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Over an interval $[a, b] $ we can define the mean value of a function $f$ as $$\overline{f} = \frac{\int_{a} ^{b} f(x) dx} {b-a}. $$ Is there a formula for the mean value of a function over $R$?

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  • $\begingroup$ Take the limit. ($b \to +\infty, a \to -\infty$) $\endgroup$ – user202729 Jan 16 '18 at 15:10
  • $\begingroup$ It is useful sometimes to consider the integral from -R to +R and then put the limit as $R\to\infty$. Note that this doesn't always exist, not even for continuous functions, and moreover if you change one of the extremes and compute the limit in an asymmetric way, this might change. $\endgroup$ – Tommaso Seneci Jan 16 '18 at 15:15
  • $\begingroup$ For example for the function $e^{-a|t|} $ the outcome with this formula is zero. It doesn't make any sense. $\endgroup$ – user440413 Jan 16 '18 at 15:18
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One way of extending it would be to take the limit symmetrically: $$\lim_{c \to +\infty}\frac{1}{2c}\int_{-c}^c f(x)\,\mbox{d}x$$ Of course, functions with a finite integral on $\mathbb{R}$ will then have a mean of $0$. Constant functions have a mean equal to their function value.

For example for the function $e^{-a|t|}$ the outcome with this formula is zero. It doesn't make any sense.

The mean of $e^{-|kt|}$ on $[a,b]$ becomes arbitrarily small if you take $b-a$ sufficiently large. Why would any finite but strictly positive number make (more) sense to you, in this case?

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  • $\begingroup$ If the result was 1 that would be more logical. Differently put is there a formula that can give symmetrical functions the average value of f(0)? $\endgroup$ – user440413 Jan 16 '18 at 15:29
  • $\begingroup$ I don't understand what you mean by "the average value of f(0)"; f(0) is just a number...? $\endgroup$ – StackTD Jan 16 '18 at 15:31
  • $\begingroup$ The average value of function f= f(0). $\endgroup$ – user440413 Jan 16 '18 at 15:35
  • $\begingroup$ I see you edited your comment and changed 0 to 1 as what you would find 'more logical'. But why 1? On a sufficiently large interval, the mean would become a lot smaller than 1... So why would it be logical to be 1 on $\mathbb{R}$? $\endgroup$ – StackTD Jan 16 '18 at 15:37
  • $\begingroup$ I edited because I made a mistake I meant 1. I know that the answer should be 1, the professor told so. It is more logical imo because it's a value that the function takes (f(0)) while 0 is not even a value of the function. Function is positive on R. Anyway I don't really care what is more logical but how to force the function to have this value as average :) $\endgroup$ – user440413 Jan 16 '18 at 15:46

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