During my projet, I encountered the following function defined for all $\displaystyle t\in[0,\frac{\pi}{2}]$ by : $$f(t)=\int_{0}^t \sqrt{\cos(x)} dx$$

and I need to prove the inequality below :

$$\forall x,y >0\ \ \ \ x+y\leq\frac{\pi}{2}\Rightarrow \frac{f(x+y)^2}{\sin(x+y)}\leq \frac{f(x)^2}{\sin(x)}+\frac{f(y)^2}{\sin(y)} $$

I don't really know if the inequality is true or not , what I know is that I want it to be true, so I can go forward in my work.


  1. Is there a closed form for the function $\displaystyle f$?.
  2. Can we prove the inequality above.

For the first question for $\displaystyle t=\frac{\pi}{2}$ we have $\displaystyle f(\frac{\pi}{2})=\sqrt{\frac{2}{\pi}}\Gamma(\frac{3}{4})^2$ (see walframalpha) and for the other values of $\displaystyle t$, mathematica made use of elliptic integral and I don't know their properties very well, Using the values for some elements I conjectured that : $$ f(t)=2 E(\frac{t}{2}|2) $$ $\displaystyle E(x,m)$ is the elliptic integral with the second kind with the parameter $\displaystyle k=m^2$

Is this equality true? can this help me solve the second question?

Update : using the definition of the elliptic integral I proved that: $$ f(t)=2 E(\frac{t}{2}|2) $$ hence the first question is solved, but I still can't use the proprieties of the eleptic integrals to prove the inequality, I think that there is no hope that the inequality is true.

Any help/comment will be greatly appreciated,Thank you.

  • $\begingroup$ This may be pedantic, but do you mean $f(t) = \int_0^t \sqrt{\cos(x)} \, d x$? $\endgroup$ – PhoemueX Feb 1 '15 at 21:18
  • $\begingroup$ yes , you're right I will correct it, thanks; $\endgroup$ – Elaqqad Feb 1 '15 at 21:23
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    $\begingroup$ Take $g(x) = \frac{f^2(x)}{\sin(x)}$ then you need to show $g(x) + g(y) \geq g(x+y)$. It suffices to show that $g'(x) \leq 1$. $\endgroup$ – Winther Feb 1 '15 at 21:50
  • $\begingroup$ @Winther thanks,your comment was very helpful in fact I proved that $g'(x)\leq 1$ how I can prove your statement : $g'(x)\leq 1 \Rightarrow g(x)+g(y)\geq g(x+y)$ $\endgroup$ – Elaqqad Feb 1 '15 at 22:08
  • $\begingroup$ I might have been to quick with that comment. On second thought I think you need to show that $g''(x) \leq 0$ (so that $g'$ is decreasing). Then $g(x+y) - g(x) = \int_x^{x+y} g'(t) dt$ and $\int_x^{x+y} g'(t) dt = \int_0^{y} g'(t+x) dt \leq \int_0^{y} g'(t) dt = g(y)$ gives the result. $\endgroup$ – Winther Feb 1 '15 at 22:19

If $g(0) = 0$, $g'(x) \geq 0$ and $g''(x)\leq 0$ then

$$g(x+y) - g(x) = \int_x^{x+y} g'(t)dt = \int_0^{y} g'(t+x)dt \leq \int_0^{y} g'(t)dt = g(y)$$

giving us

$$g(x+y) \leq g(x) + g(y)$$

If we define

$$g(x) = \frac{f^2(x)}{\sin(x)}$$

then $\lim\limits_{x\to 0}g(x) = \lim\limits_{x\to 0} \frac{x^2\cos(x)}{\sin(x)} = 0$ and to prove the inequality we need to show that $g'(x)\geq 0$ and $g''(x)\leq 0$.

Take $w = \frac{f(x)\sqrt{\cos(x)}}{\sin(x)}$ to find

$$g'(x) = 1 -\left(1 - w\right)^2$$

$$g''(x) = -2\left(1 - w\right)\cot(x)\left(\left(\frac{\tan^2(x)}{2}+1\right)w-1\right)$$

We now need to show $1 \geq w\geq \frac{1}{\frac{\tan^2(x)}{2}+1} = \frac{2\cos^2(t)}{1 + \cos^2(t)}$. The first inequality, $w\leq 1$, follows from

$$\frac{d}{dx}\left(\frac{\sin(x)}{\sqrt{\cos(x)}} - f(x)\right) = \frac{\sin^2(x)}{\cos^{3/2}(x)} \geq 0$$

and the second inequality follows form

$$\frac{d}{dx}\left(f(x) - \frac{2\cos^{3/2}(x)\sin(x)}{(1 + \cos^2(x))}\right) = \frac{4\sqrt{\cos(x)}\sin^2(x)}{(1+\cos^2(x))^2} \geq 0$$

  • $\begingroup$ How did you conclude that $\lim_{x\to 0} g(x) = \lim_{x\to 0} \frac{x^2\cos(x)}{\sin(x)}$? Confused there $\endgroup$ – jameselmore Feb 1 '15 at 23:25
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    $\begingroup$ @jameselmore I should have added more details. We have $\lim_{x\to 0} \frac{f(x)}{x} = \lim_{x\to 0} \sqrt{\cos(x)}$ by L'Hopitals rule. This gives $\lim_{x\to 0} g(x) = \lim_{x\to 0} \frac{x^2}{\sin(x)} \cdot \frac{f^2(x)}{x^2} = \lim_{x\to 0} \frac{x^2}{\sin(x)} \cdot \cos(x)$ $\endgroup$ – Winther Feb 1 '15 at 23:35
  • $\begingroup$ i don't think that we need $g(0)=0$ $\endgroup$ – Elaqqad Feb 1 '15 at 23:35
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    $\begingroup$ @Elaqqad For the argument here we do otherwise we get $g(x+y) - g(x) \leq g(y) - g(0)$ (since $\int_0^y g'(t) dt = g(y)-g(0)$) so we get $g(x+y) \leq g(x) + g(y) - g(0)$ instead of $g(x+y) \leq g(x) + g(y)$. $\endgroup$ – Winther Feb 1 '15 at 23:36

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