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Math Help - proving integral inequalities

  1. #1
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    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}


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  2. #2
    MHF Contributor Siron's Avatar
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    Re: proving integral inequalities

    You can compute the first integral by using integration by parts.
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  3. #3
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    Re: proving integral inequalities

    Quote Originally Posted by Siron View Post
    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}
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  4. #4
    MHF Contributor Siron's Avatar
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    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.
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  5. #5
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    Re: proving integral inequalities

    Quote Originally Posted by Siron View Post
    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
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  6. #6
    MHF Contributor Siron's Avatar
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    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
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  7. #7
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    Re: proving integral inequalities

    Then is the process the same for #2?
    It's quite complicated when evaluating the definite integral
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  8. #8
    MHF Contributor Siron's Avatar
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    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}}
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