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:


$$s(t) = a\cos (\omega_rt-\phi)+b$$ $$g(t) = \cos (\omega_rt)$$

$\omega_r$ 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


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 :)

  • 1
    $\begingroup$ Sparing the details isn't necessarily helpful here, since the assumptions of their integration are likely key. Can you include a link? $\endgroup$ Jul 29, 2014 at 20:37
  • $\begingroup$ Good point. I can't link because it's unpublished material but you're probably right about assumptions. The equation might actually just be some abuse of notation, I see a lot of that in my field. $\endgroup$ Jul 29, 2014 at 20:44
  • $\begingroup$ Understandable. Only thing I can guess off the top of my head is that that's not an integral per se: rather, that they're considering a generic time-average $\langle f(t) \rangle$ and assuming that $f(t)=A\cos(\omega(t+\tau))$ for all signals of interest (i.e. that there's only one dominant frequency.) But without context I can only speculate. $\endgroup$ Jul 29, 2014 at 20:47
  • $\begingroup$ Since you can't link the material directly, one thing you could do is find a published paper that has similar arguments and link that instead. That would at least give us something in the ball-park. $\endgroup$ Jul 29, 2014 at 20:56
  • 1
    $\begingroup$ I'm going through similar work and will hopefully find a useful one to link. $\endgroup$ Jul 29, 2014 at 21:06

2 Answers 2


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.

  • $\begingroup$ It's an electrical engineering paper, so probably the obvious interpretation isn't what they intend. (I speculated a bit above about what it might really mean.) $\endgroup$ Jul 29, 2014 at 20:49
  • $\begingroup$ Maybe they're just saying it's bounded by $T$? And perhaps to make things nontrivial, $\omega$ is not constant? $\endgroup$ Jul 29, 2014 at 20:52
  • $\begingroup$ Thanks for the lesson. As for $\omega$ it is the signal pulsation and always constant. I'll add a bit more context to the problem, thanks for your time $\endgroup$ Jul 29, 2014 at 20:53
  • $\begingroup$ I don't think it's a bounding thing---his statement makes me suspect they're considering time-averaged quantities, e.g. $$\langle f(t)\rangle=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2} f(t)\cos(\omega(t+\tau))\,dt.$$ But it's decidedly a guess. $\endgroup$ Jul 29, 2014 at 20:54
  • $\begingroup$ I added a second section to my initial question giving the context $\endgroup$ Jul 29, 2014 at 21:06

It turns out it's a notation issue in the paper because another domain-specific assumption is that the time over which the signals are correlated (T) is always much larger than the wavelength of the signals.

Therefore it's implied that the limit include the $1/T$ term:

$$\lim_{ T \rightarrow \infty } \frac{1}{T}\int_{T/2}^{-T/2}...$$

The B term in the original question is thus $0$.

As for the $b$, it's assumed to be a different DC offset than the one in the original signal $s(t)$ but they use the same variable.

I'm accepting Robert's answer as the original question is simply ill-posed. Thanks


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .