It is well known that a circle with radius *r* has diameter *2πr*. But is there a similar formula for the circumference of an ellipse?

This is our goal:

The circumference of an ellipse defined by $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1, \quad a>b$$ is $$4a\left(\frac{\pi}{2}\right) \sum_{k=0}^\infty \left[\frac{(2k)!}{4^k(k!)^2}\right]^2 \left(\frac{e^{2k}}{1-2k}\right),$$ where \(e = \sqrt{1-b^2/a^2}\) is the eccentricity of the ellipse.

This formula looks complicated, but it’s not too hard to derive as we’ll see.

## Step 1. An easy integral

Let $$I_n = \int_{0}^{\pi/2} \sin^n x\,dx.$$ It is easy to show (using integration by parts) that $$I_n=\left(\frac{n-1}{n}\right)I_{n-2},$$ and obviously \(I_0=\pi/2\). Then $$I_{2k} = \left(\frac{2k-1}{2k}\right) \left(\frac{2k-3}{2k-2}\right) \cdots \left(\frac{1}{2}\right) \left(\frac{\pi}{2}\right).$$ But $$(2k)(2k-2)\cdots(4)(2) = 2^k(k)(k-1)\cdots(2)(1)=2^k k!$$ and $$(2k-1)(2k-3)\cdots(3)(1) = \frac{(2k)(2k-1)\cdots(2)(1)}{(2k)(2k-2)\cdots(4)(2)} = \frac{(2k)!}{2^k k!},$$ so $$I_{2k} = \left(\frac{\pi}{2}\right)\frac{(2k)!}{4^k (k!)^2}.$$

## Step 2. A square root formula

Now we use the well-known formula $$(1+x)^\alpha = \sum_{k=0}^\infty \binom{\alpha}{k} x^k,$$ where $$\binom{\alpha}{k} = \frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}.$$ This holds for any complex \(\alpha\) and \(|x|<1\). Plug in \(-x\) and \(\alpha=1/2\) to get \begin{align}

\sqrt{1-x} &= \sum_{k=0}^\infty \frac{\frac{1}{2}(\frac{1}{2}-1)\cdots(\frac{1}{2}-k+1)}{k!} (-x)^k \\

&=\sum_{k=0}^\infty \frac{(1)(-1)(-3)\cdots(5-2k)(3-2k)}{2^k k!} (-x)^k \\

&=\sum_{k=0}^\infty \frac{(1)(1)(3)\cdots(2k-5)(3-2k)}{2^k k!} x^k \\

&=\sum_{k=0}^\infty \frac{(2k)!}{(1-2k) 4^k (k!)^2} x^k,

\end{align} which holds for all \(|x|<1\).

## Step 3. The arc length formula

Non-rigorously, from the Pythagorean theorem you can see from the above diagram that $$ds = \sqrt{dx^2+dy^2} = \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx,$$ where \(ds\) is an infinitesimal part of the circumference.

The length of the top half of the ellipse is in fact defined as $$\int_{-a}^{a} \sqrt{1+f'(x)^2}\,dx,$$ where \(f(x)=(b/a)\sqrt{a^2-x^2}\) is just the equation of the ellipse in terms of \(y\). Let \(E\) be the circumference of the entire ellipse. Then \begin{align}

E &= 2 \int_{-a}^{a} \sqrt{1+\left(-\frac{bx}{a\sqrt{a^2-x^2}}\right)^2}\,dx \\

&= 2 \int_{-a}^{a} \sqrt{\frac{a^2-e^2x^2}{a^2-x^2}}\,dx

\end{align} after some simplification using the fact that from the definition of the eccentricity \(e\) we have $$b^2=a^2(1-e^2) \Rightarrow e^2=\frac{a^2-b^2}{a^2e^2} \Rightarrow a^2-b^2=a^2e^2.$$ Finally, after applying the substitution \(x = a\sin\theta\) and noting that the integrand is even, we have $$E = 4a \int_{0}^{\pi/2} \sqrt{1-e^2\sin^2\theta}\,d\theta.$$

## Step 4. Putting it together

Since \(e<1\), the series we derived in Step 2 converges uniformly on \([-e^2,e^2]\). Therefore we can use Step 2 and Step 1 to deduce that \begin{align}

E &= 4a \int_{0}^{\pi/2} \sum_{k=0}^\infty \frac{(2k)!}{(1-2k) 4^k (k!)^2} (e^2\sin^2\theta)^k \,d\theta \\

&= 4a \sum_{k=0}^\infty \frac{(2k)!}{(1-2k) 4^k (k!)^2} e^{2k} \int_{0}^{\pi/2} \sin^{2k}\theta \,d\theta \\

&= 4a \sum_{k=0}^\infty \frac{(2k)!}{(1-2k) 4^k (k!)^2} e^{2k} \left(\frac{\pi}{2}\right) \frac{(2k)!}{4^k (k!)^2} \\

&= 4a\left(\frac{\pi}{2}\right) \sum_{k=0}^\infty \left[\frac{(2k)!}{4^k(k!)^2}\right]^2 \left(\frac{e^{2k}}{1-2k}\right) \\

\end{align} as desired.

Thanks!

In the penultimate equations of step 3, the denominator should not contain e^2.