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)
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)
Note that $\sqrt{\pi}/2=\Gamma(3/2)$.
For $0
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.