Prove $\int_1^\infty {f(x)-L\over f(x)} dx$ diverges

Question

Let $f: [1, \infty)\rightarrow \mathbb R$ be a strictly decreasing function that converges to $0$ and fix some $L \in \mathbb R^+$. I want to show that $$\int_1^\infty {f(x)-L\over f(x)} dx$$ diverges.

What I have is \begin{align*}\int_1^\infty {f(x)-L\over f(x)} dx &= \int_1^\infty1dx - L\int_1^\infty{1\over f(x)}dx \\ &= \infty - L\int_1^\infty {1\over f(x)}dx \end{align*}

Is it enough to argue that since $\lim\limits_{x\rightarrow 0} f(x) = 0$, then $1/f(x)$ diverges, and so $\int_1^\infty 1/f(x)dx$ diverges. Therefore, the original integral must diverge?

Answer 1

You cannot split the integral into those two terms since you will end up with $\infty-\infty$.

$\frac {f(x)-L} {f(x)} \to -\infty$ as $x \to \infty$ and this implies that the integral is $-\infty$.