Bounds for integral 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:
$$\int\limits_{x=0}^{\infty} e^{-\dfrac{(\sqrt{x})^2}{2n}} dx = 2n$$
 A: 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 $2n+\sqrt{2\pi}\sqrt{n}+2$.
