# Prove $\exp(\int^x_0 \frac{U(t)}{t} dt)$ is reguarly varying (Extreme Values, Regular Variation and Point Processes)

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 $$\int_0^1 \ln \left(\frac{f(x)}{f(\lambda x)}\right) {\rm d} \lambda$$.