# Average value of a function sent to infinity

1. Mar 17, 2006

### 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?

2. Mar 18, 2006

### 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.

3. Mar 18, 2006

### 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.

4. Mar 18, 2006

### benorin

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

5. Mar 18, 2006

### Hurkyl

Staff Emeritus
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.

Last edited: Mar 18, 2006
6. Mar 18, 2006

### benorin

Indeed, I took the CPV of such an integral.

7. Mar 18, 2006

### 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).

Last edited: Mar 18, 2006
8. Mar 18, 2006

### matt grime

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

9. Mar 18, 2006

### Hurkyl

Staff Emeritus
By definition,

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

10. Mar 18, 2006

### hypermonkey2

Ah, i understand now. Thanks very much!

Last edited: Mar 18, 2006