I'm trying to follow some equations in an electrical engineering paper that I'm reading. I'll spare you the details, but at one point I come across:
$$\lim_{ T \rightarrow \infty }\int_{-T/2}^{T/2} \cos (\omega_r(t+\tau)) dt$$
For the reasoning in the paper to work this integral should equal T.
I can't prove this mathematically, nor find some intuitive reasoning for it. Intuitively I would have said the answer is $0$…
I guess it could also be rewritten as two integrals:
$$\lim_{ T \rightarrow \infty }\int_{-T/2}^{-\tau} \cos (\omega_r(t+\tau)) dt + \lim_{ T \rightarrow \infty }\int_{-\tau}^{T/2} \cos (\omega_r(t+\tau)) dt$$
but it didn't get me anywhere.
I know that $\displaystyle\lim_{ T \rightarrow \infty }\int_{0}^{T} \cos(x) dx$ is undefined as sinusoidal functions never converge, but I would expect the symmetry of $\displaystyle\lim_{ T \rightarrow \infty }\int_{-T}^{T} \cos(x) dx$ to make the integral equal to $0$.
I'd appreciate if anyone could point me in the right direction.
Edit: providing some context
It's one of the terms in a signal correlation. The full problem can be stated as follows:
Let
$$s(t) = a\cos (\omega_rt-\phi)+b$$
$$g(t) = \cos (\omega_rt)$$
$\omega_r$ is constant (pulsation) and $s(t)$ is a phase delayed version of $g(t)$ with a change of amplitude and a DC offset, b.
They define the correlation of the signals as:
$$h(\tau)=(s\otimes g)(t)=\frac{1}{T}\lim_{T\rightarrow \infty }\int_{-T/2}^{T/2}s(t)\cdot g\left (t+\tau\right ) dt$$
And they state that the result of this integral is
$$h(\tau)=\frac{a}{2}\cos(\omega_rt+\phi)+b$$
Without provide further details.
I naively did:
$$h(\tau)= \frac{1}{T}\lim_{T\rightarrow \infty }\int_{-T/2}^{T/2}\left [a\cos (\omega_rt-\phi)+b \right ]\cdot \cos (\omega_r(t+\tau)) dt$$
$$= \underbrace{\frac{a}{T}\lim_{T\rightarrow \infty }\int_{-T/2}^{T/2}\left [\cos (\omega_rt-\phi)\cos (\omega_r(t+\tau)) \right ]dt}_{\text{A}} + \underbrace{\frac{b}{T}\lim_{T\rightarrow \infty }\int_{-T/2}^{T/2} \cos (\omega_r(t+\tau)) dt}_{\text{B}}$$
And trying to figure out the right integral, which must be equal to T if the term is to be equal to $\frac{bT}{T}=b$.
Maybe I'm doing something obviously wrong 🙂
Best Answer
Since $\cos$ is even, $\int_{-T}^{T} \cos(x)\; dx = 2 \int_{0}^{T} \cos(x)\; dx$, not $0$. But surely you know $\int_0^T \cos(x)\; dx = \sin(T)$?
More generally, $$\int_{-T/2}^{T/2} \cos(\omega (t + \tau))\; dt = 2\,{\frac {\sin \left( \omega\,T/2 \right) \cos \left( \omega\,\tau \right) }{\omega}} $$
This has no limit as $T \to \infty$ unless $\cos(\omega \tau)$ happens to be $0$. Either the paper is wrong, or you're misreading it.