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Thread: Integral inequality involving exp(-2x^2)

  1. #1
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    Integral inequality involving exp(-2x^2)

    Prove that $\displaystyle \frac{\pi}{2} - 1 < \int^1_0 e^{-2x^2}\,\mathrm{d}x.$

    I thought of integrating the inequality $\displaystyle 2\sqrt{1-x^2} - 1 \leq e^{-2x^2}$ from 0 to 1 but the inequality isn't strict. I'm not sure what I could do to make it strict; I think a different approach is needed.

    Any ideas? Thanks.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by dropout_expert View Post
    Prove that $\displaystyle \frac{\pi}{2} - 1 < \int^1_0 e^{-2x^2}\,\mathrm{d}x.$

    I thought of integrating the inequality $\displaystyle 2\sqrt{1-x^2} - 1 \leq e^{-2x^2}$ from 0 to 1 but the inequality isn't strict. I'm not sure what I could do to make it strict; I think a different approach is needed.

    Any ideas? Thanks.
    I don't know how you got the inequality... But I think it's a good start.

    You can prove that it is a strict inequality... The way I found looks a bit messy ><

    Prove that there isn't $\displaystyle x \in [0,1)$ such that $\displaystyle 2 \sqrt{1-x^2}-1=e^{-2x^2}$ (we can exlude 1, because it's not equal when x=1)

    Consider $\displaystyle f(x)=e^{-2x^2}-2 \sqrt{1-x^2}+1$

    - $\displaystyle f'(x)=2x \left(-2e^{-2x^2}+\frac{1}{\sqrt{1-x^2}}\right)$
    But since x is in [0,1), 2x$\displaystyle \geq 0$

    ~~~~~~~~~~
    Now what's the sign of $\displaystyle g(x)=-2e^{-2x^2}+\frac{1}{\sqrt{1-x^2}}$ ?
    Differentiate and you'll get :
    $\displaystyle g'(x)=x \left(8e^{-2x^2}+\frac{1}{(1-x^2)^{3/2}}\right)$, which is always positive, for any x in [0,1)
    Thus $\displaystyle g$ is increasing in [0,1)
    So $\displaystyle g(x) \geq g(0)=-1$ and we have $\displaystyle \lim_{x \to 1} g(x)=\infty$

    So there exists a unique value $\displaystyle \alpha \in [0,1)$ such that $\displaystyle g(\alpha)=0$ (by the intermediate value theorem, or something like that)

    and for any $\displaystyle x \in [0,\alpha] ~,~ g(x) \leq 0$ and for any $\displaystyle x \in [\alpha,1)~,~ g(x) \geq 0$

    But $\displaystyle f'(x)=2xg(x)$. So they have the same sign.
    ~~~~~~~~~~


    Hence :
    $\displaystyle \begin{array}{c|ccccc|}
    x & 0 & & \alpha & & 1 \\ \hline
    f' & & - & & + & \\ \hline
    f & & \searrow & & \nearrow & \\
    & & & f(\alpha) & & \\ \hline
    \end{array}$


    So $\displaystyle \boxed{f(x) \geq f(\alpha)}$, for any $\displaystyle x \in [0,1)$


    Now you're up to prove that $\displaystyle f(\alpha)>0$ :
    We know that $\displaystyle \alpha$ is such that $\displaystyle g(\alpha)=0$, that is to say :
    $\displaystyle -2e^{-2\alpha^2}+\frac{1}{\sqrt{1-\alpha^2}}=0 \Rightarrow \sqrt{1-\alpha^2}=\frac 12 \cdot e^{2 \alpha^2} \quad {\color{red}\star}$

    $\displaystyle f(\alpha)=e^{-2 \alpha^2}-2 \sqrt{1-\alpha^2}+1$
    By $\displaystyle {\color{red}\star}$, we have :
    $\displaystyle f(\alpha)=e^{-2 \alpha^2}-e^{2 \alpha^2}+1=\frac 1\varphi (-\varphi^2+\varphi+1)$, where $\displaystyle \varphi=e^{2 \alpha^2}$
    Note that $\displaystyle \varphi \in (0,1]$ << MISTAKE

    Hence $\displaystyle f(\alpha)=-\frac 1\varphi \left(\frac{1+\sqrt{5}}{2}-\varphi\right)\left(\varphi-\frac{1-\sqrt{5}}{2}\right)$

    We can easily see that $\displaystyle \frac{1+\sqrt{5}}{2}-\varphi > 0$, because $\displaystyle \frac{1+\sqrt{5}}{2}>1$.
    And similarly, we can see that $\displaystyle \varphi-\frac{1-\sqrt{5}}{2}>0$, because $\displaystyle \frac{1-\sqrt{5}}{2}<0$

    Thus $\displaystyle \boxed{f(\alpha)>0}$



    I hope this ... erm helps ^^



    Edit : okay, because of the mistake pointed out above, the conclusion is false... I'm too tired to think again about it, sorry ><
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  3. #3
    Moo
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    I'm going to shoot myself..................................

    If x=0, then $\displaystyle e^{-2x^2}=2 \sqrt{1-x^2}-1$


    I was trying to beat a dead horse
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  4. #4
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    Hiya Moo,

    pi/2 is the area of the semi-circle with radius 1, which we can write as the integral from -1 to 1 of sqrt(1-x^2), and since the function is even, we can write this as the integral from 0 to 1 of sqrt(1-x^2). I think a function like 2/(x^2 + 1) would work, too.

    I haven't checked your working yet (I will now), but can't we simply substitute x = 0 into the equation which results in equality? Hence the inequality can't be strict over [0,1]. I might be making a silly mistake, though.

    Thanks.

    Quote Originally Posted by Moo View Post
    I'm going to shoot myself..................................

    If x=0, then $\displaystyle e^{-2x^2}=2 \sqrt{1-x^2}-1$


    I was trying to beat a dead horse
    Ah, you noticed. It happens to the best of us!

    Quick follow-up: it turns out that even though the inequality isn't strict, we can make the integral inequality strict! The following theorem is what was needed:

    Suppose f => 0 for all x in [a,b] and f is continuous at one point c in [a,b] such that f(c) > 0. Then $\displaystyle \int^b_a f > 0.$
    Last edited by Krizalid; Mar 30th 2009 at 06:11 AM.
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