I want to prove that if $\lim_{x \to \infty} \frac{1}{x} \int^x_0 U(t) dt = \rho \in \mathbb{R}$, then $\exp(\int^x_0 \frac{U(t)}{t} dt)$ is regularly varying with $\rho$.
Looking in Seneta, it is said to use that $U \colon (0,\infty) \to (0,\infty)$ is regularly varying if and only if $\int^1_0 \log (\frac{U(x)}{U(tx)}) dt \to \rho$ as $x \to \infty$.
However, I haven't found a way to deal with $\int^x_0 \frac{U(t)}{t}dt$, since the condition is on $\frac{1}{x} \int^x_0 U(t) dt$.
Thank you.
Let's denote $f(x) = \exp \left(\int\limits_0^x \frac{U(t)}{t} {\rm d}t\right)$. Following the given theorem, one needs to prove that $\int\limits_0^1 \ln \left(\frac{f(x)}{f(\lambda x)}\right) {\rm d} \lambda \to \rho$ as $x\to\infty$. Observe that $$ \ln \left(\frac{f(x)}{f(\lambda x)}\right) = \ln f(x) - \ln f(\lambda x) = \int_0^x \frac{U(t)}{t} {\rm d}t - \int_0^{\lambda x} \frac{U(t)}{t} {\rm d}t. $$ As $\lambda \in [0, 1]$, it follows that $0 \leq \lambda x \leq x$ for all $x \geq 0$, so $$ \int_0^1 \ln \left(\frac{f(x)}{f(\lambda x)}\right) {\rm d} \lambda = \int_0^1 \left(\int_{\lambda x}^{x} \frac{U(t)}{t}{\rm d}t \right) {\rm d} \lambda. $$ Observe that it's in a double integral with over $\left\{(\lambda, t): 0\leq\lambda\leq 1, \lambda x \leq t \leq x\right\}$. By changing the order of integration, it's the same as $\left\{(\lambda, t): 0\leq t\leq x, 0 \leq \lambda \leq t/x\right\}$, so $$ \int_0^1 \ln \left(\frac{f(x)}{f(\lambda x)}\right) {\rm d} \lambda = \int_0^x \left(\int_0^{t/x} \frac{U(t)}{t} {\rm d} \lambda\right){\rm d} t = \\ =\int_0^x \frac{U(t)}{t}\left(\int_0^{t/x} {\rm d} \lambda\right){\rm d} t = \int_0^x \frac{U(t)}{t}\cdot \frac{t}{x}{\rm d} t = \frac{1}{x}\int_0^x U(t) {\rm d} t. $$ As it's given that $\frac{1}{x}\int_0^x U(t) {\rm d} t \to \rho$ as $x \to \infty$, it now can be easily seen that so does $\int_0^1 \ln \left(\frac{f(x)}{f(\lambda x)}\right) {\rm d} \lambda$.