So I'm working on improper integrals and con/divergence and want some assurance that I've done the following correctly.

$\int^∞\_{-∞}cos(\pi t)$

As far as I'm aware this is convergent if and only if 
$\int^∞\_{0}cos(\pi t)$ and 
$\int^0\_{-∞}cos(\pi t)$ are both convergent.

So starting by working out the indefinite integral...

$\int cos(\pi t)$ = $\frac{1}{\pi}sin(\pi t)+C $

So if we work on the indefinite integral $\int^0\_{-∞}cos(\pi t)$, we have...

$\int^0\_{-∞}cos(\pi t)$ = $lim\_{t->-∞}\int^0\_t cos(\pi t)$

= $lim\_{t->-∞}\frac{1}{\pi}[sin(\pi t)]^0\_t$

= $lim\_{t->-∞}\frac{1}{\pi}[0-sin(\pi t)]$

= $lim\_{t->-∞}\frac{-sin(\pi t)}{\pi}$

Since the sin function is periodic and doesn't tend towards a limit as t approaches negative infinity, then the limit does not exist. Hence the integral is divergent.

By this logic, the full integral 
$\int^∞\_{-∞}cos(\pi t)$
is also divergent and we don't need to bother calculating or testing the integral for positive infinity.

I'm still a little unsure about this topic so I'm looking for reassurance as to if I am correct, or if I have done something wrong?