I'm trying to find a proof that the following definitions of $\pi$ are equivalent:

  1. $\pi$ is the smallest positive real number such that $e^{\pi i}=-1$.

  2. $\pi$ is half the circumference of a circle with radius $1$, namely $$\pi :=\int_{-1}^{1}\frac{1}{\sqrt{1-x^2}}dx.$$

  3. $\pi$ is the area of a circle of radius $1$, namely

$$\pi :=2\int_{-1}^{1}\sqrt{1-x^2}dx.$$

The worst part is that I remember seeing a proof that the first definition was equivalent to one of the other two, yet I don't remember the proof nor where I saw it, and I now can't seem to find it. I would appreciate any help.

  • $\begingroup$ Are you allowed to use $s=r\theta$, where $s$ is the length of the arc cut off by an angle $\theta$ based at the center of a circle of radius $r$? $\endgroup$ Oct 2, 2020 at 2:45
  • $\begingroup$ Isn't 1. <=> 2. just the substitution $x=(e^{it}+1/e^{it})/2$? $\endgroup$ Oct 2, 2020 at 7:58

2 Answers 2


Here's a calculus proof that 2 is equivalent to 3.

Integration by parts gives \begin{align} \int_{-1}^1\sqrt{1-x^2}\,dx &=\left[x\sqrt{1-x^2}\right]_{x=-1}^1+\int_{-1}^1\frac{x^2}{\sqrt{1-x^2}}\,dx\\ &=\int_{-1}^1\frac{1-(1-x^2)}{\sqrt{1-x^2}}\,dx\\ &=\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}-\int_{-1}^1\sqrt{1-x^2}\,dx \end{align} and so $$2\int_{-1}^1\sqrt{1-x^2}\,dx=\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}.$$

  • $\begingroup$ Shouldn't the first $+$ be a $-$? $\endgroup$
    – Sam
    Oct 2, 2020 at 3:02

Consider the exponential function $f(z)=e^z=\sum \frac{z^k}{k!}$ which is entire and is the unique such that satisfies $f'=f, f(0)=1$ (just look at the Taylor series conditions implied by $f'=f$) so in particular we get the fundamental property that $f(z+w)=f(z)f(w)$ (both satisfy the same $g'=g, g(0)=f(w)$ so if $f(w)=0, g(z)=f(z+w)=0$ so $f=0$ contradiction, hence $f(z+w)/f(w)$ satisfies the differential equation and intial condtion of $f(z)$ so it is $f(z)$!)

But since the Taylor coefficients are real, $f(\bar z)=\bar f(z)$ so $e^{-ix}= \overline {e^{ix}}, x \in \mathbb R$ and in particular $|e^{ix}|^2=e^{ix}\overline {e^{ix}}=e^{ix}e^{-ix}=e^0=1$ so $f$ maps $i\mathbb R$ into the unit circle $S$ also as a group homomorphism; from here it is immediate that $f(i\mathbb R)=S$ (the circle doesn't have non-trivial connected subgroups for example) and actually $f$ restricted there is periodic since by compacity of $S$ the map cannot be injective and onto so there are $f(ix)=f(iy), x \ne y$ or $f(ia)=1, a=x-y \ne 0$ and choosing the smaller positive such and calling it $2b$ we get $f(2ib)=1, f^2(ib)=1$ so $f(ib)=-1$ as by the group homomorphism property any $f(ia)=1, a \ne 0$ gives a period of $f$ and $b <2b$ so it cannot be such!

We claim that $b$ is the area of the unit circle and half the circumference of same, so giving the equivalence between the 3 definitions (as for example if $c$ is the area say, we know that the $b$ above which is defined independently of $c$ is also the area so $b=c$ etc)

Since by defintion of $2b$ we get a parametrization of the unit circle by $f(it), 0 \le t \le 2b$ as $f$ is then injective on $(0,2b)$, we can apply the usual length formula ($L=\int_0^{2b}|f'(it)|dt$) noting that $|f'(it)|=1$ so getting that the circumference of the unit circle is $2b$

Similarly we easily see that $rf(e^{it})$ is a bijective map from $[\epsilon, 1]\times [0,2b)$ into the annulus with outer circle $S$ and inner circle of radius $\epsilon$ and with Jacobian $r$, so by the usual change of variable formula from $2$d calculus we get that the area of the annulus is $b(1-\epsilon^2)$ so letting $\epsilon \to 0$ we get that the area of the unit circle is $b$ and we are done!

Note that the formulas 2,3 in the OP are obtained by using different parameterizations of the unit circle and disc respectively, so if one wants to get a direct integral equivalence, just apply the change of variable formulas with respect to the parametrizations given by $f(it)=(u(t),v(t))$ and $rf(it)$ respectively


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