proving integral inequalities

I want ask how to start to tackle the problems of inequalities involving integrals:

#1 Prove that

$\displaystyle 3\sqrt{e}\leq\int_{e}^{4e}\frac{lnx}{\sqrt{x}}dx \leq6$

#2 Prove that

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

#3 Prove that

$\displaystyle 1\leq \int_{0}^{\frac{\pi}{2}} \frac{\sin{x}}{x} dx \leq \frac{\pi}{2}$

Re: proving integral inequalities

You can compute the first integral by using integration by parts.

Re: proving integral inequalities

Quote:

Originally Posted by

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

I have tried it:

$\displaystyle \int_{e}^{4e} \frac{lnx}{x}dx=2\int_{e}^{4e} lnxd\sqrt{x}$

$\displaystyle =2[ln(4e)(\sqrt{4e})-lne(\sqrt{e})]-2\int_{e}^{4e} \sqrt{x} dlnx$

$\displaystyle =2[(ln4+1)(2\sqrt{e})-\sqrt{e}]-2\int_{e}^{4e} \frac{dx}{\sqrt{x}}$

$\displaystyle =4ln4(\sqrt{e})+2\sqrt{e}-4\int_{e}^{4e} d\sqrt{x}$

$\displaystyle =4ln4(\sqrt{e})+2\sqrt{e}-4(\sqrt{4e}-\sqrt{e})$

$\displaystyle =4ln4(\sqrt{e})+2\sqrt{e}-4\sqrt{e}$

$\displaystyle =4ln4(\sqrt{e})-2\sqrt{e}$

Re: proving integral inequalities

Yes, that's correct, you can simplify it to:

$\displaystyle 2\sqrt{e}(\ln(16)-1)$

and moreover:

$\displaystyle 3\sqrt{e}\leq 2\sqrt{e}(\ln(16)-1)\leq 6$

Therefore you have proved the inequality.

Re: proving integral inequalities

Quote:

Originally Posted by

**Siron** Yes, that's correct, you can simplify it to:

$\displaystyle 2\sqrt{e}(\ln(16)-1)$

and moreover:

$\displaystyle 3\sqrt{e}\leq 2\sqrt{e}(\ln(16)-1)\leq 6$

Therefore you have proved the inequality.

But how can the first step

$\displaystyle 2\sqrt{e}(\ln(16)-1)$

reach the second step

$\displaystyle 3\sqrt{e}\leq 2\sqrt{e}(\ln(16)-1)\leq 6$

Re: proving integral inequalities

$\displaystyle 2\sqrt{e}(\ln(16)-1)\approx 5,845$ (2)

$\displaystyle 3\sqrt{e}\approx 4,946$ (1)

And $\displaystyle (1)\leq (2)\leq 6$

Re: proving integral inequalities

Then is the process the same for #2?

It's quite complicated when evaluating the definite integral

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 $\displaystyle \int \frac{\sin(x)}{x}$ in term of standard mathematical functions, similar for the second one.

I would do something with the fact that you can evaluate:

$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{dx}{1-\frac{\sin^2(x)}{2}}$