Integrability on R

1) What conditions on the integrand make it integrable over $\mathbb{R}$?

I know if a function is continuous and bounded on a closed interval $[-a,a]$ then this is enough for the function to be integrable on $[-a,a]$. But I'm not so sure if this results extends to $\mathbb{R}$? Perhaps with some type of decay conditions are required?

2) I want to prove

$$\int_{-\infty }^{\infty } \frac{(\text{cos}(t)-1)}{ t} \, dt$$

is integrable?

Is the following a valid argument. Since the integrand is an odd function, I believe the integral will be $0$ on [-a,a], so

$$\int_{-\infty }^{\infty } \frac{(\text{cos}(t)-1)}{ t} \, dt = \lim_{a\rightarrow\infty} \int_{-a}^{a} \frac{(\text{cos}(t)-1)}{ t} \, dt = \lim_{a\rightarrow\infty} 0 = 0$$

Hence since i've shown the integral is zero, it must exist, right? A type of proof by construction, I think.

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What you've shown is, that if the integral $\int_{-\infty}^\infty$ exists, then it must be $0$. However, you must allow the lower and upper limit to approach infinity independently, that is $\int_{-a}^b$ gets small as soon as $a,b\gg0$. This can be shown rigorously for this integrand whereas your simple method would erroneously also show $\int_{-\infty}^\infty x\,dx=0$! – Hagen von Eitzen Oct 18 '12 at 20:33
@Hagen von Eitzen: How do I let the limits approach infinity independently? Mind showing me an example. – aukie Oct 18 '12 at 21:25

1) That condition doesn't extend to $\mathbb{R}$ - consider $f(x)=1$, which is clearly bounded and continuous, but definitely has a divergent integral. The obvious necessary and sufficient condition is that $$\lim_{n\rightarrow\infty}\int_{-n}^{n}{f(x)dx}<\infty,$$ i.e. $f$ is integrable on every interval about $0$.
Assuming you know $f$ is integrable over some sufficiently large region, you could also show that, given $\epsilon>0$, there exists an $n>0$ such that $$\int_{-\infty}^{-n}f(x)dx+\int_{n}^{\infty}f(x)dx<\epsilon,$$ which is the kind of decay condition you were thinking of.
I don't think this is right. This is what I did proved in my post, that the integrand is integrable over all intervals around $0$. However a comment made by Hagen von Eitzen above suggests that this is not enough, since I did not allow for the limits to approach infinity independently – aukie Oct 18 '12 at 21:24
Sorry - when I said integrable, I meant that $f$ should be absolutely integrable, i.e. $|f|$ is integrable, I should have clarified that. – Peter Oct 18 '12 at 21:47