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Prove the following inequality:

$$\frac{\sqrt{\pi}}{2}\le\int_{0}^{1} \left({\log(\csc(x))}\right)^{1/3} dx$$

What should i start with? (it's not a homework but a hobby related activity)

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2 Answers 2

up vote 6 down vote accepted

Note that $\sqrt{\pi}/2=\Gamma(3/2)$.

For $0<x\le 1$ we have $\csc(x)>1/x$ so $$ \int_0^1 (\log(\csc(x)))^{1/3}~dx > \int_0^1(-\log(x))^{1/3}~dx $$ Substitute $y=-\log(x)$ to get $$ \int_0^1(-\log(x))^{1/3}~dx = \int_0^\infty y^{1/3}e^{-y}~dy=\Gamma(4/3) $$ So your inequality follows from $\Gamma(4/3)>\Gamma(3/2)$. Unfortunately $\Gamma(4/3)$ does not have a simple expression, so I don't yet see a way to show this last step except by evaluating it numerically.

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I hope you don't mind that I completed the proof by editing your post rather than saying it in the comments. It would have been too long and belongs as part of the answer anyway. –  Ragib Zaman Jul 3 '12 at 14:08
    
@Zander: thank you for the answer. –  Chris's sis Jul 3 '12 at 14:27
    
@Ragib I don't mind the edit, but I don't think it's correct. $\Gamma(x)$ has its minimum at $x=1.46\ldots<3/2$, e.g. $\Gamma(3/2)-\Gamma(1.45)>0.0005$. Your error is that $\psi(3/2)=\psi(1/2)+2>0$. –  Zander Jul 3 '12 at 14:28
    
@Zander Sorry about that. I just checked now that $\Gamma(4/3)-\Gamma(3/2) \approx 0.007$, which I believe is too close to $0$ to show it is positive without getting things getting messy. I don't think we can avoid it. –  Ragib Zaman Jul 3 '12 at 14:49
    
@Zander: I thought that if $\Gamma'(3/2)\le0$, then since $\Gamma$ is log-convex, we would have $\Gamma'(x)\le0$ for all $x\le3/2$. However, $$ \begin{align} \Gamma'(3/2)/\Gamma(3/2) &=\mathrm{H}(1/2)-\gamma\\ &=H(-1/2)+2-\gamma\\ &=2-\gamma-2\log(2)\\ &\stackrel{.}{=}0.036489973978576520559 \end{align} $$ and $$ \begin{align} \Gamma'(4/3)/\Gamma(4/3) &=\mathrm{H}(1/3)-\gamma\\ &=H(-2/3)+3-\gamma\\ &=3-\gamma-\frac12\left(\frac{\pi}{\sqrt{3}}+3\log(3)\right)\\ &\stackrel{.}{=}-0.13203378002080632300 \end{align} $$ It looks as if computation is necessary. –  robjohn Jul 3 '12 at 16:31

Proving such a thing with simple "analytic" inequalities seems hopeless, due to the combination of $\log$, $\csc$ and exponentiation by $1/3$- the integrand just seems too ugly for any simple "trick" to work. So we have to get our hands dirty.

Calculating the derivative gives $\displaystyle \frac{-\cot x}{3 (\log (\csc(x)))^{2/3}}$ and since for $x\in [0,1]$ $\cot x \geq 0$ and $\csc (x) \geq 1$ we see the derivative is always $\leq 0$, i.e the function is decreasing. Thus taking right handed Riemann sums underestimates the integral.

$$\int^1_0 \sqrt[3]{ \log(\csc (x))} dx > \frac{1}{20}\sum_{k=1} \sqrt[3]{ \log \left(\csc \left(\frac{k}{20}\right)\right)} = 0.915... $$

And since $\pi < 3.24 = 1.8^2$, we have $\displaystyle \frac{\sqrt{\pi}}{2} < 0.9$ so the desired inequality is reached.

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thank you for your solution. However, my aim is to solve it elegantly without using calculator. –  Chris's sis Jul 3 '12 at 12:58

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