Showing convergence/divergence of generalized integrals I just started working with generalized integrals on my own and it is very hard, I think. I don't quite know where to start, I can never seem to find some other integral to compare it to. I've been working on these for a while now, and I am not even close to solving them. So I didn't know where else to turn since I have no friends who study math or anything.
First (a), I wish to show that the following integrals diverges: $$\int_{0}^{\pi} \frac{\sqrt{x}}{\sin(x)}dx.$$ 
Then (b), I want to show that the following is convergent: $$\int_{0}^{\infty} \frac{\sqrt{x-\sin(x)}}{x^2} dx.$$ 
Lastly (c), I want to show that the following is divergent: $$\int_2^{\infty} \frac{\sin(1/x)}{\ln(x)} dx.$$ 
I am unsure how much I should share about what I have tried so far since nothing has worked and it would be so much to write everything down. Basically, for (a) and (c) I've been trying to find some function smaller (on the interval) than the integral I'm working on, and that diverges. For (b) I am doing the opposite. I have also tried to write the integrand $f(x)$ as a product $g(x)h(x)$ such that $h(x)$ has a finite limit A $\gt$ 0 and such that $g(x)$ converges, but again, I haven't found anything that helps me.
Any specific help about these problems would be great, but also maybe some advice on how to think when solving this type of exercises. It is especially hard when the boundaries are not from $1$ to $\infty$ since in the examples of the book they tend to compare with integrals with those boundaries.
 A: Consider the asymptotic behavior of the integrand near the singular points.  This is applied in the limit comparison test.  Specifically, suppose  $f,g:(a,b] \to \mathbb{R}^+$ are integrable away from $a$ and $\lim_{x \to a+} f(x)/g(x) = L.$  If $0 < L < \infty$ then the improper integrals of $f$ and $g$ converge or diverge together on $[a,b].$
For (a) we have
$$\tag{1}\int_{0}^{\pi} \frac{\sqrt{x}}{\sin x} \, dx = \int_{0}^{\pi/2} \frac{\sqrt{x}}{\sin x} \, dx + \int_{\pi/2}^{\pi} \frac{\sqrt{x}}{\sin x} \, dx. $$
Note that
$$ \lim_{x \to 0+}\frac{\sqrt{x}}{\sin x} \frac{1}{1/\sqrt{x}} = \lim_{x \to 0+}\frac{x}{\sin x} = 1.$$
Since $\int_0^{\pi/2}(1/\sqrt{x})$ converges, the first integral on the right-hand side of (1) converges.
However,
$$\lim_{x \to \pi-}\frac{\sqrt{x}}{\sin x} \frac{1}{1/(\pi-x)} = \lim_{x \to \pi-}\frac{(\pi - x)\sqrt{x}}{\sin x}  = \lim_{x \to \pi-}\frac{\pi - x}{\sin x}\lim_{x \to \pi-} \sqrt{x} = \sqrt{\pi}.$$
Since the integral of $1/(\pi-x)$ diverges over $[\pi/2,\pi]$ like $\ln(\pi-x)$ as $x \to \pi$, the second integral on the right-hand side of (1) diverges. Hence, the integral in (a) is divergent.
The integral (c) diverges since
$$\lim_{x \to \infty}\frac{\sin(1/x)}{\ln x} \frac{1}{1/(x \ln x)} = \lim_{x \to \infty} \frac{\sin(1/x)}{1/x} = 1,$$
and the integral of $1/(x \ln x)$ over $[2, \infty)$ diverges like $\ln \ln x$ as $x \to \infty.$
I'll leave you with (b).
