# Let $f$ be continuous and positive

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.

• For convergence, you need something to vary. What is varying here? Aug 9, 2020 at 11:22
• Hint: the denominator is a constant and you know it is finite, so you only have to show that $\int_0^1 f(x) dx$ converges. Aug 9, 2020 at 11:25
• The substitution $u=\int_0^{\infty} f(x)dx$ does not make sense. RHS is not variable , it is a constant Aug 9, 2020 at 11:52

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.
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.