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[SOLVED] Lower bound...

Juliayaho

New member
Apr 11, 2013
13
Hi guys I have a doubt.
How can I prove that
(∫ (from 0 to pi) sin^7 xdx)(∫ (from 0 to pi) sin^(7/6) xdx)^6 is at most 128
But how can I prove that the lower bound of this expression is (pi/2)^7


I think is a very interesting and not an easy question so any ideas? A guidance or something... Thanks beforehand!

[Moderator edit]: The problem appears to be to prove the following inequalities:

$$ \left( \frac{ \pi}{2}\right)^{ \! \! 7} \le \int_{0}^{ \pi}\sin^{7}(x) \, dx \cdot \left( \int_{0}^{ \pi}\sin^{7/6}(x) \, dx\right)^{ \! \! 6} \le 128.$$
 
Last edited by a moderator:

chisigma

Well-known member
Feb 13, 2012
1,704
Hi guys I have a doubt.
How can I prove that
(∫ (from 0 to pi) sin^7 xdx)(∫ (from 0 to pi) sin^(7/6) xdx)^6 is at most 128
But how can I prove that the lower bound of this expression is (pi/2)^7


I think is a very interesting and not an easy question so any ideas? A guidance or something... Thanks beforehand!

[Moderator edit]: The problem appears to be to prove the following inequalities:

$$ \left( \frac{ \pi}{2}\right)^{ \! \! 7} \le \int_{0}^{ \pi}\sin^{7}(x) \, dx \cdot \left( \int_{0}^{ \pi}\sin^{7/6}(x) \, dx\right)^{ \! \! 6} \le 128.$$
May be that You find useful the following formula...

$\displaystyle \int_{0}^{\pi} \sin^{a} x\ dx = \sqrt{\pi}\ \frac{\Gamma (\frac{a+1}{2})}{\Gamma (\frac{a}{2}+1)}$ (1)


Kind regards


$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
May be that You find useful the following formula...

$\displaystyle \int_{0}^{\pi} \sin^{a} x\ dx = \sqrt{\pi}\ \frac{\Gamma (\frac{a+1}{2})}{\Gamma (\frac{a}{2}+1)}$ (1)
Using the formula of the previous post we first can compute...


$$\int_{0}^{\pi} \sin^{7} x\ dx = \sqrt{\pi}\ \frac{\Gamma (4)}{\Gamma(\frac{7}{2}+1)} = \frac{12}{7}\ \sqrt{\pi}\ \frac{1}{\Gamma(\frac{7}{2})} = \frac{12}{7}\ \sqrt{\pi}\ \frac{2^{3}}{5 \cdot 3 \cdot \sqrt{\pi}} = \frac{96}{105}\ (1)$$

... and after the more complex...

$$\int_{0}^{\pi} \sin^{\frac{7}{6}} x\ dx = \sqrt{\pi}\ \frac{\Gamma (\frac{13}{12})}{\Gamma(\frac{7}{12}+1)} = \frac{12}{7}\ \sqrt{\pi}\ \frac{\Gamma(\frac{13}{12})}{\Gamma(\frac{7}{12})}\ (2)$$
A precise computation supplies...

$$\int_{0}^{\pi} \sin^{\frac{7}{6}} x\ dx = 1.920158338481...\ (3)$$
... so that is...

$$\int_{0}^{\pi} \sin^{7} x\ dx\ (\int_{0}^{\pi} \sin^{\frac{7}{6}} x\ dx)^{6}= 45.8251810629...\ (4)$$

A way to find an upper and lower bound of (4) is to use the asymptotic series...

$$\frac{\Gamma(x + \frac{1}{2})}{\Gamma(x)} = \sqrt{x}\ (1 - \frac{1}{8\ x} + \frac{1}{128\ x^{2}} + \frac{5}{1024\ x^{3}} - \frac{21}{32768\ x^{4}} + ...)\ (5)$$
Setting $x= \frac{7}{12}$ and considering the first two terms we obtain...

$$1.8233968...< \frac{12}{7}\ \sqrt{\pi}\ \frac{\Gamma(\frac{13}{12})}{\Gamma(\frac{7}{12})} < 2.32068393...\ (6)$$

... so that is...

$$33.60230455... < \int_{0}^{\pi} \sin^{7} x\ dx\ (\int_{0}^{\pi} \sin^{\frac{7}{6}} x\ dx)^{6} < 142.8163414... (7)$$

Of course considering more terms of (5) we improve the approximation but the lower bound we have obtained is better than $(\frac{\pi}{2})^{7} = 23.596040842...$...

Kind regards

$\chi$ $\sigma$