Proving Lower Bound of $\int_{0}^{ \pi}\sin^{7}(x) \, dx$ is $(\pi/2)^7$

  • MHB
  • Thread starter Juliayaho
  • Start date
  • Tags
    Bound Dx
In summary, the problem appears to be to prove the following inequalities:$$ \left( \frac{ \pi}{2}\right)^{ \! \! 7} \le \int_{0}^{ \pi}\sin^{7}(x) \, dx \cdot \left( \int_{0}^{ \pi}\sin^{7/6}(x) \, dx\right)^{ \! \! 6} \le 128.$$Using the formula of the previous post we first can compute...$$\int_{0}^{\pi} \sin^{7} x\ dx = \sqrt{\pi}\ \frac{\Gamma (4)}{
  • #1
Juliayaho
13
0
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)^7I 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:
Physics news on Phys.org
  • #2
Juliayaho said:
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)^7I 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$\chi$ $\sigma$
 
  • #3
chisigma said:
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

$\chi$ $\sigma$
 

FAQ: Proving Lower Bound of $\int_{0}^{ \pi}\sin^{7}(x) \, dx$ is $(\pi/2)^7$

What is the definition of a lower bound?

A lower bound is the smallest possible value or limit that a given quantity can have.

How do you prove the lower bound of an integral?

To prove the lower bound of an integral, you need to show that the integral is always greater than or equal to a specific value. This can be accomplished by using mathematical techniques such as the Mean Value Theorem or direct comparison with a known lower bound.

What is the integral of sin7(x) from 0 to pi?

The integral of sin7(x) from 0 to pi is equal to (pi/2)7.

How do you prove that (pi/2)7 is the lower bound of the integral of sin7(x) from 0 to pi?

To prove that (pi/2)7 is the lower bound of the integral of sin7(x) from 0 to pi, you can use the Mean Value Theorem to show that the integral is always greater than or equal to (pi/2)7. You can also compare the integral to the integral of a known lower bound, such as (pi/2)6.

Why is it important to prove the lower bound of an integral?

Proving the lower bound of an integral is important because it helps to establish a minimum value for a given quantity. This can be useful in various mathematical and scientific applications, such as optimization problems or calculating physical quantities. It also helps to ensure the accuracy and validity of mathematical models and calculations.

Similar threads

Replies
3
Views
1K
Replies
2
Views
1K
Replies
2
Views
2K
Replies
2
Views
2K
Replies
5
Views
2K
Replies
6
Views
2K
Replies
29
Views
2K
Back
Top