✔ 最佳答案
What is your definition for 積分?
is it 定積分? or 不定積分?
If it is the latter case,
it should be the set of function {ln|x| + C : C in R}
then by definition (in AL pure)
the indefinite integral of a integrable function f(x)
is the set { g(x) : dg/dx = f}
This is by definition, if you are not familiar with this, try to review the definition first, this is not by whatsoever inverse concept.
However, if you want to know why D(lnx) = 1/x
you must ask yourself first, what is the definition of ln.
2009-03-13 14:37:28 補充：
In AL Pure, it is usually defined as "the inverse function of exp(x),
where exp(x) defined as lim{n->inf} (1+x/n)^n."
2009-03-13 14:37:50 補充：
So, if you take this definition, you must use inverse function.
At first, you have to verify whether the definition of ln(x) make it a well-defined function. (But this is already out-syl)
2009-03-13 14:37:53 補充：
To check this, you should show that the function exp(x) is injective and continuous, so inverse function exists on its image.
(This is relatively easy)
2009-03-13 14:38:11 補充：
This just give the well-definess of ln(x), but don't have any direct information on its mapping.
We can directly find exp(1) through the computation of limit,
so we can indirectly find ln(e) by definition, but there is no direct way to find ln(2.5).
2009-03-13 14:38:25 補充：
The only indirect information known is that whenever x is positive
ln(exp(x)) = exp(ln(x)) = x
On the other hand, it is not known whether ln(x) is continuous
and differentiable.
2009-03-13 14:38:36 補充：
To do this, you should show that it is continuous first,
(otherwise the computation of limit must be done very carefully)
but how?
You don't have its direct mapping.
2009-03-13 14:38:42 補充：
The only way is to use its inverse function property, which is continuous and injective to show. (This is in mathematical analysis)
At these steps, you cannot avoid the use of inverse concept
2009-03-13 14:38:52 補充：
In AL Pure, it is too difficult to avoid the use of inverse function to discuss logarithm.
But, if you know formal power series (not the one learnt in Applied math), you can define ln(x) by a power series.
2009-03-13 14:38:58 補充：
Then it is trivial to show the derivative of ln(x) is 1/x
{But this step, we change the definition of ln, so you cannot say that ln is the inverse of exp, you must prove this instead}
2009-03-14 03:38:55 補充：
Ok,
first you need to know that (In AL Pure math, this is the definition)
the indefinite integral of a function f is defined as the set of functions
2009-03-14 03:39:36 補充：
The symbol \int f(x) dx is in fact the set { g(x) : dg(x)/dx = f(x)} (family of functions)
not just simply a one function with some formula
if you don't agree with this, your concept is not that clear
2009-03-14 03:40:39 補充：
Now, I claim that the set \int 1/x dx (denoted by A) is equal to { ln|x| + C : C in R}=B , i.e. { g(x) : dg(x)/dx = lnx } = { ln|x| + C : C in R}
LHS is the indefinite integral of 1/x
(it is not a function)
2009-03-14 03:44:02 補充：
clearly, if you differentiate every function of the form g(x) = ln|x| + C
if x>0 then it is g(x) = lnx + C , so g'(x) = 1/x
if x<0 then it is g(x) = ln(-x) + C, so g'(x) = (-1)/(-x) = 1/x
so, g(x) is in the set A
2009-03-14 03:48:05 補充：
oops, I type the set in LHS with typo,
it should be { g(x) : dg(x)/dx = 1/x }
Now I show that
if g(x) and h(x) are any two function in LHS,
consider P(x)=g(x)-h(x) , P'(x) = 1/x - 1/x = 0
so we have P(x) = C , i.e. g(x) and h(x) are differ by a constant
2009-03-14 03:50:15 補充：
but ln|x| is in LHS, so every function in LHS are differ with ln|x| by a constant,
i.e. the set in LHS is equal to RHS
If you need to consider indefinite integral, you are considering a family of functions whose derivative is the given function.
So, you need to show that both sets are equal
2009-03-14 03:51:16 補充：
For definite integral, you are considering a real number, which is the algebraic sum of area,
in neither case you are considering one function,
