# To prove the integral inequality $\int_\overline{\theta}^{\pi}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}\gt\pi$

The following inequality comes up in connection with motion in a dipole field. One has to show that $$\int_\overline{\theta}^{\pi}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}\gt\pi$$ where $$\overline{\theta}= \begin{cases}0, & 0\lt\lambda \lt1\\ \arccos(\frac{1}{\lambda}), & \lambda \gt 1\end{cases}.$$ The case $0\lt\lambda\lt 1$ is easily verified. For then $$\int_{0}^{\pi}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}+\int_{\pi/2}^{\pi}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}$$ $$\implies\int_{0}^{\pi}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}+\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\lambda \sin{\theta}}}\gt2\int_{0}^{\pi/2}d\theta=\pi.$$ However I am finding it difficult to prove the second case. Any pointers would be helpful.

Thanks.

NOTE: The inequality does not hold for all values of $\lambda > 1$, as has been noted in the following responses.

• So the question amounts to bounding $\cos(\theta)$ for $\theta \in [\arccos(1/\lambda),\pi]$ from below. On $[\arccos(1/\lambda),\pi/2]$ you have the convenient lower bound given by concavity: $\cos(\theta)$ is lies above the line between $(\arccos(1/\lambda),1/\lambda)$ and $(\pi/2,0)$. On $[\pi/2,\pi]$ you can use convexity to bound it from below by the tangent line, so in that region you have $\cos(\theta) \geq -(\theta-\pi/2)$. How good of a bound do these two give you? – Ian Nov 1 '15 at 21:03
• Actually, are you sure this result is true? For instance WA tells me that $\int_{\pi/3}^\pi \frac{1}{\sqrt{1-2\cos(x)}} dx$ (the relevant integral for $\lambda=2$) is about $2.1$. – Ian Nov 1 '15 at 21:21
• @Ian: Using concavity, as suggested on $[\arccos(\frac{1}{\lambda}), \pi / 2]$ , I obtained $\int_{\arccos(\frac{1}{\lambda})}^{\pi /2} \frac{d\theta}{\sqrt{1-\lambda \cos (\theta)}} \gt \pi - 2 \arccos (\frac{1}{\lambda})$. Still to work out in the range $[\pi/2, \pi]$. The inequality is necessarilly true, but the argument i have for this is physical. Basically, $\theta$ here is the angular position (with z-axis) of a particle moving in a dipole field. The angular position in such a field cannot have turning points. – vnd Nov 1 '15 at 21:27
• Is it possible that it does not hold for $\lambda$ too large? Just playing around on Wolfram I find that the inequality appears to fail somewhere in between $\lambda=1.2$ and $\lambda=1.3$. – Ian Nov 1 '15 at 21:29
• It should be not too difficult to integrate this exactly using elliptic integrals... maybe that would be an alternative... – tired Nov 1 '15 at 21:52

## 2 Answers

If I divide by $\lambda$ the inequality becomes: $$\int_{\underline{\theta}}^{\pi}\frac{d\theta}{\sqrt{\cos(\underline{\theta})-\cos(\theta)}}\geq \pi \sqrt{\lambda}$$

Now $$\int_{\pi/2}^{\pi}\frac{d\theta}{\sqrt{\cos(\underline{\theta})-\cos(\theta)}}\leq \int_{\pi/2}^{\pi}\frac{d\theta}{\sqrt{-\cos(\theta)}}d\theta=\int_0^{\pi/2}\frac{d\theta}{\sqrt{\sin(\theta)}}<+\infty$$ and as if $\theta \in [ \underline{\theta}, \pi/2]$, we have $\cos(\theta)-\cos(\underline{\theta})=(\theta-\underline{\theta})(-\sin(c))$ for some $c\in [ \underline{\theta}, \pi/2]$, we get $|\cos(\theta)-\cos(\underline{\theta})|\geq(\theta-\underline{\theta})\sin(\underline{\theta})$ Hence $$\int_{\underline{\theta}}^{\pi/2}\frac{d\theta}{\sqrt{\cos(\underline{\theta})-\cos(\theta)}}\leq\frac{1}{\sqrt{\sin(\underline{\theta})}} \int_{\underline{\theta}}^{\pi/2}\frac{d\theta}{\sqrt{\theta-\underline{\theta}}}= 2\frac{\sqrt{\pi/2-\underline{\theta}}}{\sqrt{\sin(\underline{\theta})}}$$

Now as $\displaystyle \sin(\underline{\theta})=\sqrt{1-\frac{1}{\lambda^2}}$, we have $\displaystyle 2\frac{\sqrt{\pi/2-\underline{\theta}}}{\sqrt{\sin(\underline{\theta})}}$ bounded as $\lambda\to +\infty$, so if I am not wrong, there is a problem.

• The sine will actually be the square root of $1-1/\lambda^2$. But that does not impede your point. In fact, you've actually proven that the integral goes to zero, because $\arccos(1/\lambda)$ goes to $\pi/2$ as $\lambda \to \infty$. – Ian Nov 1 '15 at 22:09
• Oh yes thanks, I have edited. Have you seen a problem in my computations ? – Kelenner Nov 1 '15 at 22:10
• Sorry, I should say that you proved that the integral up to $\pi/2$ goes to zero. But I do think you have proven that the result does not hold for all $\lambda$: if it did, the rescaled integral would need to blow up, but it does not. In fact I think the integral is something like $O(\lambda^{-1/2})$. – Ian Nov 1 '15 at 22:13
• Ok, thanks. Perhaps the OP can modify the hypothesis. – Kelenner Nov 1 '15 at 22:14

Your inequality is wrong in general (indicated below), and your proof for $\lambda<1$ has a small mistake. Let us start with the latter.

The problem with your proof

You divide the integral in the intervals $(0,\pi/2)$ and $(\pi/2,\pi)$ and that is fine. But since $\cos \theta<0$ in the interval $(\pi/2,\pi)$ it is impossible that you get as an integrand $1/\sqrt{1-\lambda\sin \theta}$ after your change of variables. What you get if you just shift $\theta\mapsto \theta-\pi/2$ is $1/\sqrt{1+\lambda\sin\theta}$. But this is less than $1$ on the interval. Instead, let $\theta\mapsto \pi-\theta$. Then, you will end up with $$\int_0^{\pi/2}\frac{1}{\sqrt{1-\lambda\cos\theta}}+\frac{1}{\sqrt{1+\lambda\cos\theta}}\,d\theta.$$ Then, according to the inequality $a+b\geq 2\sqrt{ab}$, you find that your integrand is bounded below by $$\frac{2}{\sqrt[4]{1-\lambda^2\cos^2\theta}}$$ which is greater than $2$ on the interval. Integrating you get a lower bound of $\pi$.

Proof that the inequality is not true for all $\lambda>1$

Here, as suggested by @Ian, it is sufficient to consider $\lambda=2$, and hence show that the integral $$\int_{\pi/3}^\pi\frac{1}{\sqrt{1-2\cos\theta}}\,d\theta<\pi.$$ In the interval $(\pi/3,\pi)$ it holds that $$1-2\cos\theta>\frac{9}{2\pi}(\theta-\pi/3).$$ Thus $$\int_{\pi/3}^\pi\frac{1}{\sqrt{1-2\cos\theta}}\,d\theta< \int_{\pi/3}^\pi\frac{1}{\sqrt{\frac{9}{2\pi}(\theta-\pi/3)}}\,d\theta=\frac{4\pi}{3\sqrt{3}}<\pi.$$