Asymptotic order of $\int_{\log_2(x)}^\infty (x^{1/t}-1-\frac{\ln(x)}{t}) dt$ Can someone help me find the asymptotic order of this integral $$\int_{\log_2(x)}^\infty (x^{1/t}-1-\frac{\ln(x)}{t}) dt$$ For a fixed x?
I need it up to a logarithmic factor of $x$, but I think the order of the integral itself is probably logarithmic, so that might just be the main term, in which case that is fine too.
I would appreciate any help, thanks.
 A: Let's assume $x>1$, so that $\ln x>0$. The change of variables $u=1/t$ turns the integral into
$$
\int_0^{(\ln 2)/\ln x} (x^u-1-u\ln x) \frac{du}{u^2}.
$$
(One could do this without the change of variables, really....) Expanding $x^u=e^{u\ln x}$ in a power series yields
$$
\int_0^{(\ln 2)/\ln x} \bigg( \sum_{k=0}^\infty \frac{(u\ln x)^k}{k!}-1-u\ln x \bigg) \frac{du}{u^2} = \int_0^{(\ln 2)/\ln x} \sum_{k=2}^\infty \frac{(u\ln x)^k}{k!} \frac{du}{u^2}.
$$
Everything is nonnegative, so we may integrate term by term:
$$
\sum_{k=2}^\infty \int_0^{(\ln 2)/\ln x} \frac{(u\ln x)^k}{k!} \frac{du}{u^2} = \sum_{k=2}^\infty \frac{(\ln x)^k}{k!} \frac{u^{k-1}}{k-1}\bigg|_0^{(\ln 2)/\ln x} = \ln x \sum_{k=2}^\infty \frac{(\ln 2)^{k-1}}{k!(k-1)}.
$$
So the order is $C\ln x$, with an explicit constant even (WolframAlpha evaluates the sum in a closed form that isn't so helpful to humans; its numerical value is about 0.392).
A: This integral is indeed a constant multiple of $\log(x)$. Here is another approach:
$$
\begin{align}
\int_{\log_2(x)}^\infty\left(x^{1/t}-1-\frac{\log(x)}{t}\right)\,\mathrm{d}t
&=\int_{\log_2(x)}^\infty\left(e^{\log(x)/t}-1-\frac{\log(x)}{t}\right)\,\mathrm{d}t\\
&=\int_{\log_2(x)}^\infty\sum_{k=2}^\infty\frac1{k!}\left(\frac{\log(x)}{t}\right)^k\,\mathrm{d}t\\
&=\sum_{k=2}^\infty\frac{\log(x)^k}{k!}\int_{\log_2(x)}^\infty\frac{\mathrm{d}t}{t^k}\\
&=\sum_{k=2}^\infty\frac{\log(x)^k}{k!}\frac1{(k-1)\log_2(x)^{k-1}}\\
&=\log(x)\sum_{k=2}^\infty\frac{\log(2)^{k-1}}{(k-1)k!}
\end{align}
$$
Where the constant
$$
\begin{align}
\sum_{k=2}^\infty\frac{\log(2)^{k-1}}{(k-1)k!}
&=1-\gamma+\mathrm{Ei}(\log(2))-\log(\log(2))-\frac1{\log(2)}\\
&\doteq0.39176599490866084389
\end{align}
$$
