# Why exponential terms like $\exp (i\omega t)$ make no contribution when averaged over a long time?

In the physics of waves, I often see expressions like $$A\exp(i\omega t) + f(t)$$ where $$A$$ is a constant, $$w$$ is the angular frequency and $$f(t)$$ is an arbitrary function that depends on time.

It also often said that when such expressions are averaged over a long time, terms like $$A\exp(i\omega t)$$ makes 'no contribution'. Why is that so?

The time average of a quantity between $$t=0$$ and $$t=T$$ is

$$\overline f = \frac{1}{T}\int_0^T f(t) dt$$

if $$f(t) = Ae^{i\omega t}$$, then $$\overline f = \frac{A}{T}\int_0^T e^{i\omega t} dt = \frac{A(e^{i\omega T}-1)}{i\omega T}$$ which goes to zero as $$T\rightarrow \infty$$. It's not that $$f$$ doesn't contribute at all, but if you integrate it over a long time, its average contribution becomes very small.

• Thanks for this great answer. – TaeNyFan Jun 21 at 5:41

$$e^{i\omega t}$$ makes a cycle. Average it over the whole cycle and it averages out. Average it over many cycles and it averages out except the fraction of the last cycle. So it mostly averages out.

If it isn't $$e^{i\omega t}$$ but something more complicated like $$|\sin(i\omega t)e^{i\omega t}|$$ then it might not average out. You have to check.

The reason is simply this

$$A\int_{0}^{\frac{2\pi}{\omega}} e^{i \omega t}=A\frac{1}{\omega}e^{i \omega t} \bigg \vert_0^{\frac{2\pi}{\omega}}=A\frac{1}{\omega}(e^{2 \pi i}-1)=A\frac{1}{\omega}(1-1)=0.$$

If one picks a longer time interval, it can be split into many smaller intervals that contribute nothing.

• What about for a long time interval like $10.5\frac{2\pi}{w}$? For the last the $0.5\frac{2\pi}{w}$ time interval, it will not be zero. – TaeNyFan Jun 19 at 11:29
• @TaeNyFan I guess I'm too late with answering this comment, since J.Murray 's answer explains this. Even if the interval is not whole, when averaging you divide by $T$ but the magnitude of the integral of the phase is always bounded. – Stratiev Jun 19 at 12:19

$$e^{i\omega t} = \cos(\omega t) + i\sin(\omega t)$$. Now when $$t=T i$$ very large both $$\cos$$ and $$\sin$$ go over $$-1$$ to $$+1$$. Integral is the sum of these values. It can be $$-1$$ or $$+1$$. The average over a long time will be $$-1$$ or $$1/T$$ which becomes very small. Therefore this does not contribute.

• Have a look at MathJax in order to improve mathematical expressions. Additionally try improving the composition of the text so that your answer is comprehensible to more people – ohneVal Jun 19 at 13:21