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

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


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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:
    $\displaystyle \int_{e}^{4e} \frac{lnx}{x}dx=2\int_{e}^{4e} lnxd\sqrt{x}$
    $\displaystyle =2[ln(4e)(\sqrt{4e})-lne(\sqrt{e})]-2\int_{e}^{4e} \sqrt{x} dlnx$
    $\displaystyle =2[(ln4+1)(2\sqrt{e})-\sqrt{e}]-2\int_{e}^{4e} \frac{dx}{\sqrt{x}}$
    $\displaystyle =4ln4(\sqrt{e})+2\sqrt{e}-4\int_{e}^{4e} d\sqrt{x}$
    $\displaystyle =4ln4(\sqrt{e})+2\sqrt{e}-4(\sqrt{4e}-\sqrt{e})$
    $\displaystyle =4ln4(\sqrt{e})+2\sqrt{e}-4\sqrt{e}$
    $\displaystyle =4ln4(\sqrt{e})-2\sqrt{e}$
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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:
    $\displaystyle 2\sqrt{e}(\ln(16)-1)$
    and moreover:
    $\displaystyle 3\sqrt{e}\leq 2\sqrt{e}(\ln(16)-1)\leq 6$

    Therefore you have proved the inequality.
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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:
    $\displaystyle 2\sqrt{e}(\ln(16)-1)$
    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
    $\displaystyle 3\sqrt{e}\leq 2\sqrt{e}(\ln(16)-1)\leq 6$
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: proving integral inequalities

    $\displaystyle 2\sqrt{e}(\ln(16)-1)\approx 5,845$ (2)
    $\displaystyle 3\sqrt{e}\approx 4,946$ (1)

    And $\displaystyle (1)\leq (2)\leq 6$
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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 $\displaystyle \int \frac{\sin(x)}{x}$ in term of standard mathematical functions, similar for the second one.
    I would do something with the fact that you can evaluate:
    $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{dx}{1-\frac{\sin^2(x)}{2}}$
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