$$\frac{1}{\Gamma(r)}\int_{\mu}^{\infty}t^{r-1}e^{-t}dt=\sum_{x=0}^{r-1}\frac{e^{-\mu}\mu^x}{x!}$$ What I have tried-
$$\frac{1}{\Gamma(r)}\int_{\mu}^{\infty}t^{r-1}e^{-t}dt=e^{-\mu}\sum_{x=0}^{r-1}\frac{\mu^x}{x!}$$ $$\frac{1}{\Gamma(r)}\int_{\mu}^{\infty}t^{r-1}\sum_{n=0}^{\infty}\frac{(-1)^nt^n}{n!}dt=e^{-\mu}\sum_{x=0}^{r-1}\frac{\mu^x}{x!}$$
But I tried interchanging the sum and the integral and it didn't make sense anymore. Any ideas?
