I did Lebesgue integration and $1/2$ appeared unexpectedly.

Question

I am interested in Lebesgue integral over $[0;1]$ of the function

$$f(x) = \sum_{n=1}^\infty n \cdot \chi_{[0;n^{-2}]}(x);$$

Here $\chi_{[0;n^{-2}]}(x)$ is $1$ if $x \in [0;n^{-2}]$ and $0$ otherwise.

So intuitively this should be $\zeta(1)$ (summand multiplied by the length of interval, all added together).

However once I wrote $$ A_i = \left[ \frac{1}{(i+1)^2} ; \frac{1}{i^2} \right ); $$

$$ \int_{[0;1]} f(x) d\mu = \sum_{i=1}^\infty \int_{A_i} f(x) d\mu = \sum_{i=1}^\infty \sum_{j=1}^i j \cdot \mu(A_i);$$

From here not everything goes as expected

$$ \int_{[0;1]} f(x) d\mu = \sum_{i=1}^\infty \frac{i(i+1)}{2} \cdot \left( \frac{1}{i^2} - \frac{1}{(i+1)^2} \right) = \frac{1}{2} \sum_{i=1}^\infty \left( \frac{i+1}{i} - \frac{i}{i+1} \right) = \frac{1}{2} \sum_{i=1}^\infty \left( \frac{1}{i} + \frac{1}{i+1} \right) = \zeta(1) - \frac{1}{2};$$

What is the problem?

Well, no. But $\zeta(1)$ was mentioned without any further comment. Not to mention what $f(0)$ could possibly be. — Lord_Farin, May 01 '13 at 10:00
@Lord_Farin It seems kind of silly, that he would define that wacky function to begin with, without attempting to sum the series in a non standard way. — Ethan Splaver, May 01 '13 at 10:01
I am not sure if this sentiment is shared by anybody, but it reminds me of "joke proofs" of $0=1$. Given that it's the beginning of May, I am wondering if this problem was not by any chance posed to the OP on April Fool's Day. — Jakub Konieczny, May 01 '13 at 10:18

Jakub Konieczny · Accepted Answer · 2013-05-01T10:00:55.433

2

The problem is that $\zeta(1) = \infty$. As far as I am able to check the computations, most of them are correct, except for the fact that some of the sums are in fact divergent. Also, just before the end, I think you mean something like: $$ \int f = \dots = \frac{1}{2} \sum_{i=1}^\infty (\frac{i+1}{i} - \frac{i}{i+1}) = \frac{1}{2} \sum_{i=1}^\infty (\frac{1}{i} + \frac{1}{i+1}) \\= \frac{1}{2} \sum_{i=1}^\infty + \frac{1}{2} \left( \sum_{i=1}^\infty \frac{1}{i} - 1\right) = \zeta(1) - \frac{1}{2} $$

The result is that $\zeta(1) - \frac{1}{2} = \zeta(1)$, which is true if you allow for infinities in your equations. It basically says that $\infty - \frac{1}{2} = \infty$. Note that we could legally sum some divergent sequences, because they converge to $+ \infty$. We definitely cannot subtract $\zeta(1)$ from both sides, to get $\frac{1}{2} = 1$.

Disclaimer: I am assuming $\zeta$ to be defined as $\zeta(t) = \sum_{n=1}^\infty \frac{1}{n^t}$. This is defined, and equal to $+\infty$ at $t= 1$. I know some complex-analysts would disagree.

edited May 01 '13 at 10:00

answered May 01 '13 at 09:50

Jakub Konieczny

12,588

If you are referring to the analytic continuation of the zeta function, it also diverges from its left sided limit. – Ethan Splaver May 01 '13 at 10:06
1

@Ethan: to me it looks as a standard accident when playing with calculus and not being careful enough. Sort of a situation when you do enough manipulations and end up with something distinct from what you started with because of divergence along the way. – Jakub Konieczny May 01 '13 at 10:06
@Ethan: I think you deleted a comment I am referring to in my previous comment. As for the analytic continuation, what I meant is that if we stick to reals, $\zeta(1) = + \infty$ as a limit, but if we extend to the complex numbers, then probably the limit no longer exists. – Jakub Konieczny May 01 '13 at 10:09
@Ethan: This is only true if you have a lot of negative terms as well! If all terms are positive, you can sum in any order you like, and the result is always the same (though sometimes infinite). – Jakub Konieczny May 01 '13 at 10:11

I did Lebesgue integration and $1/2$ appeared unexpectedly.

1 Answers1