Integral of $ \int_{-\infty}^\infty \cos (\pi t) dt$ I need to determine whether the integral 
$$ \int_{-\infty}^\infty cos \,(\pi t) \;dt$$
is convergent or divergent.  I rewrote this improper integral as 
$$ \lim \limits_{x \to{-\infty}}\int_{x}^0 cos \,(\pi t) \;dt +  \lim \limits_{x \to{\infty}}\int_0^x cos \,(\pi t) \;dt$$
I integrated
$ \lim \limits_{x \to{-\infty}}\int_{x}^0 cos \,(\pi t) \;dt$ 
to get 
$ \frac{sin \,(\pi t)}{2}$
which gives me 
$ \frac{sin \,0}{0} - \frac{sin \,t}{t}$
Obviously I can't divide by 0 so does this mean that the function is divergent, or is there some other step I can take that I'm missing?
 A: We have
$$\int_0^x \cos(\pi t)\,dt=\left[\frac{\sin(\pi t)}{\pi}\right]_0^x=\frac{\sin(\pi x)}{\pi}\ ;$$
this continues to oscillate between $1/\pi$ and $-1/\pi$ and therefore has no limit as $x\to\infty$.  Hence
$$\int_0^\infty \cos(\pi t)\,dt$$
diverges, and so does
$$\int_{-\infty}^\infty \cos(\pi t)\,dt\ .$$
A: You've definitely done something wrong with that integration and substitution, if you're getting (a) a $t$ in your final answer (hint: it disappears when you integrate over it) and (b) getting things divided by 0 and/or $t$.
$\int_0^x \cos (\pi t) dt = \sin (\pi x) / \pi$, which takes values from $-1/\pi$ to $1/\pi$ as $x$ varies. If we take the two-sided integral, then we have $\int_{-x}^x \cos (\pi t) dt = (\sin (\pi x) - \sin (-\pi x)) / \pi = 2\sin (\pi x) / \pi$, because the function we are integrating is even (and hence has the same area to the left and right of the y-axis). This also behaves sinusoidally as $x$ varies, so even if we just try to take the limit as $x \rightarrow \infty$ we can see we're not going to get a nice limit.
