More general formula for integrals

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Discussion Overview

The discussion revolves around the search for a more general formula for integrals, particularly focusing on the integral of the form int(x^k). Participants explore special cases, such as the integral of x^(-1) and the implications of using limits and substitutions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether a more general formula exists for the integral than int(x^k) = (x^(k+1))/(k+1), specifically to include cases like int(x^(-1)) = ln|x|.
  • Another participant suggests combining the formulas into a single expression, stating that int(x^k) = (x^(k+1))/(k+1) for k != -1 and int(x^(-1)) = ln(|x|).
  • Several participants propose using limits to define the integral, with one stating that the limit can handle the singularity at k = -1, while another points out that this limit leads to undefined behavior for certain values of x.
  • A participant mentions a correction to a previous equation, indicating that a "-1" was missing in the numerator of a limit expression.
  • One participant presents a general definition of the integral using Riemann sums, noting its generality but limited practical use.

Areas of Agreement / Disagreement

Participants express differing views on the generality and applicability of the proposed formulas and definitions. There is no consensus on a single, universally accepted formula for the integral.

Contextual Notes

Some participants highlight limitations in their approaches, such as the handling of singularities and the conditions under which certain formulas apply. The discussion reflects a range of mathematical reasoning without resolving the complexities involved.

echandler
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I was wondering: Is there an even more general formula for the integral than int(x^k) = (x^(k+1))/(k+1) that accounts for special cases like int(x^(-1)) = ln|x| and possibly u substitutions?
 
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You can combine both in a single formula:
"int(x^k) = (x^(k+1))/(k+1) for k!=-1, int(x^(-1))=ln(|x|)"
Apart from that... no.
 
use limits

$$\int \! x^k \, \mathrm{d}x=\lim_{a \rightarrow k+1} \frac{x^a}{a}+\mathrm{Constant}$$

That is a removable singularity. When we write it in terms of usual functions we appear to be dividing by zero, but we could define a new function without doing so. Other examples include
sin(x)/x
log(1+x)/x
(e^x-1)/x
(sin(tan(x))-tan(sin(x)))/x^7

going the other way we can define the function of two variables
$$\mathrm{f}(x,k)=\int \! x^k \, \mathrm{d}x$$
without any worry about dividing by zero
 
lurflurf said:
use limits

$$\int \! x^k \, \mathrm{d}x=\lim_{a \rightarrow k+1} \frac{x^a}{a}+\mathrm{Constant}$$
For k=-1, that limit is zero for x=0 (which does not fit to the ln), and it is undefined everywhere else. As simple example, consider x=1, where you get the limit of 1/a for a->0.
 
lurflurf said:
use limits

$$\int \! x^k \, \mathrm{d}x=\lim_{a \rightarrow k+1} \frac{x^a-1}{a}+\mathrm{Constant}$$

There was a "-1" missing in the numerator, which I added in the quoted equation above. Note that for ##k \neq -1##, the -1/a term can be absorbed into the integration constant.
 
This is a funny question !
May be, more intuitive if presented on the exponential forme, such as :
 

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Mute said:
There was a "-1" missing in the numerator, which I added in the quoted equation above.
Ah, that makes sense.
 
The general definition of the integral that I use is:

\int_a^b f(x) dx = \lim_{\text{max} \Delta x_k \to 0} \sum_{k=1}^n f(x_k^*) \Delta x_k

Not very useful, but it's definitely general.
 

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