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My advisor told me to look up the proof of the following standard estimate so that we can adapt it to the case where we are dealing with something similar but including the addition of a polynomial integrand.

I have yet to find a reference containing the proof and was wondering what the reference was or if someone knew a quick way to prove the following estimate:

How do you show $\exists c > 0$ so that $\int_u^\infty \exp(-z^2)\;\mathrm dz < \frac{c}{u} \exp{(-u^2)}$?

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Integrate by parts, $v=1/z$, $dw=ze^{-z^2}$. – André Nicolas Oct 19 '11 at 23:25
I suppose this kind of question gets asked (and answered) again and again. A fairly recent discussion on this topic can be found here – Dilip Sarwate Oct 20 '11 at 0:46
Dear @Dilip, if you're adding the tag faq, please don't forget to add an entry to the list. In this case, I had only added one since I thought the other had it all covered. If you think this should also be included, please make the necessary edits in the meta post. – J. M. Oct 20 '11 at 22:22
See… for a related question. – Mariano Suárez-Alvarez Oct 20 '11 at 23:10
Dear Mariano: As requested by @J.M. above, I have added this question also to the faq list. Actually, the answer in the comment by Andre' Nicolas and Chris Taylor's answer here are essentially the same as those given for the other question, and Henry's answer is also closely related. May I suggest that the moderators simply merge the two questions so that there is only one entry in the faq for this topic? Another point: maybe the faq entry should be for probability rather than statistics to match the tags here – Dilip Sarwate Oct 20 '11 at 23:33
up vote 2 down vote accepted

The obvious thing to do is take the derivative with respect to $u$ of $-\frac{c}{u} \exp{(-u^2)}$ which is $c\left(2+\frac{1}{u^2}\right) \exp{(-u^2)}$.

So if $u \gt 0$ and $c \ge \frac{1}{2}$ then $c \left(2+\frac{1}{u^2}\right) \gt 1$ and $c\left(2+\frac{1}{x^2}\right) \gt 1$ for $x \ge u$ so

$$\int_{u}^{\infty} \exp{(-x^2)} dx \lt \int_{u}^{\infty} c\left(2+\frac{1}{x^2}\right) \exp{(-x^2)} dx = \left[-\frac{c}{x} \exp{(-x^2)}\right]_{x=u}^\infty = \frac{c}{u} \exp{(-u^2)}.$$

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Note that

$$\int_u^\infty e^{-z^2}dz \leq \int_u^\infty \frac{z}{u} e^{-z^2}dz = \left[ -\frac{1}{2u}e^{-z^2}\right]_u^\infty = \frac{1}{2u} e^{-u^2}$$

In particular, $c=1/2$ suffices.

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I like this proof. – Nate Eldredge Oct 21 '11 at 1:44

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