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.
 A: 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.
A: 
Since the integrand is an odd function, I believe the integral will be $0$ 

In the sense of principal value, yes, for the reasons you described. But the standard definition of improper integral over $\mathbb R$ requires both $\int_0^\infty f(x)\,dx$ and $\int_{-\infty}^0 f(x)\,dx$ to converge. This is not the case with $f(x)=(\cos x-1)/x$, and therefore the integral $\int_{-\infty}^\infty f(x)\,dx$ diverges. To see what the problem is, write 
$$
\int_0^\infty \frac{\cos x -1}{x}\,dx = 
\int_0^1 \frac{\cos x -1}{x}\,dx + \int_1^\infty \frac{\cos x }{x}\,dx - 
\int_1^\infty \frac{ 1}{x}\,dx
$$
and observe that the first two integrals on the right converge while the last one does not.

What conditions on the integrand make it integrable over $\mathbb R$?

For the property of being integrable (in the improper Riemann sense) there is no  necessary and sufficient condition that isn't tautological. One can give some sufficient conditions: for example, if $f$ is continuous and the function $(x^2+1)f(x)$ is bounded, then $f$ is integrable on $\mathbb R$. 

Perhaps with some type of decay conditions are required?

Decay conditions may be sufficient (see above), but they are not necessary. Even an unbounded function may be integrable on $\mathbb R$. For example, the function
$$f(x)=\begin{cases} 2^n \quad & n\le x\le  n+ 4^{-n} ,\ n=1,2,3,\dots \\
0 & \text{otherwise} \end{cases}$$ 
is unbounded but integrable on $\mathbb R$, with $\int_{-\infty}^\infty f(x)\,dx = 1$.
