Show $\int_0^\infty f\left(x+\frac{1}{x}\right)\,\frac{\ln x}{x}\,dx=0$ if $f(x)$ is a bounded non-negative function

Solution 1:

Just substitute $u = \frac{1}{x}$ and get

$$\int_\infty^0 f\left(u+\frac{1}{u}\right) \frac{\ln u}{u}\,du.$$

Note the switch of integral limits.

I want to know why must $f(x)$ be a bounded non-negative function?

It need not be, all you need is that the integral exists, be it as a Lebesgue integral or an improper Riemann integral. Since the factor $\frac{\ln x}{x}$ tends to $0$ for $x\to\infty$ almost fast enough to be integrable, a fairly slow decay of $f$ at infinity is sufficient to guarantee the existence of the integral, e.g. if $\lvert f(t)\rvert \leqslant C\cdot(\log t)^{-(2+\delta)}$ for some positive constants $C$ and $\delta$ (and $f$ is measurable), the integral exists as a Lebesgue integral. Also, the integral exists as a Lebesgue integral if $f$ is Lebesgue-integrable. As an improper Riemann integral, it also exists if $f$ oscillates appropriately, without needing a decreasing amplitude, e.g. for $f = \sin$ or $f = \cos$.

You can relax that condition, if you interpret the integral as

$$\lim_{\varepsilon\searrow 0} \int_{\varepsilon}^{1/\varepsilon} f\left(x+\frac{1}{x}\right) \frac{\ln x}{x}\,dx,$$

a principal value integral. Then you only need $f$ to be locally integrable, the substitution shows that these integrals are $0$ for all $\varepsilon > 0$.