Studying Mathematics Thread

[polymath257]

Ok, now I'm confused about the following:

The improper integral of 1/x from -e to e is said to diverge yet can be assigned a CPV value of 0.

Intuitively, the 0 answer makes sense (since symmetrically opposite areas, even infinite, should completely cancel each other out), but it's considered to be a problematic answer. Can anyone ELI5 why 0 is not always the right answer here? What does definite integral really mean then?

After some reading, I think I have my answer:

Intuitively speaking, there is nothing wrong with the thinking that, even if infinite, symmetrical and opposite arithmetic areas cancel each other out. And this accords well with the CPV. However, mathematically speaking, the way definite integrals are defined, symmetry isn't assumed by the definition. So when dealing with improper integrals (like the one above), one can arbitrarily choose whatever values they want close to 0 (one to the left of 0 and the other to its right), and so the values need not be opposites of each other. Since that's the case, when calculating the limit, it could be anything depending on the arbitrary values chosen as the bounds closest to 0 (e.g., 0 in the case of a and -a, ln(2) in the case of a and -2a, ln(3) in the case of a and -3a).

A roughly similar situation happens with integrals from -infinity to infinity.

So, what is the integral from -infinity to infinity of x dx?

If you do it symmetrically, from -b to b, and take the limit, the answer is 0.

If you do it from -b to b+1/b, and take the limit as b goes to infinity, you will get 2.

If you do it from -b to b+1, you will get infinity.

Play around with the function 2x/(x^2 +1). This function goes to 0 at infinity, but the doubly infinite integral is still undefined since the positive and negative are both infinite.

The symmetric integral is 0.

The integral going from -b to b^2 gives 2.

The integral from -b to b^3 gives 3.

Such is the issue with divergent integrals. They are like conditionally convergent series: you can get any value out of them if you play enough.