# proving integral inequalities

• December 15th 2011, 04:42 AM
maoro
proving integral inequalities
I want ask how to start to tackle the problems of inequalities involving integrals:
#1 Prove that
$3\sqrt{e}\leq\int_{e}^{4e}\frac{lnx}{\sqrt{x}}dx \leq6$

#2 Prove that
$\frac{\pi}{2}\leq\int_{0}^{\frac{\pi}{2}}$ $\frac{dx}{\sqrt{1-\frac{1}{2} \sin^2{x}}}$ $\leq\frac{\pi}{\sqrt{2}}$

#3 Prove that
$1\leq \int_{0}^{\frac{\pi}{2}} \frac{\sin{x}}{x} dx \leq \frac{\pi}{2}$

• December 15th 2011, 07:33 AM
Siron
Re: proving integral inequalities
You can compute the first integral by using integration by parts.
• December 15th 2011, 08:23 AM
maoro
Re: proving integral inequalities
Quote:

Originally Posted by Siron
You can compute the first integral by using integration by parts.

I have tried it:
$\int_{e}^{4e} \frac{lnx}{x}dx=2\int_{e}^{4e} lnxd\sqrt{x}$
$=2[ln(4e)(\sqrt{4e})-lne(\sqrt{e})]-2\int_{e}^{4e} \sqrt{x} dlnx$
$=2[(ln4+1)(2\sqrt{e})-\sqrt{e}]-2\int_{e}^{4e} \frac{dx}{\sqrt{x}}$
$=4ln4(\sqrt{e})+2\sqrt{e}-4\int_{e}^{4e} d\sqrt{x}$
$=4ln4(\sqrt{e})+2\sqrt{e}-4(\sqrt{4e}-\sqrt{e})$
$=4ln4(\sqrt{e})+2\sqrt{e}-4\sqrt{e}$
$=4ln4(\sqrt{e})-2\sqrt{e}$
• December 15th 2011, 09:04 AM
Siron
Re: proving integral inequalities
Yes, that's correct, you can simplify it to:
$2\sqrt{e}(\ln(16)-1)$
and moreover:
$3\sqrt{e}\leq 2\sqrt{e}(\ln(16)-1)\leq 6$

Therefore you have proved the inequality.
• December 15th 2011, 09:10 AM
maoro
Re: proving integral inequalities
Quote:

Originally Posted by Siron
Yes, that's correct, you can simplify it to:
$2\sqrt{e}(\ln(16)-1)$
and moreover:
$3\sqrt{e}\leq 2\sqrt{e}(\ln(16)-1)\leq 6$

Therefore you have proved the inequality.

But how can the first step
$2\sqrt{e}(\ln(16)-1)$
reach the second step
$3\sqrt{e}\leq 2\sqrt{e}(\ln(16)-1)\leq 6$
• December 15th 2011, 09:15 AM
Siron
Re: proving integral inequalities
$2\sqrt{e}(\ln(16)-1)\approx 5,845$ (2)
$3\sqrt{e}\approx 4,946$ (1)

And $(1)\leq (2)\leq 6$
• December 15th 2011, 09:20 AM
maoro
Re: proving integral inequalities
Then is the process the same for #2?
It's quite complicated when evaluating the definite integral
• December 15th 2011, 09:33 AM
Siron
Re: proving integral inequalities
No, It won't be the same, because indeed as you note they're very complex to evaluate and moreover you can't express $\int \frac{\sin(x)}{x}$ in term of standard mathematical functions, similar for the second one.
I would do something with the fact that you can evaluate:
$\int_{0}^{\frac{\pi}{2}} \frac{dx}{1-\frac{\sin^2(x)}{2}}$