6
$\begingroup$

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.

Questions

  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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.