# What is the relation between $\Gamma(a)$ and circles?

Looking through my calc textbook, it states that $$\int_0^\infty x^{a-1} e^{-x} \text{d}x =\Gamma(a)$$ As I have read ahead, I can understand most of the fairly basic concepts behind this function, the graph shape, common points, integral behavior, etc. In my studies, I want to look more into the individul points and those relations to $\pi$. I was first looking over this question when introduced to the idea that $$\Gamma(\frac{1}{2})=\sqrt{\pi}$$ from the amazing WA. And I think it is painfully obvious that a circle has a ratio of $$\frac{\text{circumference}}{\text{diameter}}=\pi$$ So you can see why I might be stunned with this fact. When asking my teacher about it, he said he teaches what he knows... and he didn't teach that part. As that answer didn't exactly fit my description of helpful, I am turning to the ultimate SE for assistance. My investigations of the factorial function will most likely occur here because my teacher is having problems answering my more recent questions, so please don't be questionable of asking so many of them. I am new here on the Stack, so I am sorry if my question is too open-ended. Thanks for any help!

• $\pi$ comes up in all sorts of unexpected places. For example $\sum\frac1{n^2}=\pi^2/6$. If you look at the proof that $\Gamma(1/2)=\sqrt{\pi}$ you'll see where it comes up. But that doesn't make it intuitive. Oct 31, 2015 at 23:43
• Also, $\Gamma(x)$ is a continuous function that diverges to infinity, so it's ultimately going to take every value greater than it's minimum. Oct 31, 2015 at 23:45
• @GregoryGrant One could see my confusion though, yes?
– user285523
Oct 31, 2015 at 23:46
• @GregoryGrant I actually understand that relation though, based on rotation around the complex plane with a radius of 1... and Taylor Series, but that is beside the point.
– user285523
Oct 31, 2015 at 23:49
• Have a look at my answer to this closed question: math.stackexchange.com/q/1045648 Nov 1, 2015 at 0:04

The area of a circle with radius $r$ is just $r^2$ times the area of the unit circle, by homothety.

So the area of the circle is the square of the radius times a universal constant, given by: $$\begin{eqnarray*}2\int_{-1}^{1}\sqrt{1-x^2}\,dx &=& 4\int_{0}^{1}\sqrt{1-x^2}\,dx = 2\int_{0}^{1}x^{-1/2}(1-x)^{1/2}\,dx\\ &=& 2\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)} = \color{red}{\Gamma(1/2)^2},\end{eqnarray*}$$

by exploiting the properties of Euler's beta function.

On the other hand, by the definition of the $\Gamma$ function and through a change of variable:

$$\Gamma\left(\frac{1}{2}\right) = \int_{0}^{+\infty}\frac{dx}{e^x\sqrt{x}} = 2\int_{0}^{+\infty}e^{-z^2}\,dz = \int_{-\infty}^{+\infty}e^{-z^2}\,dz$$ we get a Gaussian integral, whose square is straighforward to compute through Fubini's theorem.

Moreover, at page $5$ of the wonderful book by Keith Ball, An Elementary Introduction to Modern Convex Geometry, you may find an interesting derivation of the measure of the $n$-dimensional unit ball through Gaussian integrals. That is a very convincing way to check that $\pi$ and the values of the $\Gamma$ function are clearly linked.

$$\Gamma\left(\frac{1}{2}\right)=\int_0^\infty x^{-\frac{1}{2}}e^{-x}dx.$$ Substitution of $u = x^{\frac{1}{2}}$ yields:

$$\Gamma\left(\frac{1}{2}\right)= 2\int_0^\infty e^{-u^2} \, du.$$

Now taking the square of this one obtains:

$$\Gamma^2\left(\frac{1}{2}\right)= 4 \int_0^\infty e^{-u^2} \, du \int_0^\infty e^{-v^2} \, dv = 4 \int_0^\infty \int_0^\infty e^{-u^2-v^2} \, du \, dv.$$

Now polar coordinates can be introduced, i.e.

