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Math Help - Integral inequality involving exp(-2x^2)

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

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

    I thought of integrating the inequality 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 \frac{\pi}{2} - 1 < \int^1_0 e^{-2x^2}\,\mathrm{d}x.

    I thought of integrating the inequality 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 x \in [0,1) such that 2 \sqrt{1-x^2}-1=e^{-2x^2} (we can exlude 1, because it's not equal when x=1)

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

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

    ~~~~~~~~~~
    Now what's the sign of g(x)=-2e^{-2x^2}+\frac{1}{\sqrt{1-x^2}} ?
    Differentiate and you'll get :
    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 g is increasing in [0,1)
    So g(x) \geq g(0)=-1 and we have \lim_{x \to 1} g(x)=\infty

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

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

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


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


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


    Now you're up to prove that f(\alpha)>0 :
    We know that \alpha is such that g(\alpha)=0, that is to say :
    -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}

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

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

    Thus \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 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 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 \int^b_a f > 0.
    Last edited by Krizalid; March 30th 2009 at 07:11 AM.
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