Integral inequality (2)

[Grigorios Kostakos]

For \(0\leqslant{k}\leqslant1\) prove that \[\displaystyle\int_{0}^{2\pi}{\sqrt{1-k^2\sin^2{x}}\,dx}\leqslant2\pi\sqrt{1-\frac{k^2}{2}}\,.\]

[Papapetros Vaggelis]

Good evening.

Let \(\displaystyle{I=\int_{0}^{2\pi}\sqrt{1-k^2\cdot \sin^2 x}\,dx}\)

From Cauchy-Schwarz inequality we have

\(\displaystyle{I\leq \sqrt{\int_{0}^{2\pi}\left(1-k^2\cdot \sin^2 x\right)\,dx}\cdot \sqrt{\int_{0}^{2\pi}\,dx}=2\pi\sqrt{1-\frac{k^2}{2}}}\) because,

\(\displaystyle{\sqrt{\int_{0}^{2\pi}\,dx}=\sqrt{2\pi}}\)

and

\(\displaystyle{\int_{0}^{2\pi}\left(1-k^2\cdot \sin^2 x\right)\,dx=}\)

\(\displaystyle{=\int_{0}^{2\pi}\left[1-\frac{k^2}{2}\cdot\left(1-\cos (2x)\right)\right]\,dx}\)

\(\displaystyle{=\int_{0}^{2\pi}\left[1-\frac{k^2}{2}+\frac{k^2}{2}\cdot \cos (2x)\right]\,dx}\)

\(\displaystyle{=\left[\left(1-\frac{k^2}{2}\right)\cdot x+\frac{k^2}{4}\cdot \sin (2x)\right]_{0}^{2\pi}}\)

\(\displaystyle{=2\pi\left(1-\frac{k^2}{2}\right)}\).

[Tolaso J Kos]

A nice result , using the above , is the following:

Given an ellipse $\mathcal{C}: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we know that the length of it is given by the elliptic integral:

$$\ell = 4a \int_{0}^{\pi/2}\sqrt{1-\epsilon^2 \sin^2 t} \, {\rm d}t$$

where $\epsilon$ is the eccecentrity then $\displaystyle \ell \leq 2a\pi \sqrt{1- \frac{\epsilon^2}{2}}$ holds.

Proof:

We just apply the Cauchy - Schwartz inequality hence:

\begin{align*}
\ell &= 4a \int_{0}^{\pi/2} 1\cdot \sqrt{1-\epsilon^2 \sin^2 t}\, {\rm d}t \\
&\leq 4a\left ( \int_{0}^{\pi/2}1 \, {\rm d}t \right )^{1/2} \left ( \int_{0}^{\pi/2}\left ( 1-\epsilon^2 \sin^2 t \right )\, {\rm d}t \right )^{1/2} \\
&=2a \pi \sqrt{1- \frac{\epsilon^2}{2}}
\end{align*}

The upper bound is very close to the length of the ellipse when $\epsilon \rightarrow 0$ . In fact the error is $\mathcal{O}(\epsilon^4)$ as $\epsilon \rightarrow 0$.