Let $P(X=n) = \lambda^ne^{-\lambda}/n!$ which is a Poisson distribution. I need to find the cumulative distribution function.
The first think that comes to my mind is summation across n from $-\infty$ to $\infty$. But the answer is in terms of integration of $\lambda$.
$P(X\le n) = 1-\int_{0}^{\lambda} (x^ne^{-x}/n!)dx$
How do we change that integration over x into integration over $\lambda$?
The Poisson distribution takes values $n \in \{0,1,2,\dotsc\}$ with probability $P(X=n) = e^{-\lambda} \lambda^n / n! =: p_n(\lambda) $, so the cumulative distribution function is $$ P(X \leq n) = \sum_{k=0}^n e^{-\lambda} \frac{\lambda^k}{k!} = \sum_{k=0}^n p_k(\lambda) . $$ This sum has a rather weird property (which admittedly one would not spot unless one is familiar with particular differential equations): if we differentiate with respect to $\lambda$, we find that $$ p_k'(\lambda) = \frac{d}{d\lambda} e^{-\lambda} \frac{\lambda^k}{k!} = - e^{-\lambda} \frac{\lambda^k}{k!} + e^{-\lambda} \frac{\lambda^{k-1}}{(k-1)!} = -p_k + p_{k-1} , $$ and $p_0'(\lambda) = 0$. Thus, $$ \frac{d}{d\lambda} P(X \leq n) = \sum_{k=0}^n p_n'(\lambda) = 0 + \sum_{k=1}^n (p_{k-1}(\lambda)-p_k(\lambda)) . $$ If we write out the sum, we see that it telescopes, leaving only $-p_n$. Hence, by the Fundamental Theorem of Calculus, $$ P(X \leq n) = P(X \leq n)(\lambda=0) - \int_0^{\lambda} p_n(x) \, dx . $$ The first term is $1$ since a Poisson distribution with parameter $0$ takes the value $0$ with probability $1$, the second is the integral given in the answer.
Of course everyone is wondering why you would want to do this. I think the simplest reason is, if you want to estimate the probability of being larger than $n$, it is much easier to do this with the expression $\int_0^{\lambda} p_n $ than the infinite sum $\sum_{k=n+1}^{\infty} p_k$.