More general formula for integrals

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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