How can calculate this integral $$\lim_{N \to \infty} \int_{x=0}^{\infty}f(x) \frac{x f'\left(\frac{x}{1-1/N}\right)}{f\left(\frac{x}{1-1/N}\right)} dx$$ where $f(x)$ is a probability density function?
If there exist a function $g(x)$ such that $$\Bigg|f(x) \frac{x f'\left(\frac{x}{1-1/N}\right)}{f\left(\frac{x}{1-1/N}\right)}\Bigg|< g(x)$$ for all $N$, then I can use the Dominated Convergence Theorem. But can I calculate the integral without such an assumption? I am thinking maybe the probability function $f(x)$ has some characteristics that could help. Any idea?
Not really a mathematically rigorous answer, but too long for a comment. Change variables $x/(1-1/N)=\tau$. We get $$ \lim_{N \to \infty} \int_{x=0}^{\infty}f(x) \frac{x f'\left(\frac{x}{1-1/N}\right)}{f\left(\frac{x}{1-1/N}\right)} dx=\lim_{N \to \infty} \left[(1-1/N)^2\int_{x=0}^{\infty}f((1-1/N)\tau) \frac{\tau f'\left(\tau\right)}{f\left(\tau\right)} d\tau\right]\ . $$ Then morally the function $f((1-1/N)\tau)$ can be Taylor-expanded for $N\to\infty$ as $$ f((1-1/N)\tau)\approx f(\tau)-\frac{1}{N}f'(\tau)\tau+\frac{1}{2N^2}f''(\tau)\tau^2-\ldots\ , $$ therefore the sought limit should be $$ \int_0^\infty d\tau\ \tau f'(\tau)=-\int_0^\infty d\tau\ f(\tau)=-1\ , $$ where one uses integration by parts, and assumes that $f(\tau)\to 0$ for $\tau\to\infty$ and $\int_0^\infty d\tau f(\tau)=1$ by normalization of the pdf. This result can be easily checked explicitly in a couple of cases, like $f(x)=\exp(-x)$ or $f(x)=\frac{2/\pi}{1+x^2}$.