Integral of sin and cos

Starting from FT relation of delta function, I can write the followings:
$$ \int_{-\infty}^{\infty} \cos{\alpha x} dx = 0 $$
$$ \int_{-\infty}^{\infty} \sin{\alpha x} dx = 0 $$
The question is how am I supposed to prove those equations, sin and cos are stable oscillating functions.

 

In the same argument we will face problem for the cosine as the symmetric integral doesn't always lead to zero.

About this averaging thing, I have heard many times in literature they always say that. My current guess for this is
$$ \lim_{T\to \infty} \frac{1}{T} \int_{-T}^T \sin{\alpha x} dx = 0 $$
because the integral must be "some value" but finite, and divided by very large T will result zero? Is my assumption correct?

 

I wrote that to verify if my understanding of "averaged out to 0" was correct.
But you said that those integrals are undefined, what did you mean by that? Those integrals are derived from delta function, does this mean delta function is undefined too?

 

Would you please elaborate more about this? Because to me it's clear from the math that
$$ \int_{-\infty}^{\infty} e^{i\alpha x} dx = \int_{-\infty}^{\infty} (\cos{\alpha x}+i \sin{\alpha x}) dx = \delta (\alpha) = 0 $$
if ##\alpha \neq 0 ##.
And what does it mean with Riemann sense?

 

No we didn't. But after a short search improper integrals are those involving infinite upper and/or lower limits. Also if there is undefined value of the integrand within integration interval such as integral of (x^2-4)/(x-2) with undefined value at x = 2. Am I wrong?
But are you talking about this because the limit of the integrals in question is +- infinity?

 

What I know about delta function is that ## \delta(0) = undefined ## (just as undefined as 0/0), ## \delta(x) = 0 ## elsewhere. If not so, what is the correct representation of delta function, if there is any?

 

Ok somehow what I conclude from these talks is that delta function is kind of not a proper function (sorry if I use mathematically wrong term).