Im not sure about a limit of an integral. I would like to prove that there is a solution for this integral for d, and this solution is unique. The integral is:
$$\beta = \int_d^{\infty}(x-d)f(x) dx$$
We know that $0<\beta<\mathbb{E}[X]$, and f(x) is the density function of X. X is a nonnegative random variable, $\mathbb{E}[X] < \infty$. I have tried the following: Let be $g(d) = \int_d^{\infty}(x-d)f(x) dx$.
First lets see $g(0) = ?$. Its easy, i think $g(0) = E[X]$. Next I have showed that $g(d)$ is a monotone decreasing and continous function. Its differentiable, so its continous, and its easy to prove, that its monotone decreasing.
So the last thing to show is $$\lim_{d \rightarrow \infty} \int_d^{\infty}(x-d)f(x) dx = -\infty $$ If the limit is $-\infty$, there would be a unique solution for $g(d)$. Unfortunately i cant prove this step. Does anybody have ideas for this problem?