Power functions and $\lim\limits_{x \rightarrow +\infty}{\displaystyle\frac{\int_{0}^{x}{f(t)\mathrm{d}t}}{x f(x)}}=\displaystyle\frac{1}{1 + \beta}$

Solution 1:

In general, $f$ will be “almost” a power function. More precisely, it will be regularly varying with index $\beta$, i.e., of the form $f(x)=x^\beta\cdot\ell(x)$ where $\ell$ is a slowly varying function, such as $\ell(x):=\log(1+x)$. This is in essence the content of Karamata's theorem (see Theorems 1.5.11 and 1.6.1 of Bingham, Goldie, and Teugel's book “Regular variation”): Karamata's theorem states that a positive, locally bounded function $f:[0,\infty)\to[0,\infty)$ is a regularly varying function with index $\beta\ge0$ if and only if $$\frac{x^{\sigma+1}f(x)}{\int_0^xt^\sigma\,f(t)\,\mathrm dt}\xrightarrow[x\to\infty]{}\sigma+\beta+1\tag{$C_\sigma$}$$ holds for some $\sigma>-(\beta+1)$. In this case, $(C_\sigma)$ holds for all $\sigma\ge-(\beta+1)$.

The representation theorem (Theorem 1.3.1 in the book) says that we can write any slowly varying function $\ell$ in the form $$\ell(x)=c(x)\exp\left(\int_0^x\varepsilon(u)\frac{\mathrm du}u\right)$$ for some $c$ measurable with $c(x)\to c\in(0,\infty)$ and $\varepsilon(x)\to0$ as $x\to\infty$.