A conjecture about the rationallity of a definite integral

Click For Summary

Discussion Overview

The discussion revolves around a conjecture regarding the rationality of a definite integral, specifically whether the integral of a function being rational implies that the integral of the product of the function and its variable is also rational. The scope includes mathematical reasoning and exploration of counterexamples.

Discussion Character

  • Exploratory, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant proposes the conjecture that if \(\int_0^1 f(x) dx \in \mathbb{Q}\), then \(\int_0^1 x f(x) dx \in \mathbb{Q}\) under the assumption that \(f(x)\) is integrable.
  • Another participant asserts that the conjecture is false and provides a counterexample using the function \(f(x) = \pi x \sin(\pi x)\), noting that \(\int_0^1 x f(x) dx\) results in an irrational number.
  • Subsequent replies express surprise at the counterexample and inquire about the method used to find it, indicating that the search for counterexamples was not straightforward for them.
  • One participant describes their thought process in deriving the counterexample, detailing their calculations and adjustments to the function to reach an irrational result.

Areas of Agreement / Disagreement

Participants do not reach consensus; there is a clear disagreement regarding the validity of the conjecture, with one participant providing a counterexample that challenges the initial claim.

Contextual Notes

The discussion highlights the complexity of the conjecture and the difficulty in finding counterexamples, indicating that assumptions about the functions involved may affect the outcomes.

Damidami
Messages
93
Reaction score
0
Is it true that

[itex]\int_0^1 f(x) dx \in \mathbb{Q} \Rightarrow \int_0^1 x f(x) dx \in \mathbb{Q}[/itex]

?

(Suppose [itex]f(x)[/itex] integrable as needed)

I thought of this conjecture yesterday and still couldn't prove it, I tried using integration by parts to relate it to the original, but didn't work.

Any ideas? Or counterexamples?
Thanks!
 
Physics news on Phys.org
It is false. Try
[tex]f(x) = \pi x \sin(\pi x),\ xf(x) = \pi x^2 \sin(\pi x)[/tex]
 
Wow, thanks!

I couldn't find a counterexample, it seems it wasn't so trivial to find one (to me)

How did you find it so fast?
 
Damidami said:
Wow, thanks!

I couldn't find a counterexample, it seems it wasn't so trivial to find one (to me)

How did you find it so fast?

I just started with sin(πx) which gave 2/π so I changed it so πsin(πx). That gave 2. Then I figured multiplying by x would get π involved in the answer so I tried xπsin(πx) which gave 1, so I tried x2πsin(πx) which gave an irrational.
 
LCKurtz said:
I just started with sin(πx) which gave 2/π so I changed it so πsin(πx). That gave 2. Then I figured multiplying by x would get π involved in the answer so I tried xπsin(πx) which gave 1, so I tried x2πsin(πx) which gave an irrational.

Brilliant! Thank you.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K