# Improper integral done two different ways

1. Feb 11, 2016

I think it was Gauss who calculated a limit in two different ways, getting -1/2 one way and infinity the other. As he didn't see the error, he wrote sarcastically, "-1/2 = infinity. Great is the glory of God" (In Latin). Anyway, it appears that Wolfram Alpha could do the same thing, as I asked it to calculate the integral of x/(x2-1) dx from x=0 to 2, which it said diverged... presumably having found the limit of the integral from 0 to 1, then from 1 to 2, and concluding that it diverged as soon as an infinity appeared. However, asking it (Wolfram α) to calculate the integral of 1/2*integral of ln |u| du from u = -1 to 3, it comes up with (ln(27)-4)/2, i.e., a finite result, presumably having subtracted the two identical limits before evaluating them. (I don't have "Pro" so I can only guess what path it took.) The two results should be equal, (integration by substitution). I am inclined to accept the finite result, but is there something I am missing here? Thanks in advance.

2. Feb 11, 2016

### HallsofIvy

Gauss got two different results for a limit by (intentionally) applying two different rules for convergent sequences to a series he knew was not convergent. Here, you are taking what is often called the "Cauchy principle value" of a divergent integral by applying a technique that, properly, only applies to convergent integrals.

In general, if an integrand, f(x), has a singularity a x= b, the integral from a to c, with a< b< c, is given by $\lim_{\alpha\to 0} \int_a^{b- \alpha} f(x) dx+ \lim_{\beta\to 0} \int_{b+\beta}^c f(x)dx$. Specifically, we do not take the two limits at b, from above and below, at the "same rate".

The "Cauchy principle value" is given by $\lim_{\epsilon\to 0} \int_a^{b- \epsilon} f(x) dx+ \lim_{\epsilon\to 0} \int_{b+\epsilon}^c f(x)dx$ where we do take the same limit in both integrals. It can be shown that if the first integral converges then so does the second and the give the same value. However, the second, the "Cauchy principal value" may converge when the first does not. But only the first is the true "improper integral".

3. Feb 11, 2016

Thanks, HallsofIvy. Your response was illuminating, but I still have certain unresolved issues here.

I understand that the rate of approach to a number may be different from above and below, hence the limits cannot be added. But in this case, the integral of x/(x2-1) from 0 to 2 can be broken up into

the integral of x/(x2-1) from 0 to 1 + the integral of x/(x2-1) from 1 to 2;

because the integral from a to b = the negative of the integral from b to a, I can state this as

the integral of x/(x2-1) from 0 to 1 - the integral of x/(x2-1) from 2 to 1;

This becomes

[limit ln |x2 -1| from x=0 to 1 - limit ln |x2 -1| from x=2 to 1]/2

Because of the absolute value, we could state this as

[lim ln |u| from x = 1 to 0) - (lim ln |u| from x = 3 to 0)]/2

That is, both limits are approaching zero from the same direction. Therefore the possibility that they approach at the same rate has more credence; if it were so, then I could split the above into

[lim ln |u| from x = 1 to 0) - (lim ln |u| from x = 1 to 0)]/2 - (ln |u| from x = 3 to 1)/2 = -ln(3)/2

(which would agree with Wolfram’s finite assessment).

However, you state that the series diverges, which means that there is something invalid in the above. I would be grateful if you could tell me why you know it is divergent, and which step of the above is flawed.

4. Feb 12, 2016

### pwsnafu

This makes no sense. Limits are not interested in "approach the same rate" (unless you are talking about ratios of limits). The point of limits is that they generate the same answer irrespective of "rate", hence the default is that they approach at different rates. There are times when we use principal value integrals, but this extremely rare.

Honestly, you seem to be thinking more in terms of order notation than limit theory.

5. Feb 12, 2016