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

Let $$f: [0,+\infty) \to [0,+\infty)$$ be differentiable, $$f' > 0$$. Prove that $$\forall \epsilon > 0: \lim_{t\to\infty}\dfrac{1}{t^2}\int_{0}^{t}\dfrac{\left(f(x)\right)^{1+\epsilon}}{f'(x)}\mathrm dx =+\infty$$

I used L'Hôpital's rule and got $$\lim\limits_{t\to\infty}\dfrac{1}{t^2}\int_{0}^{t}\dfrac{\left(f(x)\right)^{1+\epsilon}}{f'(x)}\mathrm dx =\dfrac{1}{2}\displaystyle\lim_{t\to\infty}\dfrac{\left(f(t)\right)^{1+\epsilon}}{tf'(t)}$$ If $$\lim\limits_{t\to\infty}f'(t)$$ exists, then we can prove the above statement using L'Hôpital's rule. But if $$\lim\limits_{t\to\infty}f'(t)$$ does not exist, I don't know how to proceed

• What is $f^{1+\epsilon}$? – Wojowu May 26 '19 at 11:45
• @Wojowu I interpreted it as $f^{1+\epsilon}(x):=f(x)^{1+\epsilon}$. – Maximilian Janisch May 26 '19 at 11:53
• Note that $\lim_{t\to\infty} \frac{f(t)}{tf’(t)}=0$ for $f(t)=\exp(t)$. The $\epsilon$ is definitely needed. – Maximilian Janisch May 26 '19 at 12:01
• @TonyS.F. There is a reason \displaystyle and the likes should not be used in titles. – StubbornAtom May 26 '19 at 12:51
• @StubbornAtom my edit was just to clarify what was meant by $f^{1+\epsilon}$... – TSF May 26 '19 at 13:41

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 $$\lim_{s\to\infty}\Phi_{g}(s)=\infty$$.
• This is absolutely epic! One should note though that $\lim_{x\to\infty} f(x)$ need not be $\infty$. I don’t think that you need this, but your function $g$ is only defined on something like $[f(0), \lim_{t\to\infty}f(t)[$ (remark: somebody posted an answer here that, I think, works if $f$ doesn’t blow up; in the worst case you could incorporate this answer). – Maximilian Janisch May 26 '19 at 15:05