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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. – Gastón Burrull Jun 4 '12 at 8:01
@Gastón Burrull: it's a problem from high school. I think that we may avoid it. I hope so .. – OFFSHARING Jun 4 '12 at 8:01
Hmm a really hard definite integral in high school is so strange – Gastón Burrull Jun 4 '12 at 8:02
@Gastón Burrull: i agree with you. I see no way to solve it. – OFFSHARING Jun 4 '12 at 8:04
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 . – Gastón Burrull Jun 4 '12 at 8:05

1 Answer 1

up vote 8 down vote accepted

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 $(\int_0^1 \! \sqrt{x} e^{-x^2} \, dx)^2 \le \int_0^1 \! x \, dx \cdot \int_0^1 \! e^{-2x^2} \, dx$? – user12014 Jun 4 '12 at 8:23
@PZZ: Cauchy-Schwarz inequality yields what you suggest and what I wrote. – Did Jun 4 '12 at 8:39
Oh, of course. I should have seen that. – user12014 Jun 4 '12 at 8:44
@did: great job! Thanks. – OFFSHARING Jun 4 '12 at 8:44
Using the fact that $e^x \ge 1 + x$ we have that $\int_{0}^{1} \sqrt{x}\space e^{-x^2}dx \leq \int_{0}^{1} \frac{\sqrt{x}}{1+x^2}dx \leq \int_{0}^{1} \frac{\sqrt{x}}{1+x^3}dx=\frac{\pi}{6}$. – OFFSHARING Jul 2 '12 at 17:07

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