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I need to prove that the following inequality holds:

$$\int_{0}^{1} \sqrt{x}\space e^{-x^2}dx \leq \frac{\pi}{6}$$

No progress on it, yet. Any suggestion is welcome. Thanks.

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    I think that yo must use gamma function.2012-06-04
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    @Gastón Burrull: it's a problem from high school. I think that we may avoid it. I hope so ..2012-06-04
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    Hmm a really hard definite integral in high school is so strange2012-06-04
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    @Gastón Burrull: i agree with you. I see no way to solve it.2012-06-04
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    Maybe with a direct use of this identity $$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$$ you probably must evaluate value exactly .2012-06-04
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    @Gastón Burrull: OK. Thanks.2012-06-04
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    Also you can use a upper bound by taking a convenient partition of a upper riemman sum. For example the maximum of your function is so close to $\pi/6$2012-06-04
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    You can also just drop the $\sqrt{x}\leq 1$ in the interval and then $\int_{0}^{1} e^{-x^2}dx =\frac{2}{\sqrt{\pi}}\text{erf}(1)=0.746...<\frac{\pi}{6}$.2012-06-04
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    @NickKidman $\frac {\pi} 6 \approx 0.523\dots$, so the $\sqrt{x}$ is crucial.2012-06-04
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    @MichaelBoratko: Seems to be true, my bad.2012-06-04

1 Answers 1

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If Cauchy-Schwarz inequality is in one's toolkit, one can write $$ \left(\int_0^1\sqrt{x}\mathrm e^{-x^2}\mathrm dx\right)^2\leqslant\left(\int_0^1x\mathrm e^{-2x^2}\mathrm dx\right)\cdot\left(\int_0^1\mathrm dx\right)=\left[-\tfrac14\mathrm e^{-2x^2}\right]_0^1=\tfrac14(1-\mathrm e^{-2}). $$ Hence $$ I=\int_0^1\sqrt{x}\mathrm e^{-x^2}\mathrm dx\leqslant\tfrac12\sqrt{1-\mathrm e^{-2}}\lt\tfrac12\lt\tfrac\pi6. $$ Edit: The numerical approximation of $I$ above is not so bad since the bound $\mathrm e^{-x^2}\geqslant\mathrm e^{-x\sqrt{x}}$ for every $x$ in $(0,1)$ yields the lower bound $I\geqslant\frac23(1-\mathrm e^{-1})\approx0.4214$, to be compared with the upper bound $\tfrac12\sqrt{1-\mathrm e^{-2}}\approx0.4619$ (while the appearance of $\frac\pi6\approx0.5236$ in the picture remains a mystery to me is convincingly explained by @Chris in a comment below).

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    How did you get the first inequality with Cauchy-Schwarz? Doesn't it yield $(\int_0^1 \! \sqrt{x} e^{-x^2} \, dx)^2 \le \int_0^1 \! x \, dx \cdot \int_0^1 \! e^{-2x^2} \, dx$?2012-06-04
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    @PZZ: Cauchy-Schwarz inequality yields what you suggest *and* what I wrote.2012-06-04
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    Oh, of course. I should have seen that.2012-06-04
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    @did: great job! Thanks.2012-06-04
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    @did: i just found an elementary solution that explains why $\frac{\pi}{6}$ is there.2012-06-06
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    Using the fact that $e^x \ge 1 + x$ we have that $\int_{0}^{1} \sqrt{x}\space e^{-x^2}dx \leq \int_{0}^{1} \frac{\sqrt{x}}{1+x^2}dx \leq \int_{0}^{1} \frac{\sqrt{x}}{1+x^3}dx=\frac{\pi}{6}$.2012-07-02