TITLE: Estimating an integral
QUESTION [6 upvotes]: This question is a more specific question related to https://mathoverflow.net/questions/69900/asymptotics-for-the-number-of-ways-to-sum-primes-such-that-the-sum-is-n 
I am looking for a lower bound (that is tight as possible) for the integral $\int_{2}^n e^{\sqrt{x/\log{x}}} dx.$ Unfortunatelly, $e^{\sqrt{x/\log{x}}}$ is not integrable, so one has to bound it first and then evalute the integral of the bounded function.  For example $$\int_{2}^n e^{\sqrt{x/\log{x}}} \geq \int_{2}^n e^\sqrt[3]{x} dx = O(n^{\frac{2}{3}}e^{\sqrt[3]{n}}) $$
The bound obtained above is inferior to $e^{\sqrt{n/\log{n}}}$, so I should either find a better bound for the integrated function, or perhaps use another (unknown to me) trick to bound the stated integral.
Any suggestions are appreciated!

REPLY [3 votes]: If one writes $I_n$ for the integral from $2$ to $n$, obvious nonasymptotic bounds are
$$
\exp\left(\sqrt{(n-1)/\log(n)}\right)\le I_n\le (n-2)\exp\left(\sqrt{n/\log(n)}\right),
$$
for every $n\ge3$.
In particular, $\displaystyle\log\left(I_n\right)=(1+o(1))\sqrt{n/\log(n)}$.