Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 .. – I'm an artist 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. – I'm an artist 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
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).

share|cite|improve this answer
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. – I'm an artist Jun 4 '12 at 8:44
Using the fact that $e^x \ge 1 + x$ we have that $\int_{0}^{1} \sqrt{x}\space e^{-x^2}dx \leq \int_{0}^{1} \frac{\sqrt{x}}{1+x^2}dx \leq \int_{0}^{1} \frac{\sqrt{x}}{1+x^3}dx=\frac{\pi}{6}$. – I'm an artist Jul 2 '12 at 17:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.