A conjecture about the rationallity of a definite integral

[Damidami]

Is it true that

\int_0^1 f(x) dx \in \mathbb{Q} \Rightarrow \int_0^1 x f(x) dx \in \mathbb{Q}

?

(Suppose f(x) 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!

 

[LCKurtz]

It is false. Try
f(x) = \pi x \sin(\pi x),\ xf(x) = \pi x^2 \sin(\pi x)

 

[Damidami]

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?

 

[LCKurtz]

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.

 

[Damidami]

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.