1. **Problem statement:** Show that for integer $k \geq 1$, $$\int_\mu^\infty \frac{1}{\Gamma(k)} z^{k-1} e^{-z} \, dz = \sum_{x=0}^{k-1} \frac{\mu^x e^{-\mu}}{x!}$$ This relates the cumulative distribution functions (cdfs) of the gamma and Poisson distributions. 2. **Recall definitions and properties:** - The gamma function for integer $k$ satisfies $\Gamma(k) = (k-1)!$. - The integrand $\frac{1}{\Gamma(k)} z^{k-1} e^{-z}$ is the probability density function (pdf) of a Gamma distribution with shape $k$ and rate 1. - The right side is the sum of Poisson probabilities with parameter $\mu$ up to $k-1$. 3. **Approach:** Use integration by parts $k-1$ times or verify the derivative of the proposed antiderivative. 4. **Verification by differentiation:** Define $$F(z) = -e^{-z} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1)!} z^{k-j-1}$$ We want to show $$\frac{d}{dz} F(z) = z^{k-1} e^{-z}$$ 5. **Differentiate $F(z)$:** $$\frac{d}{dz} F(z) = -\frac{d}{dz} \left(e^{-z} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1)!} z^{k-j-1} \right)$$ Using product rule: $$= -\left(-e^{-z} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1)!} z^{k-j-1} + e^{-z} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1)!} (k-j-1) z^{k-j-2} \right)$$ $$= e^{-z} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1)!} z^{k-j-1} - e^{-z} \sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-1)!} (k-j-1) z^{k-j-2}$$ 6. **Simplify the second sum:** Note that $$\frac{(k-j-1)}{(k-j-1)!} = \frac{1}{(k-j-2)!}$$ So the second sum becomes $$\sum_{j=0}^{k-1} \frac{\Gamma(k)}{(k-j-2)!} z^{k-j-2}$$ Shift index $m = j+1$ in the second sum to align powers: $$\sum_{m=1}^k \frac{\Gamma(k)}{(k-m-1)!} z^{k-m-1}$$ 7. **Rewrite the difference:** $$\frac{d}{dz} F(z) = e^{-z} \left( \frac{\Gamma(k)}{(k-0-1)!} z^{k-1} + \sum_{j=1}^{k-1} \frac{\Gamma(k)}{(k-j-1)!} z^{k-j-1} - \sum_{m=1}^k \frac{\Gamma(k)}{(k-m-1)!} z^{k-m-1} \right)$$ The sums cancel except the first term of the first sum and the last term of the second sum. The last term of the second sum at $m=k$ is zero because factorial of negative number is undefined and the term vanishes. Thus, $$\frac{d}{dz} F(z) = e^{-z} \frac{\Gamma(k)}{(k-1)!} z^{k-1} = e^{-z} z^{k-1}$$ since $\Gamma(k) = (k-1)!$. 8. **Conclusion:** The derivative of $F(z)$ is the integrand, so $$\int_\mu^\infty \frac{1}{\Gamma(k)} z^{k-1} e^{-z} \, dz = F(\infty) - F(\mu) = 0 - F(\mu) = e^{-\mu} \sum_{j=0}^{k-1} \frac{\mu^j}{j!}$$ because the term at infinity vanishes. This matches the right side, proving the identity. **Final answer:** $$\int_\mu^\infty \frac{1}{\Gamma(k)} z^{k-1} e^{-z} \, dz = \sum_{x=0}^{k-1} \frac{\mu^x e^{-\mu}}{x!}$$