1. Statement of the problem. We are given the function $$g(r)=\frac{r}{2\pi h}\int_0^{\infty} e^{-rt}\,dt$$ We assume $r>0$ so the integral converges. 2. Formula and rules used. For $r>0$, the basic integral is $$\int_0^{\infty} e^{-rt}\,dt=\frac{1}{r}.$$ This comes from evaluating an improper exponential integral; the parameter must satisfy $r>0$. 3. Intermediate work. Substitute the integral result into $g(r)$ to get $$g(r)=\frac{r}{2\pi h}\cdot\frac{1}{r}.$$ Show cancellation of $r$: $$g(r)=\frac{\cancel{r}}{2\pi h}\cdot\frac{1}{\cancel{r}}=\frac{1}{2\pi h}.$$ 4. Final answer and comment. Therefore $g(r)=\frac{1}{2\pi h}$ for all $r>0$. This shows $g(r)$ is independent of $r$ under the convergence assumption.