Asymptotic expansion of the integral $\int_2^\infty \frac{x^t}{\ln(t)} dt$ for $x \to 1$ If we define
$$F(x)=\int\limits_{2}^{\infty}\frac{x^t}{\ln(t)}dt$$
I'm interested in the asymptotic expasion of $F$ as $x$ approaches 1. I'm pretty sure this integral has no elementary anti-derivative so I can't venture down that route. I can't think of any other hard and fast methods to determine the behaviour of this function so I am wondering if there are any other methods of attack to try?
 A: For simplicity let $-\ln x = 1/\lambda$, so that you're interested in the behavior of
$$
\int_2^\infty e^{-t/\lambda} \frac{dt}{\ln t}
$$
as $\lambda \to \infty$.  Making the change of variables $t/\lambda = u$ yields
$$
\int_2^\infty e^{-t/\lambda} \frac{dt}{\ln t} = \lambda \int_{2/\lambda}^\infty e^{-u} \frac{du}{\ln u + \ln \lambda}. \tag{1}
$$
The denominator behaves like $\ln \lambda$ as $\lambda \to \infty$, so we might expect that
$$
\int_{2/\lambda}^\infty e^{-u} \frac{du}{\ln u + \ln \lambda} \approx \frac{1}{\ln \lambda} \int_0^\infty e^{-u}\,du = \frac{1}{\ln \lambda}.
$$
Let's prove this.  Define
$$
f_\lambda(u) = \frac{\ln \lambda}{\ln u + \ln \lambda}.
$$
We have
$$
f'_\lambda(u) = - \frac{\ln \lambda}{u(\ln u + \ln \lambda)^2} < 0,
$$
so since $f_\lambda\left(1/\sqrt{\lambda}\right) = 2$ and $f_\lambda(\infty) = 0$ we must have
$$
0 < f_\lambda(u) \leq 2 \tag{2}
$$
for all $u \geq 1/\sqrt{\lambda}$.  So, if we write
$$
\int_{2/\lambda}^\infty e^{-u} f_\lambda(u)\,du = \int_{2/\lambda}^{1/\sqrt{\lambda}} e^{-u} f_\lambda(u)\,du + \int_{1/\sqrt{\lambda}}^\infty e^{-u} f_\lambda(u)\,du \tag{3}
$$
then the first integral has the bound
$$
0 < \int_{2/\lambda}^{1/\sqrt{\lambda}} e^{-u} f_\lambda(u)\,du < \int_{2/\lambda}^{1/\sqrt{\lambda}} f_\lambda(2/\lambda)\,du = \left(\frac{1}{\sqrt{\lambda}} - \frac{2}{\lambda}\right) \frac{\ln \lambda}{\ln 2} \tag{4}
$$
and thus tends to zero as $\lambda \to \infty$.  For the second integral,
$$
\int_{1/\sqrt{\lambda}}^\infty e^{-u} f_\lambda(u)\,du = \int_0^\infty e^{-u} f_\lambda(u) \chi_{[1/\sqrt{\lambda},\infty)}\,du,
$$
where $\chi_{[a,b]}$ is the characteristic function on the interval $[a,b]$, we know from $(2)$ that the integrand is bounded by
$$
\left|e^{-u} f_\lambda(u) \chi_{[1/\sqrt{\lambda},\infty)}\right| \leq 2e^{-u} \in L^1[0,\infty)
$$
and
$$
\lim_{\lambda \to \infty} e^{-u} f_\lambda(u) \chi_{[1/\sqrt{\lambda},\infty)} = e^{-u}
$$
pointwise on $(0,\infty)$, so by the dominated convergence theorem we may conclude that
$$
\lim_{\lambda \to \infty} \int_{1/\sqrt{\lambda}}^\infty e^{-u} f_\lambda(u)\,du = \lim_{\lambda \to \infty} \int_0^\infty e^{-u} f_\lambda(u) \chi_{[1/\sqrt{\lambda},\infty)}\,du = \int_0^\infty e^{-u}\,du = 1.
$$
Combining this with $(3)$ and $(4)$ we get, by the definition of $f_\lambda(u)$,
$$
\lim_{\lambda \to \infty} \ln \lambda \int_{2/\lambda}^\infty e^{-u} \frac{du}{\ln u+\ln\lambda} = \lim_{\lambda \to \infty} \int_{2/\lambda}^\infty e^{-u} f_\lambda(u)\,du = 1.
$$
In other words,
$$
\int_{2/\lambda}^\infty e^{-u} \frac{du}{\ln u+\ln\lambda} \sim \frac{1}{\ln \lambda}
$$
as $\lambda \to \infty$.  Combining this with $(1)$ yields
$$
\int_2^\infty e^{-t/\lambda} \frac{dt}{\ln t} \sim \frac{\lambda}{\ln \lambda}
$$
as $\lambda \to \infty$, or, since $\lambda = -\frac{1}{\ln x} \sim \frac{1}{1-x}$ as $x \to 1^-$,

$$
\int_2^\infty \frac{x^t}{\ln t}\,dt \sim \frac{1}{(x-1)\ln(1-x)}
$$
  as $x \to 1^-$.

A: My approach has been totally empirical : I computed the numerical value $f(x)$ for  $0.990 \leq x \leq 0.999$ and I tried several emprical models for fitting the data. The simplest I found is $$f(x)=\frac{0.652876}{(1-x)^{0.802491}}$$ which does a pretty good job (at least in my opinion). Extrapolated at $x=0.9999$, this gives a value of $1058.75$ for an "exact" value of $1194.03$. 
The same thing was done for $0.9990 \leq x \leq 0.9999$; it leads to $$f(x)=\frac{0.433212}{(1-x)^{0.859902}}  $$
I hope this helps in spite of its very empirical nature.
