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Suppose $f$ is Riemann integrable on $[1,b]$ for every $b>1,$ and that $\int_1^\infty |f| <\infty.$ Show

$$\lim_{n\to \infty} \int_1^\infty f(x^n)\, dx = 0.$$

Unfortunately, I'm not even sure where to begin this problem. I know that because $\int_1^\infty |f| <\infty,$ the improper integral $\int_1^\infty f $ converges. Because of the exponent, I considered applying the root test for integrals, but I am rather lost. Thanks in advance

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  • $\begingroup$ You could try whether a substitution helps. $\endgroup$ – Daniel Fischer Nov 6 '16 at 21:51
  • $\begingroup$ You don't want to say "$f$ converges". You want to say the integral of $f$ converges. $\endgroup$ – zhw. Nov 6 '16 at 23:28
  • $\begingroup$ good catch @zhw. thanks $\endgroup$ – sequenceDerivative Nov 6 '16 at 23:42
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Using substitution $x^n = t$

$$\int_1^{\infty} f(x^n)dx = \int_1^{\infty}\frac1n \frac{t^{1/n}}t f(t)dt$$

$\frac1n \frac{t^{1/n}}t f(t)$ converges to $0$ and is absolutely bounded by $|f|$ so by the dominated convergence theorem the integral converges to $0$

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  • $\begingroup$ thank you for your response. But in the substitution, why isn't it $n \frac{t}{t^{1/n}}$. $\endgroup$ – sequenceDerivative Nov 6 '16 at 22:15
  • $\begingroup$ @sequenceDerivative Because $x = t^{1/n}$, so $dx = d(t^{1/n}) = \frac{1}{n} t^{\frac{1}{n}-1}\,dt$. $\endgroup$ – Daniel Fischer Nov 6 '16 at 22:36
  • $\begingroup$ of course, thanks @DanielFischer, I was not thinking clearly there. So that makes sense why the integrand converges to 0. But the second part of the answer still eludes me. I'm unfamiliar with the dominated convergence theorem. Is there an alternate way to finish the argument? $\endgroup$ – sequenceDerivative Nov 6 '16 at 22:43
  • $\begingroup$ @sequenceDerivative Here you can just use that for $t\geqslant 1$ we have $$\biggl\lvert \frac{1}{n} t^{\frac{1}{n}-1} f(t)\biggr\rvert \leqslant \frac{1}{n}\lvert f(t)\rvert.$$ $\endgroup$ – Daniel Fischer Nov 6 '16 at 22:53

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