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I once heard on the internet that $(\frac{1}{2})!=\frac{\sqrt{\pi}}{2}$ so now, I'm wondering what $(\frac{1}{4})!$ is equal to.

My attempt:

Since $(\frac{1}{2})!=\frac{\sqrt{\pi}}{2}$ and since $\frac{1}{4}=\frac{1}{2}\div2$ then: $$\begin{align} (\frac{1}{4})!=\frac{\sqrt{\pi}}{2}\div2= \\ \frac{\sqrt{\pi}}{4} \end{align}$$ Is my assumption correct? If not, what is the true answer?

By the way, I asked this question just because I am curious.

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    Soo... $\left(\frac{x}2\right)!=\frac12\cdot x!$, you think? As in, $4!=24$ and $2!=2$ hence $2=\frac12\cdot24$... For a more coherent approach, please see WP on the Gamma function. – 2017-01-22
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    Short answer: we don't know if $(1/4)!$ can be expressed using other known constants, and it is believed to not be so. (unless you want to use special functions) – 2017-01-22
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    it is $$\Gamma\left(\frac{5}{4}\right)$$ – 2017-01-22

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Your logic is incorrect. That is not how the factorial nor the Gamma function behave. Take for example $4!$. Since $2=4\div2$, you seem to think that $2!=4!/2$, but a quick check says this is wrong.

Thanks to expressions for values of the gamma function and Wikipedia, it is known that

$$(1/4)!=\Gamma(5/4)=\frac12\pi^{1/4}K\left(\frac1{\sqrt2}\right)^{1/2}\approx0.90640247705$$

where $K(x)$ is the elliptic $K$ function (complete elliptic integral of the first kind).

In general, for non-integer $x$, we usually extend the factorial as follows:

$$x!=\int_0^\infty t^xe^{-t}\ dt$$

for $x>-1$. Other forms may be given in the first link, and for your specific problem, many forms are given in the Wikipdia.

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    Actually, it's $\Gamma(5/4)$. – 2017-01-22
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From here:

We have $$(1/4)! =\Gamma (5/4) =\Gamma (1/4)\frac {(4 (1)-3)!!!!}{4^1} =\Gamma (1/4)\frac {1}{4} \approx 0.90640$$

Also in general for any $x$, $$(x)! \neq \frac {(2x)!}{2} $$ Hope it helps.