The order of magnitude of $\int_{2}^{n}\frac{1}{n^{\frac{1}{s-1}}(\log n)^{\frac{s-2}{s-1}}}dx$? I am wondering is it true that
$$\int_{2}^{n}\frac{1}{\sqrt{x\log x}}dx=O(\sqrt{\frac{n}{\log n}})?$$
I have a feeling that it is true, but have no idea about how to prove it.
Edit: mathworker21's comment indicates
$$\int_{2}^{n}\frac{1}{n^{\frac{1}{s-1}}(\log n)^{\frac{s-2}{s-1}}}dx$$
has the order of magnitude of $(n/\log n)^{\frac{s-2}{s-1}}$
for a fixed integer $s\ge 3$ as $n$ tends to infinity.
 A: Let $F(n)$ be the integral in question. Change variables by $x=e^y$. Then $dx=e^y \, dy$, and we have
$$ F(n) = \int_{\log{2}}^{\log{n}} y^{-1/2} e^{-y/2} e^y \, dy = \int_{\log{2}}^{\log{n}} y^{-1/2} e^{y/2} \, dy . $$
Integrating by parts,
$$ F(n) = \left[ 2y^{-1/2} e^{y/2} \right]_{\log{2}}^{\log{n}} + \int_{\log{2}}^{\log{n}} y^{-3/2} e^{y/2} \, dy = 2\sqrt{\frac{n}{\log{n}}} + C + \int_{\log{2}}^{\log{n}} y^{-3/2} e^{y/2} \, dy $$
Integrating the remaining integral term by parts,
$$ \int_{\log{2}}^{\log{n}} y^{-3/2} e^{y/2} \, dy = \left[ 2y^{-3/2} e^{y/2} \right]_{\log{2}}^{\log{n}} + 3\int_{\log{2}}^{\log{n}} y^{-5/2} e^{y/2} \, dy , $$
and both of these terms can be bounded by $K(\log{n})^{-3/2} n^{1/2} = o(n^{1/2}(\log{n})^{-1/2}) $ for some $K$. Hence
$$ \int_2^n \frac{dx}{\sqrt{x\log{x}}} \sim 2\sqrt{\frac{n}{\log{n}}} . $$

More generally, for
$$ G(n) = \int_2^n \frac{dx}{x^{\alpha} (\log{x})^{\beta}} , $$
 we can do exactly the same thing, which gives
$$ G(n) = \int_{\log{2}}^{\log{n}} y^{-\beta} e^{(1-\alpha)y} \, dy = \frac{1}{1-\alpha} (\log{n})^{-\beta} n^{1-\alpha} - \frac{1}{1-\alpha} (\log{2})^{-\beta} 2^{1-\alpha} + \frac{\beta}{1-\alpha} \int_{\log{2}}^{\log{n}} y^{-\beta-1} e^{(1-\alpha) y} \, dy $$
If $\alpha > 1$, the constant term dominates, and we can find the limit using the incomplete $\Gamma$ function. If $ \alpha = 1$, the integral can be done exactly. If $\alpha < 1$, we can integrate by parts again and do exactly the same as before to show that
$$ G(n) \sim \frac{1}{1-\alpha} (\log{n})^{-\beta} n^{1-\alpha} , $$
the only change being in which endpoint we take when estimating the final integral if $\beta < -2$.
A: You could even compute exactly the integral.
Almost as Chappers did, first let $x=e^{t^2}$  to make
$$I=\int\frac{dx}{\sqrt{x\log x}}=2\int e^{t^2}\,dt=\sqrt{2 \pi }\, \text{erfi}\left(\frac{t}{\sqrt{2}}\right)$$ Back to $x$
$$I=\sqrt{2 \pi }\, \text{erfi}\left(\frac{\sqrt{\log (x)}}{\sqrt{2}}\right)$$ and have a look here for the asymptotics of $\text{erfi}(z)$ for large values of the argument.
