# Test $\int_{0}^{\infty}\frac{e^{ix}}{\log(x)}, \int_{0}^{\infty}\frac{\cos(x)}{x^p}, \int_{0}^{\infty}\cos(x^2)$ for convergence

I need to test these integrals for convergence $\int_{0}^{\infty}\frac{e^{ix}}{\log(x)}, \int_{0}^{\infty}\frac{\cos(x)}{x^p}, \int_{0}^{\infty}\cos(x^2)$ with $p\in\mathbb{R}$. However I suck horribly at this.

Let me start with $\int_{0}^{\infty}\frac{e^{ix}}{\log(x)}$. Am I right in testing its series representation for absolute convergence, i.e. $\sum_{n=0}^{\infty}\left|\frac{e^{in}}{\log(n)}\right|$?

Am I right in arguing that $\int_{0}^{\infty}\frac{\cos(x)}{x^p}$ diverges for $p\geq1$ since $0\leq\frac{\cos(x)}{x^p}\leq1/x$? How would I go about $p<1$?

For $\int_{0}^{\infty}\cos(x^2)$ I think I integrate by parts, yes? Since our function is continuous I can rewrite it as $$\int_{0}^{1}\cos(x^2)dx + \int_1^{\infty}\cos(x^2)dx.$$

The first term should be constant and go towards $0$ as $x$ approaches $\infty$. The second one gives me a bit more trouble. I assume I need partial integration here, but I'm not seeing how.

As regards the first one, the function $\frac{e^{ix}}{\log(x)}$ is not integrable in a neighbourhood of $1$. Why?
For the second one, you are right, it diverges for $p\geq 1$ in $[0,1]$. For $0<p<1$ note that, by integration by parts, $$\int_1^{+\infty}\frac{\cos(x)}{x^p} dx=\left[\frac{\sin(x)}{x^p}\right]_1^{+\infty}+p\int_{1}^{+\infty}\frac{\sin(x)}{x^{p+1}} dx.$$ Can you take it from here? What about the case when $p\leq 0$?
For the third one, let $t=x^2$ then $$\int_1^{+\infty}\cos(x^2)dx=\frac{1}{2}\int_1^{+\infty}\frac{\cos(t)}{\sqrt t}dt.$$ What may we conclude?