# Proving that $\frac{1}{\Gamma(r)}\int_{\mu}^{\infty}t^{r-1}e^{-t}dt=\sum_{x=0}^{r-1}\frac{e^{-\mu}\mu^x}{x!}$ [duplicate]

$$\frac{1}{\Gamma(r)}\int_{\mu}^{\infty}t^{r-1}e^{-t}dt=\sum_{x=0}^{r-1}\frac{e^{-\mu}\mu^x}{x!}$$ 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?

• Do you recognize the Poisson probability terms in the sum? Have a look here And please: Is this a question from a course or textbook? If so, please add the self-study tag & read its wiki. Commented Mar 1, 2021 at 10:40
• Discussed at math.stackexchange.com/q/467341/321264 and its linked threads. Commented Mar 1, 2021 at 10:56
• You can find more closely related threads with this site search.
– whuber
Commented Mar 1, 2021 at 15:59