big)$ can be expressed by $\pi_2$, while $\Gamma\big(\tfrac13\big)$ by $\pi_3$, and $\Gamma\big(\tfrac14\big)$ by $\pi_4$ using the formulas in the first section.

Answer 2

No explicit evaluations of other gamma values are known. See the recent paper by Raimundas Vidūnas, Expressions for values of the gamma function. (available online)

The author writes

Values of the gamma function are of broad interest. ... An easy consequence of the reflection formula is $\Gamma(\frac{1}{2}) = \sqrt \pi$. ... No explicit evaluations of other gamma values are known.

Unfortunately, per the conditions of your question, only $\Gamma(\frac{1}{2})$ is known. The form of known representations of the gamma function and terms are discussed soon after.

Some gamma terms occur as values of hypergeometric functions at special points and as period integrals. ... Conversely, some gamma values at rational points can be expressed in terms of the elliptic integrals.

Nonetheless, some beautiful relations between specific gamma values exist (see the paper above).

$$\frac{\Gamma(\frac{1}{2}) \Gamma(\frac{2}{3})}{\Gamma(\frac{3}{4}) \Gamma(\frac{5}{12})} = \frac{\sqrt{\sqrt 3 +1}}{2^{1/4} 3^{3/8}}$$

$$\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac4{15}\right)}{\Gamma\left(\frac13\right)\Gamma\left(\frac2{15}\right)}=\frac{\sqrt2\,\,\sqrt[20]3}{\sqrt[6]5\,\sqrt[4]{5-\frac{7}{\sqrt{5}}+\sqrt{6-\frac{6}{\sqrt{5}}}}}=\frac{\phi \,\, \sqrt[20]3 \,\, \sqrt{\!\sqrt 3 \cdot \sqrt[4] 5-\phi^{3/2}}}{\sqrt 2 \,\, \sqrt[24] 5}$$

Hope this helps :)