Integral of sin and cos

[blue_leaf77]

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.

 

[mathman]

Strictly speaking these integrals are undefined. The sine integral can be "proved" as the limit of finite symmetric integrals. Both integrals are limits of integrals which fluctuate and average out to 0.

 

[blue_leaf77]

The sine integral can be "proved" as the limit of finite symmetric integrals.

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?

 

[mathman]

The limit after dividing by T is zero, as you have noticed. What has that got to do with the original question?

 

[blue_leaf77]

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?

 

[pwsnafu]

There is no real function with the properties of the Dirac delta. It is a Schwartz distribution. That integral is not really an integral in the Riemann (or Lebesgue) sense.

 

[blue_leaf77]

There is no real function with the properties of the Dirac delta.

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?

 

[pwsnafu]

Just to make sure we are on the same page, do you know what improper integration is? Do you understand why the first integral is undefined?

Secondly, and this is absolutely critical, the Dirac delta is not defined as $\delta(0)=\infty$, $\delta(x) = 0$ elsewhere. If that is what you were taught you need to forget it. That is a heuristic not a definition. So the right hand side of what you what you wrote is undefined as well.

 

[blue_leaf77]

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?

 

[pwsnafu]

The reason why I bring this up is that
$\int_{-\infty}^{\infty} f(t) \, dt = \lim_{x\to-\infty} \lim_{y \to \infty} \int_{x}^{y} f(t) \, dt = \lim_{y\to\infty} \lim_{x \to -\infty} \int_{x}^{y} f(t) \, dt$.
You need the same answer independent of the order you take the limit.

And equivalent form is
$\int_{-\infty}^\infty f(t) \, dt = \lim_{x\to-\infty} \int_{x}^0 f(t) \,dt + \lim_{y\to\infty}\int_{0}^y f(t) dt$.
This may be more clear. You need to the two integrals to exist independent of each other. Does this happen in your equation?

 

[blue_leaf77]

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?

 

[pwsnafu]

Correct definition: for a continuous function $f:\mathbb{R}\to\mathbb{R}$, the Dirac delta of $f$ at $a$ is the number $f(a)$ and we write $\langle \delta, f \rangle = f(a)$.

A more common notation is $\int_{-\infty}^{\infty} f(x) \delta(x) \, dx = f(a)$, however, this is an abuse of notation. The left hand side is not an integral at all.

The Fourier transform of the Dirac is defined as $\langle \hat\delta, f\rangle = \langle\delta, \hat f \rangle$ which exists provided $f$ has a valid Fourier transform.

 

[blue_leaf77]

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).