Belonging to an $ L^p$ space... so I'm currently tussling with this particular problem. I have $$f(x)={e^{-|x|}-1 \over x} $$ $$f_\varepsilon(x)={e^{-|x|}-{\sin(\varepsilon x) \over \varepsilon x} \over x},\quad x \in \mathbb R, \ \varepsilon > 0$$ 
I need to determine whether they belong to $L^1(\mathbb R), L^2(\mathbb R)$. The question is, must I necessarily brute-force it via the $\int_\mathbb R|f|^pdx<\infty$ condition or is there a smarter way to go about this? Can I determine the answer by looking for something in the functions?
 A: Hints:
$1).\ $ If $0\leqslant   t \leqslant  |x|,\ \left|{e^{-|t|}-1 \over t}\right |={1-e^{-|t|} \over t}=-\sum^{\infty}_{k=1}\frac{(-1)^kt^k}{k!}$ and this series converges uniformly on $[1,x]$ so it may be integrated termwise:
$\displaystyle\int^x_0 {e^{-t}-1 \over t}dt=-\sum^{\infty}_{k=1}\frac{(-1)^kx^{k+1}}{(k+1)!}=-\sum^{\infty}_{k=2}\frac{(-1)^kx^{k}}{k!}=-e^{-x}+1+x$ so the integral diverges.
This shows that  $\displaystyle\int^0_{-\infty} {e^{-|t|}-1 \over t}dt$ diverges as well.
Since $\displaystyle\int^1_{-1} {e^{-|t|}-1 \over t}dt$ converges, putting the pieces together, we get that the integral diverges. 
$2).\ $ Now,  $\displaystyle \int^{\infty}_1\left|{e^{-|t|}-1 \over x}\right|^2dx$ and $\displaystyle \int_{-\infty}^{-1}\left|{e^{-|t|}-1 \over x}\right|^2dx$ clearly converge so all you need consider is $\left|{e^{-|t|}-1 \over x}\right|^2$ on $[-1,1]$.
$3).\ $ Finally, split up $\mathbb R$ and analyze as above to consider the three integrals whose integrand is $f_{\epsilon.}$
