# Limit of certain integrals related to an absolutely integrable function

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

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

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 $0$
• thank you for your response. But in the substitution, why isn't it $n \frac{t}{t^{1/n}}$. – sequenceDerivative Nov 6 '16 at 22:15
• @sequenceDerivative Because $x = t^{1/n}$, so $dx = d(t^{1/n}) = \frac{1}{n} t^{\frac{1}{n}-1}\,dt$. – Daniel Fischer Nov 6 '16 at 22:36
• @sequenceDerivative Here you can just use that for $t\geqslant 1$ we have $$\biggl\lvert \frac{1}{n} t^{\frac{1}{n}-1} f(t)\biggr\rvert \leqslant \frac{1}{n}\lvert f(t)\rvert.$$ – Daniel Fischer Nov 6 '16 at 22:53