Average value of alternating current for a long period

Question

I came across the following in my physics text-book while reading alternating currents:

Average value of current is given by: $$I_{av}=\frac{\int_{t_1}^{t_2}{I(t)dt}}{t_2-t_1}$$ Over a long period of time, the denominator tends to a large value, and the numerator a finite one.

$\therefore$ The average value of the alternating current is zero for a long period of time.

Is this valid? How can we say that the area under the graph of a function is finite over a long period? Usually alternating current is a sine function. Since it alternates between positive and negative areas, doesn't it become indeterminate for indefinite time periods?

Answer 1

No, because the "accumulated area" might oscillate but it will constantly be suppressed by the 1/time term. You see this if you just do the integration. Setting $t_1=0$:

$ \frac{1}{t_2}\int_{0}^{t2} \sin(t)dt = \frac{1}{t_2}(1 - \cos(t_2)) \leq \frac{2}{t_2} $

where the last inequality if because cosine is bounded by -1 and 1. Clearly, if we take the limit as $t_2 \to \infty$ this quantity goes to $0$.

But if you offset your sine from $0$, then you will get a non-zero answer:

$ \frac{1}{t_2}\int_{0}^{t2} (\sin(t) + a)dt = \frac{1}{t_2}(1 - \cos(t_2)) + a \to a $ as $t_2\to \infty$

Answer 2

We can, for the sake of demonstration, assume the alternating current to be of the form

$$I=I_0\sin(\omega t)$$

Now, $\int_0^{\infty} I\:\mathrm dt$ is divergent, but we can assing a value of $1/\omega$ to it, by employing Cesaro summation. Thus we get

$$\lim_{T\to \infty}\frac{\int_0^T I\:\mathrm dt}{T}=\lim_{T\to\infty}\frac 1 {\omega T}=0$$