Let $f$ be continuous and positive

Question

Let $f$ be continuous and positive, and assume $\int_0^\infty f(x)\mathrm dx$ converges. I'm supposed to prove the below integral converges:

$$\int_1^\infty \frac{f(x)}{\int_0^\infty f(t)\mathrm dt} \mathrm dx $$ I attempted $u$-substitution with $u=\int_0^\infty f(x)\mathrm dx$, but that didn't take me anywhere.

Answer 1

That's just $\int_1^\infty f(x)\,dx$ divided by the constant $\int_0^\infty f(x)\,dx$. Since everything is positive, and therefore $\int_{1+\varepsilon}^Mf(x)\,dx$ is "increasing" and bounded by $\int_0^\infty f(x)\,dx$, we have convergence.

Answer 2

This is trivial: $$\int_0^ \infty f(t)dt$$ converges by assumption and thus can be pulled outside the integral. Then, it suffices to note that $$\int_1^\infty \frac{f(x)}{\int_0^ \infty f(t)dt}dx= \frac{1}{\int_0^\infty f(t)dt} \int_1^\infty f(x) dx $$ $$\leq \frac{1}{\int_0^\infty f(t)dt} \int_0^\infty f(x) dx < \infty$$

where we used that $f$ is positive in the step to justify the inequality.