Is this improper integral answer correct?

Question

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?

Answer 1

$$\int^∞_{-∞}\cos(\pi t) \,dt= \int^∞_{0}\cos(\pi t)\,dt + \int^0_{-∞}\cos(\pi t)\,dt$$

Indeed, you are indeed correct that $\int^0_{-\infty}cos(\pi t)\,dt$ diverges.

And as you stated at the start, the integral is convergent if and only if split integrals are both convergent. Since you showed one of the two integrals in divergent, the entire integral is divergent.

Answer 2

let us take an odd function, that is $f(-x)=-f(x)$. Now define the integral: $$I(n)=\int_{-n}^nf(x)dx=\int_{-n}^0f(x)dx+\int_0^nf(x)dx$$ $$I(n)=\int_n^0f(-x)(-dx)+\int_0^nf(x)dx$$ $$I(n)=\int_n^0f(x)dx+\int_0^nf(x)dx$$ $$I(n)=-\int_0^nf(x)dx+\int_0^nf(x)dx=0$$ so as $n\to\infty$ ,$I(n)=0$ even if over the positive or negative regions the integral is not convergent, however for the case of an even function these positive and negative regions do not "cancel" out and so it is required that both the positive and negative regions converge independently