Partition a divergent integral into finite values

[LikeMath]

Hi there,

I am reading an article, but I faced the following problem, and I am wondering if it is well known fact.

If the integral of a function on some interval is infinity, can we partition this interval into countable disjoint (in their interiors) subintervals such that the integral on each interval is 1/2 for example?

Thanks in advance
Likemath

 

[mathman]

Hi there,

I am reading an article, but I faced the following problem, and I am wondering if it is well known fact.

If the integral of a function on some interval is infinity, can we partition this interval into countable disjoint (in their interiors) subintervals such that the integral on each interval is 1/2 for example?

Thanks in advance
Likemath

It looks like it is doable. You could use any positive finite number. (I assume you mean the integral over each piece = 1/2, so that the total is the sum of an infinite number of 1/2's).

 

[LikeMath]

Thank you, but how can I convince my self that it is doable?

 

[mfb]

Counterexample:
$f:[0,3] \to R$
$f(x)=\frac{1}{1-x}$ for $0\leq x<1$
$f(x)=-1$ for $1\leq x \leq 2$
$f(x)=\frac{1}{x-2}$ for $2 < x \leq 3$

The ugly part is the middle section: You cannot combine it with any interval of the other two sections, as this would diverge.

It is true for a large class of functions, however: if $\int_a^x f(x') dx'$ is well-defined for every x in your interval (a,b), it is continuous and you can find appropriate intervals with the intermediate value theorem. This directly gives a way to count them, too, of course.