Show these functions are/ aren't Lebesgue Integrable? How do I go about showing this? 
I have recently been learning the comparison test, MCT, Fatou's Lemma and DCT for Lebesgue integrals, but have been struggling with the details of the proofs.
1) f is 0 a.e. so is integrable to 0
2) not integrable, by showing integral tends to infinity. Is there a cleaner approach?
3) Unsure how to approach, possibly comparison?
4) Modulus is less than 1, so comparison test.
5) Comparison test. 
6) and onwards, unsure how to approach
Any help would be greatly appreciated. How much detail is necessary in these sort of proofs? Is there a special set of functions or things I should notice which will help me in future?
Thank you!
 A: (iii) $\int_1^\infty |f| = 1 + 1/2 + 1/3 + \cdots $
(vi) $x^ne^{-x} \to 0$ for every $n.$
(vii) Surely you have some ideas here. Try $\alpha = 0,-1,-2$ for example.
(viii) the integrand is greater than $1/(\sqrt 2 \cdot x)$ on each $[2n\pi -\pi/4,2n\pi + \pi/4].$
(ix) $f(x) \sim 1/x$ as $x\to \infty.$
A: Assuming that integrable means integrable over the whole domain :
(i)Yes. $f=0$ a.e.
(iv)Yes. $f$ is bounded, and the domain is bounded.
(iii)  Although $\lim \int_0^x f(y) dy$  exists, the Lebesgue integral is defined as $\int f^+ - \int f^-$ and neither of these is finite.
(ii) For $\pi  /2>x>\pi /3$ we have $\tan x= \cot (\pi /2-x)>(1/2)/\sin (\pi /2 -x)>(1/2)/(\pi /2-x)$.
For (v) and (vii) consider that $\log x<<x<<e^x$ as $x\to \infty.$ Use comparison test to $e^{-x}$ or to $x^{\alpha}.$ Note that (vii) depends on the value of $\alpha$.
(ix) $(\sin (1/x)/(1/x)\to 1$ as $x\to \infty$.
(viii) For $n\in N,$ for $u=2 \pi n,$ and  $v=u+\pi /2$ we have $f(x)=(\cos x)/(1+x)\geq 0$ and $\int _u^v f(x) dx> 1/(1+v).$ See my remarks above, concerning (iii).
(vi) $e^{-x}\to 1$ as $x\to 0$ .Use comparison to $\int_u^1 \log x \; dx$  as $u\to 0.$
