Is there any intuitive way to think about the gamma function?

Is there a way to realize the gamma function intuitively? My first (and probably correct) guess is no, because, for example, $\Gamma(\frac 12)=\sqrt{\pi}$ doesn't make any intuitive sense at all. Also, to me it seems like negative integers would be more likely candidates to have defined values than fractions, but negative integers aren't defined with the gamma function. I saw this question which pointed out "Euler's reflection formula":

$$\Gamma(z)\Gamma(1-z)=\frac {\pi}{\sin(\pi z)}$$

This may offer some inspiration, but most likely not. Thank you for any responses to this question!

Edit: I saw the marked duplicate to this question. While it is good, I think it focuses more on the mathematical rigor than any intuition (when I say this I mean something that could be taught to a first or second year calculus student) that may exist.

• possible duplicate of Intuition for the definition of the Gamma function? Jun 7, 2015 at 4:01
• Jun 7, 2015 at 4:02
• Also also related Jun 7, 2015 at 4:58
• @columbus8myhw (and half of the world) please add some relevant text to your links (i.e., not "see this", "this page") Can you see why Zev Chonoles's links are much more useful than yours? Jun 7, 2015 at 5:58
• Oh, OK… My link is an answer I made a while back trying to explain some of the intuition behind the gamma function. Jun 7, 2015 at 6:00

$\Gamma\bigg(\dfrac12\bigg)=\sqrt\pi~$ doesn't make any intuitive sense at all.
Speak for yourself ! All factorials of argument $\dfrac1n$ are connected to geometric shapes of the form
$x^n+y^n=1$. We have $~\displaystyle\int_0^\infty e^{-x^n}~dx~=~\bigg(\dfrac1n\bigg){\large!}\quad$ and $\quad\displaystyle\int_0^1\sqrt[m]{1-x^n}~dx~=~\frac{\bigg(\dfrac1m\bigg){\large!}~\bigg(\dfrac1n\bigg){\large!}}{\bigg(\dfrac1m+\dfrac1n\bigg){\large!}}$
Connecting the latter integral to the superellipse $x^n+y^m=1$ is trivial. As for the former, just integrate along the two axes of coordinates. I wrote about all of this here, in my answer to this question.
$\Gamma(z)\Gamma(1-z)=\dfrac\pi{\sin(\pi z)}$
Read the article on the Basel problem, and then see my answer to this question. Similarly, we also have $\Gamma\bigg(\dfrac12+z\bigg)\Gamma\bigg(\dfrac12-z\bigg)=\dfrac\pi{\cos(\pi z)}~.$