Question

Integrating discontinuous functions

Answer 1

\[\Large f(x) = \frac{x}{2} \text{ if x is rational} \\ \Large f(x)= \frac{x}{3} \text{if x is irrational}\]

Can we integrate this?

Answer 2

@dan815

Answer 3

we have to argue that the integral will be bounded by these 2 areas we know that much

Answer 4

\[\Large \frac{x}{3} \le f(x) \le \frac{x}{2} \implies \frac{x^2}{6} \le \int\limits f(x) dx \le \frac{x^2}{4} \]

Answer 5

its weird though i feel like as a limiting thing

Answer 6

u wont ever actually go to the irrational line

Answer 7

like its not a true irrational number until its approximated to infinite decimals

Answer 8

but until then u are always on the rational line

Answer 9

The issue here is that we have infinitely many discontinuities. If you had finitely many discontinuities, then you can just break up the interval into finite subintervals and calculate the integral of each piece. But we can't do that here.

Answer 10

Another interesting and related function might be \[f(x) = x \text{ when x is rational} \\ f(x) = x^n \text{ when x is irrational, smallest possible n to make rational }\]

Answer 11

it would be an interesting problem, it questions the idea of infinity and irrationals like are we gonna say there are as many infinite numbers as the numbers that lead to infinity

Answer 12

@jim_thompson5910 I agree we can't integrate it as is, but I don't see why we can't say the integral has a lower and upper bound on what it could possibly be, or why not some sort of average of the ratio of rationals to irrationals?