Any other idea to solve this limit?

Question

I'm not that good at integrating, sometimes the algebra comes after me.

That said, I'd like to solve this problem from the analysis point of view.

Knowing that $|\alpha| \neq |\beta|$, find $$\lim_{t \to \infty} \frac{1}{t} \int_{0}^{t} \cos (\alpha t) \sin (\beta t) dt. $$

I know the limit is zero coz I did integrate and simplify as usual, but I'd like to use something like the fact that if $\lim_{x \to \infty} u(x)=0$, and $v$ is bounded, then $\lim_{x \to \infty} u(x) v(x) =0$, however in this case writing $v(x)= \int_{0}^{x} \cos (\alpha t) \sin (\beta t) dt $ is not bounded, in fact $|v(x)|\leq C t$ so even using the squeeze theorem the limit of $v(x)u(x)$ would be 1. (Why is that?).

I'd gladly hear some techniques in analysis to tackle this.

Thanks.

Answer 1

I am not comfortable with your unique $t$ notation. It is preferable to write :

$$\lim_{T \to \infty} \frac{1}{T} \int_{0}^{T} \cos (\alpha t) \sin (\beta t) dt.$$

Solution : write the integrand under the form :

$$\tfrac12(\sin((\beta+\alpha)t)+\sin((\beta-\alpha)t))$$

and decompose your initial integral into 2 integrals.

Then use the fact that

$$\int_0^{T}\sin(\gamma t)dt=-\dfrac{1}{\gamma}(\cos(\gamma T)-1)$$

It is now easy to conclude...

Answer 2

If you don't want to compute explicitly your integral, at some point you have to use that your integrand is periodic and that it's integral is $0$ on a period $T$. Thus, the integral on $[0,t]$ with $t>T$ is reduced to an integral on an interval contained in $[0,T]$ and $$ \frac{1}{t} \int_0^t \cos(\alpha s)\sin(\beta s)\,\mathrm{d} s \leq \frac{1}{t} \int_0^T |\cos(\alpha s)\sin(\beta s)|\,\mathrm{d} s \leq \frac{T}{t}$$ which converges to $0$ when $t\to\infty$.