Show that $\int_{\mu}^{\infty} \frac{1}{\Gamma(k)} z^{k-1} e^{-z}dz = \sum_{x = 0}^{k-1} \frac{\mu^x}{x!e^\mu}$.

Question

I'd like to show that

$$\int_{\mu}^{\infty} \frac{1}{\Gamma(k)} z^{k-1} e^{-z}dz = \sum_{x = 0}^{k-1} \frac{\mu^x}{x!e^\mu}$$ where $k = 1, 2, 3...$ and $\mu$ is a real number.

I've done integration by parts and a pattern seems to be emerging, namely, the value of the antiderivative seems to be $$-z^{k-1}e^{-z} -(k-1)[-z^{k-2}e^{-z}...]$$

I'm not sure where to go from here, though.

Answer 1

We want to show

$$\int_{\mu}^{\infty} \frac{1}{\Gamma(k)} z^{k-1} e^{-z}dz = \sum_{x = 0}^{k-1} \frac{\mu^x}{x!e^\mu}$$

So, the easiest way is to use the series representation of the incomplete Gamma function:

$$ \int_{\mu}^{\infty} \frac{1}{\Gamma(k)} z^{k-1} e^{-z}dz = \frac{1}{\Gamma(k)} \int_{\mu}^{\infty} z^{k-1} e^{-z}dz = \frac{1}{\Gamma(k)} \Gamma(k,\mu)$$

$$ \Gamma(s,x) = (s-1)!e^{-x} \sum_{i=0}^{s-1}\frac{x^i}{i!}$$

Hence, we can modify the original equation as:

$$ \int_{\mu}^{\infty} \frac{1}{\Gamma(k)} z^{k-1} e^{-z}dz = \frac{1}{\Gamma(k)} (k-1)!e^{-\mu} \sum_{i=0}^{k-1}\frac{\mu^i}{i!}$$

Also, we know that for an integer $k$, we have:

$$ \Gamma(k) = (k-1)!$$

Therefore, we get

$$ \int_{\mu}^{\infty} \frac{1}{\Gamma(k)} z^{k-1} e^{-z}dz = \sum_{i=0}^{k-1}\frac{\mu^i}{i!e^{\mu}}$$

Answer 2

\begin{align} \int_\mu^\infty z^{k-1}e^{-z+\mu}dz &= \int_0^\infty (\mu+u)^{k-1}e^{-u}du \\ &= \int_0^\infty \sum_{x=0}^{k-1}{k-1\choose x}u^{k-1-x}\mu^xe^{-u}du \\ &= \sum_{x=0}^{k-1}{k-1\choose x}\mu^x\int_0^\infty u^{k-1-x}e^{-u}du \\ &= \sum_{x=0}^{k-1}\dfrac{\Gamma(k)}{\Gamma(x+1)\Gamma(k-x)}\mu^x\Gamma(k-x) \\ &= \sum_{x=0}^{k-1}\dfrac{\Gamma(k)}{x!}\mu^x \\ \color{blue}{\int_{\mu}^{\infty} \frac{1}{\Gamma(k)} z^{k-1} e^{-z}dz} &= \color{blue}{\sum_{x = 0}^{k-1} \frac{\mu^x}{x!e^\mu}} \end{align}