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Show that

$$\frac{1}{2}-\frac{1}{2e}<\int_0^{+\infty}e^{-x^2}dx<1+\frac{1}{2e}$$

I know that one way to do this is to evaluate the integral in the middle, and then compare these three numbers. I wonder how can we do this without explicitly compute the integral?

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  • $\begingroup$ Where can the e in the bound possible come from i wonder... $\endgroup$
    – Lost1
    Jan 4, 2014 at 1:20
  • $\begingroup$ Integration by parts would seem to be one of the many magic bullets... $\endgroup$
    – Igor Rivin
    Jan 4, 2014 at 1:27

1 Answer 1

Reset to default
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Proof for the upper bound: $$\int_0^{\infty}e^{-x^2}dx =\int_{0}^{1}e^{-x^2}dx+\int_{1}^{\infty} e^{-x^2} dx <\int_{0}^{1}1 dx+\int_{1}^{\infty} x e^{-x^2} dx =1+\frac{1}{2e}$$ where the last integral is easily calculated by substituting $u=-x^2$.

Proof for the lower bound: $$\int_0^{\infty}e^{-x^2}dx > \int_{0}^{1}e^{-x^2}dx > \int_{0}^{1} xe^{-x^2} dx =\frac{1}{2}-\frac{1}{2e}$$ where the last integral is easily calculated by substituting $u=-x^2$.

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  • $\begingroup$ Oh nice. This is actually quite straight forward... $\endgroup$
    – Lost1
    Jan 4, 2014 at 1:30
  • $\begingroup$ This is pretty awesome! Thanks!Really like this proof $\endgroup$
    – Steven
    Jan 4, 2014 at 1:42

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