I am interested in the asymptotic behaviour of the following integral as $m \rightarrow \infty$:
$$\int_2 ^{\sqrt{m}} \frac{m-x^2}{x \mbox{ ln } x}\mbox{d}x $$
Specifically, I already know that the first term behaves like $\mathcal{O}(m \mbox{ ln } (\mbox{ ln } m))$ because we can just evaluate the integral, but I'm not sure about the other part. Is the entire integral still $\mathcal{O}(m \mbox{ ln } (\mbox{ ln } m))$?
Thanks in advance.
Just for your curiosity.
Let $x=e^y$ to make $$I=\int \frac{m-x^2}{x \,\log(x)}\,dx=\int\frac{m-e^{2 y}}{y}\,dy=\int\frac{m}{y}\,dy-2\int\frac{e^{2 y}}{2y}\,dy$$ that is to say $$I=m \log (y)-\text{Ei}(2 y)$$ where appears the exponential integral function.
Now, the asymptotics $$\text{Ei}(t)=e^t \left(\frac{1}{t}+\frac{1}{t^2}+O\left(\frac{1}{t^3}\right)\right)$$ Since we are bargaining, let me use a lower bound equal to $e$ Back to $x$, then $$J=\int_e^{\sqrt m} \frac{m-x^2}{x \,\log(x)}\,dx=m \log \left(\log \left(\sqrt{m}\right)\right)-\text{Ei}\left(2 \log \left(\sqrt{m}\right)\right)+\text{Ei}(2)$$ Just play now with the asymptotics.