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May 23rd, 2017, 10:33 AM   #1
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Smile not definitely integrable function example plz

As you know, some continuous functions doesn’t have definite integral.
But in spite of the fact that the function does not definitely integral, if the result of the area below the curvature has significance, we alternate it with approximation as through midpoint rule, trapezoidal rule, or Simson’s rule.

I wanna collect such functions whose integral-approximations are used as a replacement of definite integral in the field.
(so it implies such function is neither Riemann integrable nor Lebesque integrable)

I have googled it for many times, however, failed to find any usage. What I got is only about lists of approximation methods listed above.
So I decided to ask your aid.
Don’t need to write down all function, just let me know name of function or where I can find.
Thank you.

Last edited by akuraangran; May 23rd, 2017 at 10:57 AM.
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May 23rd, 2017, 04:25 PM   #2
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If a function is continuous and non-integrable, at least one of the integral's limits is infinite. The function could be a non-zero constant, for example.
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May 24th, 2017, 03:18 AM   #3
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Thanks for your response, but I am really sorry to say that is a little bit different from what I want to know. Non-zero constant function can yield the Riemann-integral (definite integral), especially in the R^n space with standard topology. Given that integral is defined in closed sub-set and non-zero constant is continuous in R^n, it need not to be approximated.

The function that I want to find might be somewhat like exp(x)/x, ln(x^n1 +c1)/(x^n2 + c2).

Although continuous function, in closed set of R, its definite integral does not exist. So, alternative estimate (as listed 3 methods) is used on its behalf.

Thanks for your time.

Last edited by skipjack; May 24th, 2017 at 08:21 AM.
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May 24th, 2017, 04:44 AM   #4
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Quote:
Originally Posted by akuraangran View Post
Thanks for your response, but I am really sorry to say that is a little bit different from what I want to know. Non-zero constant function can yield the Riemann-integral (definite integral), especially in the R^n space with standard topology.
No, it cannot. The integral of a constant, a, over set X is a times the measure of that set. The set R (or R^n) does not have finite measure.

Quote:
Originally Posted by akuraangran View Post
Given that integral is defined in closed sub-set and non-zero constant is continuous in R^n, it need not to be approximated.

The function that I want to find might be somewhat like exp(x)/x, ln(x^n1 +c1)/(x^n2 + c2).

Although continuous function, in closed set of R, its definite integral does not exist. So, alternative estimate (as listed 3 methods) is used on its behalf.

Thanks for your time.

Last edited by skipjack; May 24th, 2017 at 08:25 AM.
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May 24th, 2017, 06:46 AM   #5
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Quote:
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No, it cannot. The integral of a constant, a, over set X is a times the measure of that set. The set R (or R^n) does not have finite measure.
I concede that there is something that may cause equivocalness. What I mean is integral of a constant and domain of the function is the subset of R (or R^n) like [1 , 10] not all R. I operate the summation in just a bounded closed subset.
Then simply, if f = 3, then integration is done with a range 1 to 10, then
its definite integral is 3x +c and Riemann value is 27.

But the functions I mentioned above as exp(x)/x and ln(x^n1 +c1)/(x^n2 + c2), are obviously continuous in [1,10] respect to x. However, they don't have definite-integral forms well matched to them.
What I search for is some of those kind of functions and the value of integral is quite useful and may need more accurate approximation for it, if possible.

Thanks for your comment.

Last edited by skipjack; May 24th, 2017 at 08:28 AM.
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May 24th, 2017, 08:47 AM   #6
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They may not have "definite integral forms" in terms of very familiar functions, such as sin(x), but why not allow the use of other functions, such as the gamma function?
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