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How to prove that $$\int_0^{\pi/2}\frac{(a-\gamma)\cdot\sin^2\theta+1}{\sqrt{1+a\cdot \sin^2\theta}}d\theta\cdot\int_0^{\pi/2}\frac{\sin^2\theta}{\sqrt{1+a\cdot\sin^2\theta}}d\theta>1,\quad \forall a>0,$$ where $\gamma = 2-\frac{16}{\pi^2}$? Any ideas on how to prove this inequality? I found that the LHS is first increasing and then decreasing, and equality holds when $a=0$ and $a\to\infty$. But differentiating might not be a good idea.

junjiema2
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  • With given value of $\gamma$, you can show that the numerator of the left-hand integrand is always positive. This is also the case for its denominator; and the numerator and denominator of the right-hand integrand. Integrating positive functions yields a positive result, so you know that the "LHS>0". Maybe you can find a similar reasoning to prove "LHS>1". – Karlo Aug 19 '17 at 00:05
  • You can rewrite the LHS as $\left(a-\gamma\right)\left(I+{\pi\over2}\right)I$, with $I=\int\limits_0^{\pi\over 2}\dfrac{\sin^2\theta}{\sqrt{1+a\sin^2\theta}}$, so the remaining question is how to calculate the integral $I$. – Karlo Aug 19 '17 at 00:13
  • I think you derivation is incorrect... – junjiema2 Aug 19 '17 at 00:30
  • Numerically, I do not see where is the problem. For sure, if you exclude differentiation, I suppose that the problem could be difficult. – Claude Leibovici Aug 19 '17 at 02:13
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    @Karlo. This is "quite simple" elliptic integral. – Claude Leibovici Aug 19 '17 at 02:15
  • I want to find a proof of it, not numerical evaluation. I didn't exclude differentiation, but I thought the derivative could be quite nasty... – junjiema2 Aug 19 '17 at 03:27
  • It might be a good thing to have a bit of context. That essentially is an inequality for the derivative of a complete elliptic integral of the second kind, so I guess it can be proved through differential equations as usual (have a look at pages 134 and 137 here: https://drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view) What is the source of the problem? – Jack D'Aurizio Aug 19 '17 at 12:36
  • @Jack D'Aurizio The problem originates from my research. I work in Electronic Engineering. The elliptical integrals are results of taking expectations over functions of certain bivariate Gaussian random variables. – junjiema2 Aug 20 '17 at 16:18
  • Well, to know which random variables are considered might be useful. Maybe the given inequality can be proved in a geometric fashion, like here: https://math.stackexchange.com/questions/2337419/property-of-standard-normal/2337468#2337468 – Jack D'Aurizio Aug 20 '17 at 16:42
  • Thanks for the hint. But I guess it is difficult to have some geometric interpretation for my problem. My original problem is kind of difficult to explain in a few sentences, and the inequality above is a small part of the problem... – junjiema2 Aug 20 '17 at 16:50

1 Answers1

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The following inequalities can be proved through the techniques outlined in my notes (pages 134-137), i.e. by exploiting strong convexity and second-order differential equations: $$\forall a>0,\qquad \int_{0}^{\pi/2}\frac{\sin^2\theta}{\sqrt{1+a\sin^2\theta}}\,d\theta \geq \frac{\pi}{2\sqrt{4+3a}}$$ $$\forall a>0,\qquad \int_{0}^{\pi/2}\frac{1}{\sqrt{1+a\sin^2\theta}}\,d\theta =\frac{\pi}{2\,\text{AGM}(1,\sqrt{a+1})}\geq \frac{\pi}{1+\sqrt{a+1}}$$

They are enough to prove the claim in a right neighbourhood of the origin, which is a good starting point, at the very least.

Jack D'Aurizio
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  • Hi Jack. Thanks a lot for the reply. I'm out now and will check your answer tomorrow. I'm wondering if it's possible to reach you via email? – junjiema2 Aug 19 '17 at 13:50
  • @junjiema2: you can, but what is the purpose of doing that? Can't you just comment here? – Jack D'Aurizio Aug 19 '17 at 14:04
  • I looked through your note, but still have no idea on how you derived the first inequality. Could you please provide more details? Thanks you very much. @Jack D'Aurizio – junjiema2 Aug 20 '17 at 19:30