$u = r \cos \phi$

$v = r \sin \phi$

$du\,dv = r\,dr\,d\phi$ ($\phi \in [0,2 \pi]$)

After simplification the desired result is obtained.

The relation between $\Gamma$ function and $\pi$ is because the circle (which is parametrized ideally in polar coordinates) can solve this integral easy.

In general, all factorials of argument $a=\dfrac1n$ are intimately related to geometric shapes

described by algebraic equations of the form $x^n+y^n=r^n,~$ which yield the surface

$\displaystyle\int_0^r\sqrt[n]{r^n-x^n}~dx~=~r^2\displaystyle\int_0^1\sqrt[n]{1-t^n}~dt~=~r^2{2a\choose a}^{-1}.~$ Notice also that the same polar

coordinate approach used to evaluate the Gaussian integral can also be applied to the more

general $~n!~=~\displaystyle\int_0^\infty e^{-x^a}~dx~=~\displaystyle\bigg[~\int_{\mathbb R_+^2}e^{-\big(x^a+y^a\big)}~dx~dy~\bigg]^\tfrac12,~$ which makes the connection

to the aforementioned geometric shapes self-evident.

It is not just for the circle in the plane, there is a consistent version that gives the volume of the sphere in $\mathbb R^3$ and then in higher dimension. If $\omega_n$ is the $n$-volume of the unit ball (sphere together with its interior) in $\mathbb R^n,$ as a shorthand $$\omega_n = \frac{\pi^{n/2}}{\left( \frac{n}{2} \right)!}$$ The factorial is to be interpreted as $$\omega_n = \frac{\pi^{n/2}}{ \Gamma\left( 1 + \frac{n}{2} \right)}$$ The easiest proof is induction on odd $n$ by polar coordinates, then induction on even $n$ by polar coordinates.

Given $n$-volume $V_n(R) = \omega_n R^n,$ we set up in polar coordinates to find $\omega_{n+2},$ meaning the $n+2$-ball of radius $1.$ In polar coordinates in the disc of radius $1,$ if we are at a point distance $r$ from the origin, the radius of the $n$-dimensional ball over that point is $\sqrt{1 - r^2}.$ We are integrating that, $$\omega_{n+2} = \int_0^{2 \pi} \int_0^1 \omega_n \left(\sqrt{1 - r^2} \right)^n r \, dr \, d \theta$$ $$\omega_{n+2} = 2 \pi \omega_n \int_0^1 \left(1 - r^2 \right)^{n/2} \, r \, dr$$ $$\omega_{n+2} = 2 \pi \omega_n \; \left. \frac{-1}{n+2} \, \left(1 - r^2 \right)^{(n+2)/2} \, \right|_{r=0}^{r=1}$$ $$\color{blue}{ \omega_{n+2} = \frac{2 \pi \; \omega_n}{n+2}}$$

For example, in dimension $1,$ the $1$-volume of the segment of radius $1$ is its length, $2,$ so $\omega_1 = 2.$ Then $\omega_3 = \frac{4 \pi}{3},$ which you have seen as the sphere volume (radius $R$) being $\frac{4}{3} \pi R^3.$

Do these fit together? $$\color{blue}{ \omega_{n+2} = \frac{2 \pi \; \omega_n}{n+2}}$$ $$\omega_n = \frac{\pi^{n/2}}{ \Gamma\left( 1 + \frac{n}{2} \right)}$$ Proof by induction (by $2$):

$$\omega_{n+2} = \frac{\pi^{(n+2)/2}}{ \Gamma\left( 1 + \frac{n+2}{2} \right)} = \frac{ \pi \pi^{n/2}}{ \frac{n+2}{2} \Gamma\left( \frac{n+2}{2} \right)} = \frac{ \pi \pi^{n/2}}{ \frac{n+2}{2} \Gamma\left( 1 + \frac{n}{2} \right)} = \frac{ \pi \omega_n}{ \frac{n+2}{2} } = \frac{2 \pi \; \omega_n}{n+2}$$

• Your answer is equally good, but I had to pick 1 so sorry. +1 anyways
– user285523
Nov 1, 2015 at 19:54