- #1
CuriousParrot
- 11
- 0
Suppose one needs to evaluate a definite integral over a singularity, like: [itex]-\int_{-1}^3 \frac{1}{x^2} dx[/itex]
The textbook way to do so is to split the integral into two parts around the singularity and take the limit, like so:
[tex]\lim_{b\rightarrow 0} -\int_{-1}^b \frac{1}{x^2} dx[/tex]
and:
[tex]\lim_{c\rightarrow 0} -\int_{c}^3 \frac{1}{x^2} dx[/tex]
and then add the results of the two expressions above. In this case, the integral is not convergent, and one obtains an infinite result from those evaluations.
So far, so good. But here's a question:
What's wrong with just using the same limit for both integrals? Like this:
[tex]-\int_{-1}^3 \frac{1}{x^2} dx = \lim_{c\rightarrow 0} \Bigr( -\int_{-1}^c \frac{1}{x^2} dx -\int_{c}^3 \frac{1}{x^2} dx ~\Bigr)[/tex]
[tex]= \lim_{c\rightarrow 0}\Bigr(\frac{1}{x}\Bigr|_{-1}^c + \frac{1}{x}\Bigr|_c^3~\Bigr) [/tex]
The 1/c terms cancel each other out, and a finite result remains. In fact, the divergent terms would always cancel out when trying to integrate over a singularity in this way. Clearly, this is a wrong answer, so an invalid step has been performed above. But what part was wrong, and why? Perhaps it is an error to apply the same limit process to both sides of the split integral at the same time. If so, why can't you do that?
The textbook way to do so is to split the integral into two parts around the singularity and take the limit, like so:
[tex]\lim_{b\rightarrow 0} -\int_{-1}^b \frac{1}{x^2} dx[/tex]
and:
[tex]\lim_{c\rightarrow 0} -\int_{c}^3 \frac{1}{x^2} dx[/tex]
and then add the results of the two expressions above. In this case, the integral is not convergent, and one obtains an infinite result from those evaluations.
So far, so good. But here's a question:
What's wrong with just using the same limit for both integrals? Like this:
[tex]-\int_{-1}^3 \frac{1}{x^2} dx = \lim_{c\rightarrow 0} \Bigr( -\int_{-1}^c \frac{1}{x^2} dx -\int_{c}^3 \frac{1}{x^2} dx ~\Bigr)[/tex]
[tex]= \lim_{c\rightarrow 0}\Bigr(\frac{1}{x}\Bigr|_{-1}^c + \frac{1}{x}\Bigr|_c^3~\Bigr) [/tex]
The 1/c terms cancel each other out, and a finite result remains. In fact, the divergent terms would always cancel out when trying to integrate over a singularity in this way. Clearly, this is a wrong answer, so an invalid step has been performed above. But what part was wrong, and why? Perhaps it is an error to apply the same limit process to both sides of the split integral at the same time. If so, why can't you do that?