A conjecture about the rationallity of a definite integral

In summary, the conjecture that if the integral of a function f(x) from 0 to 1 is rational, then the integral of xf(x) from 0 to 1 is also rational is false. A counterexample is f(x) = πxsin(πx) which gives an irrational value for the integral of xf(x). This was found through a systematic approach of trying different functions with sin(πx) and multiplying by x to involve π in the answer.
  • #1
Damidami
94
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!
 
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  • #2
It is false. Try
[tex]f(x) = \pi x \sin(\pi x),\ xf(x) = \pi x^2 \sin(\pi x)[/tex]
 
  • #3
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?
 
  • #4
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.
 
  • #5
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.
 

1. What is a conjecture?

A conjecture is a statement or proposition that is believed to be true, but has not yet been proven or verified.

2. How does this conjecture relate to definite integrals?

This conjecture specifically deals with the rationality of the value obtained from a definite integral. It suggests that the result of certain definite integrals may always be rational numbers.

3. What evidence or research supports this conjecture?

There have been various studies and experiments conducted on this conjecture, but there is no concrete evidence or proof to support it. It remains a hypothesis that has not yet been verified.

4. Are there any counterexamples to this conjecture?

Yes, there are several known counterexamples to this conjecture. These are definite integrals whose results are irrational numbers, disproving the idea that all definite integrals have rational values.

5. What is the significance of this conjecture?

If this conjecture were to be proven true, it would have significant implications in mathematics and science. It could potentially lead to new insights and applications in fields such as calculus, number theory, and physics.

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