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I'm trying to prove divergence of the integral $\int_1^{\infty}\frac{\left|\cos{x^2}\right|}{x^q}dx$, where $0<q<1$, by applying the mean-value theorem to $\int_{n\pi}^{(n+1)\pi}\frac{\left|\cos{x^2}\right|}{x^q}dx$: there exists $c_n\in [n\pi,(n+1)\pi]$ such that $\int_{n\pi}^{(n+1)\pi}\frac{\left|\cos{x^2}\right|}{x^q}dx=\left|\cos{c_n^2}\right|\int_{n\pi}^{(n+1)\pi}\frac{1}{x^q}dx$. So if I can show that $\inf{\left|\cos{c_n^2}\right|}\neq 0$, we would have the divergence, by comparison with $\int_1^{\infty}\frac{1}{x^q}dx$. It isn't apparent to me how to prove that bit.

I know that the divergence can also be proved by noting that $\left|\cos{x^2}\right|\geq\cos^2{x^2}=\frac{1+\cos{2x^2}}{2}$ and using a test akin to Dirichlet's test for series to prove the convergence of $\int_1^{\infty}\frac{\cos{2x^2}}{x^q}dx$.

But I'd like to know if my idea can be finished.

Thank you.

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You can avoid the issues with the mean value theorem by proceeding like this: Change variables from $x$ to $x^2$ as one of the answerers suggested, so that one is attempting to show divergence of ${\displaystyle \int_1^{\infty} {|\cos(x)| \over x^r}\,dx}$, where now ${1 \over 2} < r < 1$. Next, note that one has $$\int_{2\pi}^{\infty} {|\cos(x)| \over x^r}\,dx = \sum_{n=1}^{\infty} \int_{2n\pi}^{2(n+1)\pi}{|\cos(x)| \over x^r}\,dx$$ Since the denominator is monotone increasing, this is bounded below by $$\sum_{n=1}^{\infty} \int_{2n\pi}^{2(n+1)\pi}{|\cos(x)| \over (2\pi(n+1))^r}\,dx$$ $$= {1 \over (2\pi)^r}\sum_{n=1}^{\infty} {1 \over (n+1)^r}\int_{2n\pi}^{2(n+1)\pi}|\cos(x)|\,dx$$ Since $\cos(x)$ has period $2\pi$, the integral above has some fixed value $C > 0$, so the above is equal to $$C {1 \over (2\pi)^r}\sum_{n=1}^{\infty} {1 \over (n+1)^r}$$ Since $r < 1$ this diverges.

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Careful. Showing $\cos{c_n^2}\neq 0 $ does not show divergence. (Consider $\cos{c_n^2}=n^{-(n!)}$)

For your idea to work you need to show that there exists $c_n$ such that $c_n\leq \int_{n\pi}^{(n+1)\pi}|\cos x^2|dx$ and $c_n$ does not go to zero too quickly. That is we need to bound the integral from below. (Using inequalities rather than the equality given by the mean value theorem will suffice.)

Then $$\int_{n\pi}^{(n+1)\pi}\frac{|\cos x^2|}{x^q}dx\geq c_n\cdot \int_{n\pi}^{(n+1)\pi}\frac{1}{((n+1)\pi)^q}dx=\frac{c_n\pi}{((n+1)\pi)^q}$$ Where the inequality follows since $\frac{1}{((n+1)\pi)^q}\leq x^q $ on this interval, and the average of $\cos x^2$ is greater than $c_n$.

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Like Zarrax has pointed out, we could look at ${\displaystyle \int_1^{\infty} {|\cos(x)| \over x^r}\,dx}$ where ${1 \over 2} < r < 1$.

We could argue that if ${\displaystyle \int_1^{\infty} {\cos^2(x) \over x^r}\,dx}$ diverges then by the comparison test our integral diverges, since ${|\cos(x)| \over x^r} \geq {\cos^2(x) \over x^r}$.

Showing divergence of ${\displaystyle \int_1^{\infty} {\cos^2(x) \over x^r}\,dx}$ is simple if we note the fact that: $$\frac{\cos^2(x)}{x^r}=\frac{1}{2x^r}+\frac{\cos(2x)}{2x^r}$$

But ${\displaystyle \int_1^{\infty} \frac{\cos(2x)}{2x^r}\,dx}$ converges by Dirichlets test and ${\displaystyle \int_1^{\infty} \frac{1}{2x^r}\,dx}$ obviously diverges hence $$\int_1^{\infty} {\cos^2(x) \over x^r}\,dx$$ diverges as needed.

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