Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

How can one prove that assuming $n>1$ there exists an $c>0$ such that:

$$S_n = \int\limits_{x=0}^{\infty} e^{-\dfrac{(\sqrt{x}-1)^2}{2n}} dx <cn$$

I can plug the integral into WA (assuming $n$ is real) and get the result that way but is there a nice way to derive the inequality by hand? In fact it seems the integral is less than $6n$ for $n \geq 1$ and my guess is that asymptotically $S_n \sim 2n$ (from plugging in $\lim_{n\rightarrow \infty} S_n/(2n)$ into WA again).

The similar integral, according to WA: $$\int\limits_{x=0}^{\infty} e^{-\dfrac{(\sqrt{x})^2}{2n}} dx = 2n$$

share|cite|improve this question
Note that your second one is just $\int_0^\infty \exp(-\frac x{2n}) dx$, then substitute $u=\frac x{2n}$ for an exact result. – Ross Millikan Feb 14 '13 at 15:41
Try to use Laplace's method. Here is a related problem. – Mhenni Benghorbal Feb 14 '13 at 15:48
Are you interested in the behaviour for large $n$? Very small $n$? – André Nicolas Feb 14 '13 at 16:21
Using basically the same method, I can show that for any positive $\epsilon$, there is an $N$ such that if $n\ge N$, then the integral is $\lt 2(1+\epsilon)n$. Actually, in the post I gave away a factor of $e$ in the answer above, we can get $9+(6/e)n$ for free. Can probably get the $6n$ absolute bound, or better, just a matter of tightening estimates. – André Nicolas Feb 14 '13 at 17:13
up vote 3 down vote accepted

We find a crude bound in the case $n \ge 1$.

We have $(\sqrt{z}-1)^2=z-2\sqrt{z}+1$. Note that if $z\ge 9$, then $\sqrt{z}\le \frac{z}{3}$. So if $z \ge 9$, then $e^{-(\sqrt{z}-1)/2n}\lt e^{-z/6n}$.

Integrating from $z=9$ to $\infty$, we get $6ne^{-9/6n}$, which is less than $6n$. And the integral from $0$ to $9$ is $\lt 9$, so our integral is $\lt 9+6n$. In particular, if $n\ge 1$, this is less than $15n$.

Added: The above showed the existence of a $c$, more specifically that we can take $c=15$. In the calculation, we casually gave away a lot. To get a (much) better estimate, it is convenient to get rid of the annoying square root. So let $\sqrt{z}-1=\sqrt{n}\, w$. Then $dz=2(nw+\sqrt{n})\,dw$, and we want $$\int_{-1/\sqrt{n}}^\infty 2(nw+\sqrt{n})e^{-w^2/2}\,dw.$$ One can get an explicit expression for the integral of the $2nwe^{-w^2/2}$ part: It is $2ne^{-1/2n}$, and in particular less than $2n$.

We can also find an explicit expression for $\int_0^\infty 2\sqrt{n}e^{-w^2/2}\,dw$. It is $\sqrt{2\pi}\sqrt{n}$.

For $\int_{1/\sqrt{n}}^0 2\sqrt{n}e^{-w^2/2}\,dw$, the interval has length $1/\sqrt{n}$ and the integrand is bounded by $2\sqrt{n}$, so the integral is $\lt 2$.

Putting things together, we get the bound $2n+\sqrt{2\pi}\sqrt{n}+2$.

share|cite|improve this answer

Your Answer


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