Proving $\int_0^{\infty}f(x) $ converges using Lagrange

Question

Let $f:[0,∞) \to \Bbb R$ be a differentiable such that $f(x) > 0$ for every $x \in [0, \infty)$ and a positive function. Assume there exits an $0 < L < \infty$ such that $$\lim_{x\to \infty}\,[\ln(f(x))]'=-L.$$ Prove that $\int_0^{\infty}f(x) $ converges.

We were hinted that using Lagrange would do the trick but I fail to see how applying $g(x) =\ln(f(x))$ would help or by using the limit definition. Please help solve.

Answer 1

Since $\lim\limits_{x\to \infty}\,[\ln(f(x))]' = \lim\limits_{x\to \infty}\, \dfrac{f'(x)}{f(x)}=-L < 0$, $$\forall \varepsilon \in (0,L), \exists M > 0: \forall x \ge M, \frac{f'(x)}{f(x)} \le -L + \varepsilon < 0.$$ Fix any $\varepsilon \in (0,L)$. Multiply both sides by $f(x)$. (We can do so since $f(x) > 0$ for all $x \in [0,\infty)$.) $${f'(x)} \le (-L + \varepsilon) f(x) < 0 \tag{*} \label{*}$$ From this, it's clear that $f$ is (strictly) decreasing on $[M,\infty)$.

Source: https://ltcconline.net/greenl/courses/107/Series/INTTEST.HTM

Therefore, the convergence of $\int_M^\infty f(x) dx$ is equivalent to the convergence of the infinite sum $\sum\limits_{n \ge M} f(n)$. (Since $f$ is differentiable on $[0,\infty)$, it's continuous on the closed and bounded interval $[0,M]$, so $\int_0^M f(x) dx$ is well-defined.)

To tidy things up, write $N = \min\{ n \in \Bbb{N} \mid n \ge M\}$ so that $\sum\limits_{n \ge M} f(n) = \sum\limits_{n=N}^\infty f(n)$.

Apply Grönwall's inequality (with $u(t) = f(t))$, $\beta(t) = -L + \varepsilon$ and $I = [N,\infty) \;(\subseteq [M,\infty))\;$) to see that $$ f(t)\leq f(N)\exp {\biggl (}\int _{N}^{t}(-L + \varepsilon)\,{\mathrm {d}}s{\biggr )} = f(N) \exp((-L + \epsilon)(t - N)) \tag{#} \label{#}$$ for all $t \in [N,\infty)$.

Put $t = N+1$ in \eqref{#} to get the relation $$f(N+1) \le f(N) e^{-(L-\epsilon)} \tag3 \label3.$$ To conclude that the series $\sum\limits_{n=N}^\infty f(n) < \infty$ by induction, we need \eqref{3} for all natural numbers $n \ge N$. $$\forall n \ge N, f(n+1) \le f(n) e^{-(L-\epsilon)} \tag4 \label4.$$ To justify \eqref{4}, replace $I$, $N$ and $t$ with $[n,\infty)$, $n$ and $n+1$ respectively in \eqref{#}.

Now, we see that the sequence $(f(n))_{n \ge N}$ is in fact bounded by a geometric progression with common ratio $e^{-(L-\varepsilon)} \in (0,1)$. $$0 \le \sum_{n=N}^\infty f(n) \le f(N) \sum_{n = 0}^\infty e^{-(L-\varepsilon)} = \frac{f(N)}{1 - e^{-(L-\varepsilon)}} < \infty.$$

This shows that $\sum\limits_{n=N}^\infty f(n) < \infty$, so $\int_M^\infty f(x) dx < \infty$, and hence $\int_0^\infty f(x) dx < \infty$.