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

1. ## 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.

2. Hello,
Originally Posted by dropout_expert
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 :
$\begin{array}{c|ccccc|}
x & 0 & & \alpha & & 1 \\ \hline
f' & & - & & + & \\ \hline
f & & \searrow & & \nearrow & \\
& & & f(\alpha) & & \\ \hline
\end{array}$

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 ><

3. 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

4. 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.

Originally Posted by Moo
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 $\int^b_a f > 0.$