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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.

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  • $\begingroup$ For convergence, you need something to vary. What is varying here? $\endgroup$ Aug 9, 2020 at 11:22
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    $\begingroup$ 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. $\endgroup$ Aug 9, 2020 at 11:25
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    $\begingroup$ The substitution $u=\int_0^{\infty} f(x)dx$ does not make sense. RHS is not variable , it is a constant $\endgroup$ Aug 9, 2020 at 11:52

2 Answers 2

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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.

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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.

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