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?
Best Answer
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 $$\int_{-\infty}^\infty \cos(\pi t)\,dt\ .$$