I know some integrals can't have undefined integrals, but why? And how, for example, can be proved that the area under the hyperbola $y=\frac{1}{x}$ is $\ln(x)$?
How integrals are computed?
-
0Beyond merely memorizing the derivatives of elementary functions and then adapting, some broad real-analytic methods are $u$-substitution, by-parts integration, and differentiating under the integral (aka Feynman's trick); sometimes many of these are needed multiple times; sometimes they only work with special definite integrals; sometimes other multivariable methods are needed or useful (eg the Gaussian integral); sometimes substitutions are very tedious (Euler substitutions for elliptic integrals). Complex analysis delivers the very useful tools of contour integration and residue formulas... – 2012-06-19
4 Answers
The area under the hyperbola ($y=\dfrac{1}{x}$) between point M and N is $A(x)$ .
The area under the hyperbola ($y=\dfrac{1}{x}$) between point M and L is $A(x+h)$ .
The difference of Area can be defined as $A(x+h)-A(x)$.
If $h\rightarrow0 $ then the difference area will be rectangle thus we can write:
$ A(x+h)-A(x)\approx \frac{1}{x} h$
$ \lim_{h \to 0} \frac{A(x+h)-A(x)}{h}=\frac{1}{x}$
$ \lim_{h \to 0} \frac{A(x+h)-A(x)}{h}=\frac{dA(x)}{dx}$
$\frac{dA(x)}{dx}=\frac{1}{x}$
$\int dA(x)=\int \frac{dx}{x}$
$A(x)=\int \frac{dx}{x}$
$x=e^t=1+\frac{t}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots$
$\frac{dx}{dt}=e^t=1+\frac{t}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots=x$
$\frac{dx}{dt}=x$
$\int \frac{dx}{x}=\int dt$
$\int \frac{dx}{x}=t$
$\ln x=\ln e^t=t \ln e=t$
$\int \frac{dx}{x}=\ln x$
$A(x)=\int \frac{dx}{x}=\ln x$
Every function that is finitely discontinuous has an indefinite integral. The problem is: Some integrals can't be written in the form of elementary functions.
-
0i agree with you lhf. – 2012-06-20
You must use the fundamental theorem of calculus, essentially that differentiation and integration are inverse to each other.
It can be shown that the derivative of ln$(x)$ wrt $x$ is $\frac{1}{x}$ and so ln$(x)$ is a suitable function to be taken for the integral of $\frac{1}{x}$ upto addition of a constant.
-
0Oops :p edited. – 2012-06-19
If the function being integrated has singularities, the integral can only be defined by breaking the integral up into limits approaching the singularity from whichever sides are appropriate. So the reason it is fine for some functions and not fine for others has to do with whether or not the limit of the integral as you approach the singularity exists or not.