# Improper integral

1. Sep 20, 2005

### aceetobee

Can someone explain why the following improper integral diverges?

Integral 1/x dx from -1 to 1

I know if you break it up the individual integrals (from -1 to 0 and 0 to 1) diverge to negative infinity and infinity, whose sum is indeterminant in general, but the symmetry of the integral suggests it "should be zero".

Thanks!

2. Sep 21, 2005

### aceetobee

OK... well, a little research and I think I answered my own question.

It really is an ill-definied integral, because when broken up into the limit of two separate integrals, there are an infinite number of ways this can be done, with one side approaching zero at a different speed than the other.

I guess the Cauchy Principle Value would be zero, but there are other possiblilities, so it diverges.

Please correct me if I'm wrong on this...

3. Sep 21, 2005

### HallsofIvy

If $\int_b^c f(x)dx$ is "improper" because f(a) is not defined, with b< a< c, then the integral is DEFINED as:
$$lim_{x_1->a^-}\int_b^{x_1}f(x)dx+ lim_{x_2->a^+}\int_{x_2}^cf(x)dx$$.

Since the anti-derivative of 1/x is ln|x|, neither of those limits exists when a= 0.

The Cauchy Principal Value, on the other hand is:
$$lim_{x_1->a}\left(\int_b^{x_1}f(x)dx+ \int_{x_1}^cf(x)dx\right)$$.
Since the limit is taken after both integrals are done, we can cancel the "ln|x1|" terms before the limit and just have ln|c|-ln|b|.

Last edited by a moderator: Sep 21, 2005
4. Sep 21, 2005

### Galileo

That's totally correct.