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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

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

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    $\begingroup$ 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). $\endgroup$ – Maximilian Janisch May 26 '19 at 15:05

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