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Does $$\int^{\infty}_{0}\frac{\cos ^2\left(3x+1\right)}{\sqrt{x}+1} dx$$converge/diverge?

Around $\infty$, this function behaves like:

$$\frac{\cos ^2(3x)}{\sqrt{x}} dx$$

but this does that tell me anything about its convergence or divergence? How can I proceed from here? Something I do notice that that $\cos^2(3x)$ oscillates between $0$ and $1$, and so I think it diverges. Am I right?

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  • $\begingroup$ this integral does not converge $\endgroup$ Mar 5 '17 at 16:22
  • $\begingroup$ Why would you expect this to converge? The integral of $\frac 1{\sqrt{x}}$ diverges, and $\cos^2(3x+1)$ is uniform and periodic. (Massage this argument a bit and you have a proof) $\endgroup$ Mar 5 '17 at 17:24
  • $\begingroup$ @BrevanEllefsen exactly $\endgroup$ Mar 5 '17 at 20:28
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Perhaps this approach is the most explicit $$ I\equiv\int^{\infty}_{0}\frac{\cos ^2\left(3x+1\right)}{\sqrt{x}+1} dx $$ the integrand function is always positive so if we show that the integral still diverges when done on a subset of $\mathbb{R}^+$ we are done, i'm taking as a subset the one that makes $$ \cos{(3x+1)}\ge\frac{1}{\sqrt2} $$ It's easy to show that the subset requested is this: $S={\cup}_{k=1}^{\infty}\;\left[A_k,B_k\right]$, with $A_k$ and $B_k$ given by: $$ A_k=\frac{2 \pi k}{3}-\frac{1}{3}-\frac{\pi }{12} $$ $$ B_k=\frac{2 \pi k}{3}-\frac{1}{3}+\frac{\pi }{12} $$

since the integrand function is positive it holds: $$ S \subset \mathbb{R} \implies \int_S\cdots\le\int_{\mathbb{R}^+}\cdots $$ we have: \begin{multline} I\ge\sum_{k=1}^{\infty}\int_{A_k}^{B_k}\frac{\cos ^2\left(3x+1\right)}{\sqrt{x}+1} dx\ge{1\over2}\sum_{k=1}^{\infty}\int_{A_k}^{B_k}\frac{1}{\sqrt{x}+1} dx\ge{1\over2}\sum_{k=1}^{\infty}\int_{A_k}^{B_k}\frac{1}{\sqrt{B_k}+1} dx=\\ ={1\over2}\sum_{k=1}^{\infty}\frac{B_k-A_k}{\sqrt{B_k}+1}={\pi\over 12}\sum_{k=1}^{\infty}\frac{1}{\sqrt{B_k}+1} \end{multline} and $$ \frac{1}{\sqrt{B_k}+1}\approx {\sqrt{3}\over \sqrt{2\pi k}}\implies\sum_{k=1}^{\infty}\frac{1}{\sqrt{B_k}+1}\to\infty $$ by direct comparison: $$ I \ge \sum_{k=1}^{\infty}\frac{1}{\sqrt{B_k}+1}\to\infty \implies I\to\infty $$ so the integral is divergent.

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  • $\begingroup$ Odd that this was downvoted. I like this approach. $\endgroup$ Mar 5 '17 at 23:19
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Note \begin{eqnarray} &&\int\frac{\cos ^2\left(3x+1\right)}{\sqrt{x}+1} dx\\ &=&\frac12\int\frac{1+\cos\left(6x+2\right)}{\sqrt{x}+1} dx\\ &=&\frac12\int\frac{1}{\sqrt{x}+1} dx+\frac12\int\frac{\cos\left(6x+2\right)}{\sqrt{x}+1} dx\\ &=&\frac12\int\frac{1}{\sqrt{x}+1} dx+\frac12\int\frac{1}{\sqrt{x}+1} d\sin\left(6x+2\right)\\ &=&\frac12\int\frac{1}{\sqrt{x}+1} dx+\frac12\sin(6x+2)\frac{1}{\sqrt{x}+1}+\frac12\int\frac{\sin(6x+2)}{2\sqrt x(\sqrt{x}+1)^2} dx. \end{eqnarray} Now $$\left|\int_0^\infty\frac{\sin(6x+2)}{2\sqrt x(\sqrt{x}+1)^2} dx\right|\le\int_0^\infty\frac{1}{2\sqrt x(\sqrt{x}+1)^2} dx<\infty $$ and hence $\int_0^\infty\frac{\sin(6x+2)}{2\sqrt x(\sqrt{x}+1)^2} dx$ converges. Clear $\int_0^\infty\frac{1}{\sqrt{x}+1} dx$ diverges, so the improper integral diverges.

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