Showing that the integral of sin(πt)dt is divergent?

Question

Having trouble showing why this integral diverges.

$\int_\infty^{-\infty} sin(πt)dt$

How do I write this integral as a limit?

Answer 1

$$ \int^{2n}_{-n}\sin(\pi t)dt=-\frac{1+(-1)^n}{\pi} $$ and that alternates between $0$ and $-2/\pi$ as you take the limit $n\to\infty$ ($n$ is odd in the above).

Answer 2

As

$$\lim_{n\to+\infty}\int_{2n}^{2n+\frac{1}{2}}\sin(\pi t)dt=\frac{1}{\pi}\;(\neq 0)$$

$$\int_0^{+\infty}\sin(\pi t)dt$$ diverges by Cauchy test and so does

$$\int_{-\infty}^{+\infty}\sin(\pi t)dt.$$