# How might one prove the following is either possible or impossible? [closed]

$$\int_0^{\infty} g(u)\,du = 0$$ $$\int_0^{\infty} ug(u)\,du = 0$$ where $$g(u)$$ is continuous over the whole interval and $$g(u)$$ $$\neq 0$$ for at least some values of $$u$$ in the interval.

Does this hold if $$g(u)$$ diverges as $$u \to 0$$ but the integral still converges?

• @Surb If it equals zero, it in particular converges Commented Jun 2, 2023 at 14:35
• Commented Jun 3, 2023 at 6:41

Another option is to use probability distributions. Take $$g(u) = f_X(u) - f_Y(u)$$ where $$f_X \neq f_Y$$ have positive support, then
\begin{align} \int_0^\infty g(u) \ du & = \int_0^\infty f_X(u) - f_Y(u) \ du = 0 \\ \int_0^\infty ug(u) \ du & = \int_0^\infty uf_X(u) - uf_Y(u) \ du = E(X) - E(Y). \end{align} Just pick distributions with the same mean.
This can be taken from any orthogonal polynomial systems with a proper weight, for example, you may check that $$g(u)=\left((\pi-2)u^2-\sqrt\pi u-\frac{\pi-4}{2}\right)e^{-u^2}$$ works