Prove that $\forall \epsilon > 0$: $\lim_{t\to\infty}t^{-2}\int_{0}^{t}[(f(x))^{1+\epsilon}/f'(x)]\,\mathrm dx =+\infty$

Solution 1:

Partial solution. It is easy to see that $ f $ has an inverse $ g $. We will solve the problem with the additional hypothesis that $ g $ is a polynomial function.

For the inverse function theorem, since $f^\prime(x)> 0$ for all $ x \in [0, \infty) $ then $f:[0,\infty)\to [f(0),\infty)$ has an inverse global $g:[f(0),\infty)\to [0,\infty)$. Notice that we consider $ f^\prime(0) $ as is derivable at right of zero. In addition we have the equalities $$ \begin{matrix} g(f(x))=x \quad & \quad f(g(y))=y \quad & \quad g(y)=x \quad & \quad f(x)=y\\ \\ g^\prime(y)=\frac{1}{f^\prime (x)} \quad & \quad g^\prime(y)=\frac{1}{f^\prime (g(y))} \quad & \quad g(s)=t \quad & \quad f(t)=s \end{matrix} $$ By the formula of change of variables we have \begin{align} \int_0^{t} \big( f(x) \big)^{1+\epsilon}\frac{1}{f^\prime(x)} \mathrm{d}x =& \int_{g(f(0))}^{g(s)} \big( f(x) \big)^{1+\epsilon}\frac{1}{f^\prime(x)} \mathrm{d}x, \\ =& \int_{f(0)}^{s} \big( f(g(y)) \big)^{1+\epsilon}\frac{1}{f^\prime(g(y))}g^\prime(y) \mathrm{d}y, \\ =& \int_{f(0)}^{s} \big( y \big)^{1+\epsilon}\cdot g^{\prime}(y)\cdot g^{\prime}(y)\mathrm{d}y, \end{align} Now, check for every polynomial function $ g(y)=a_ny^n+\ldots+a_1 y+a_0$ that $$ \Phi_g(s)= \frac{1}{g(s)^2}\int_{f(0)}^{s} \big( y \big)^{1+\epsilon}\cdot g^{\prime}(y)\cdot g^{\prime}(y)\mathrm{d}y, = \frac{1}{t^2}\int_0^{t} \big( f(x) \big)^{1+\epsilon}\frac{1}{f^\prime(x)} \mathrm{d}x $$ is an function such that $\lim_{s\to\infty}\Phi_{g}(s)=\infty$.