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I am trying to figure out how to compute the limit

$$\lim_{t \rightarrow \infty} \int_{-1}^1 \frac{\cos^2(t^3x^{10})}{tx^2 + 1} \, dx.$$

If I exchange the limit and integral (when is this allowed?) then I get the integral of 0. This seems too easy. What should I be looking for?

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  • $\begingroup$ If you can't use the big convergence theorems then try direct estimation. $\endgroup$
    – RRL
    Dec 7, 2020 at 19:28

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Exchanging integral and limit is allowed here: Your integrand is bounded by 1 so you can use dominated convergence theorem.

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  • $\begingroup$ But that theorem appears in a much later chapter of my text using "measure" (whatever that is). I do see that the absolute value of the integrand is uniformly bounded by 1 on the x-interval [-1,1] though. So how can I use this to prove that either I can exchange the limit and integral or else that the answer is 0? $\endgroup$
    – E2R0NS
    Dec 7, 2020 at 17:44
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Note that

$$\left|\int_{-1}^1 \frac{\cos^2(t^3x^{10})}{tx^2 + 1} \, dx\right|\leqslant\int_{-1}^1 \frac{|\cos^2(t^3x^{10})|}{tx^2 + 1} \, dx \leqslant 2 \int_0^1\frac{1}{tx^2 + 1} \, dx = \frac{2}{\sqrt{t}}\int_0^{\sqrt{t}} \frac{du}{1+u^2}$$

Try to finish from here.

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  • $\begingroup$ You wrote one inequality and two equal signs. Shouldn't the first equal sign be an inequality too since you are saying that the numerator is less or equal to one on [-1,1] for any t greater than 0? $\endgroup$
    – E2R0NS
    Dec 7, 2020 at 19:41
  • $\begingroup$ Yes - it was a typo that I fixed now. $\endgroup$
    – RRL
    Dec 7, 2020 at 20:11

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