Average value of a function sent to infinity

[hypermonkey2]

I was wondering if it would be possible to find the average value of a function with the only condition that x is element of R. For example, could we say that f(x)=4 has an average value of 4 since no matter what values we give for a or b in the integral from a to b of f(x)/(b-a) (b is not equal to a of course), we will obtain an answer of 4? However, if this is true, could it be for other functions as well, perhaps more complicated or colourful? What conditions need to be satisfied?

 

[benorin]

Recall that the average value of a function which is Riemann integrable on some interval (a,b) is given by:

$$f_{ave} = \frac{1}{b-a}\int_a^b f(x) dx$$​

I suppose that one could define an average value of a function over all of $$\mathbb{R}$$ by, say

$$f_{ave} = \lim_{L\rightarrow \infty}\frac{1}{2L}\int_{-L}^{L} f(x) dx$$​

so that for f(x)=4 we would have


$$f_{ave} = \lim_{L\rightarrow \infty}\frac{1}{2L}\int_{-L}^{L} f(x) dx = \lim_{L\rightarrow \infty}\frac{1}{2L}\int_{-L}^{L} 4 \, dx = \lim_{L\rightarrow \infty}\frac{1}{2L}\left[ 4x\right] _{x=-L}^{L} = \lim_{L\rightarrow \infty}\frac{1}{2L}\left[ 4L-(-4L)\right] = 4$$

as expected.

 

[matt grime]

But if one defines the average as

$$\frac{1}{3L}\int_{-2L}^Lf(x)dx$$

then one should check that the answers agree, or for any other way of averaging. I imagine there will be some condition on the type of f for which this makes sense.

 

[benorin]

Sure, I took the Cauchy Principal Value of the limit whose upper and lower bounds approached their repective infinities seperately.

 

[Hurkyl]

The bounds in the CPV don't go separately: they go together in a particular way. To go separately, you'd need something like:

$$\lim_{\substack{a \rightarrow -\infty \\ b \rightarrow +\infty}} \frac{1}{b-a}\int_a^b f(x) dx$$


In general, this really does matter. For example:

$$CPV \int_{-\infty}^{+\infty} \sin x \, dx = 0$$

even though

$$\int_{-\infty}^{+\infty} \sin x \, dx$$

does not exist.

 

[benorin]

Indeed, I took the CPV of such an integral.

 

[hypermonkey2]

Tell me a little more of how the Cauchy Principal Value theorem works for sin(x) integrated from -infinity to positive and how it gives zero. When i evaluate it, i find that it can simply range from -2 to 2. (if you please).

 

[matt grime]

please is such a nice word, isn't it?

 

[Hurkyl]

By definition,

$$CPV \int_{-\infty}^{+\infty} f(x) \, dx = \lim_{L \rightarrow +\infty} \int_{-L}^{L} f(x) \, dx$$

 

[hypermonkey2]

Ah, i understand now. Thanks very much!