# Showing a Definite Integral Equivalent to a Poisson Approximation

I was trying to prove this equality:

$$\sum_{i=0}^{n} e^{-\lambda} \frac{\lambda^{i}}{i!}= \frac{1}{n!} \int_{\lambda}^{\infty} e^{-x} x^{n} dx$$

This is my attempt so far:

$$\frac{1}{n!} \int_{\lambda}^{\infty} e^{-x} x^{n} dx$$ Using integration by parts with $u = x^{n}, \ du = n \cdot x^{n-1} dx, \ v = -e^{-x}, \ dv = e^{-x} dx$:

$$\frac{1}{n!} \bigg[ \frac{e^{-x}}{n+1} \ x^{n+1} \bigg\vert_{\lambda}^{\infty} - \int_{\lambda}^{\infty} -n \cdot x^{n-1} e^{-x} dx \bigg]$$

$$\require{cancel} = \frac{1}{n!} \bigg[\Big( \cancelto{0}{- \frac{x^{n}}{e^{\infty}}} + x^{n} e^{-\lambda}\Big) + n \int_{\lambda}^{\infty} x^{n-1} e^{-x} dx \bigg]$$

Using integration by parts with $u = x^{n-1}, \ du = (n-1) \cdot x^{n-2} dx, \ v = -e^{-x}, \ dv = e^{-x} dx$: $$= \frac{1}{n!} \bigg[ x^{n} e^{-\lambda} + n \left(-x^{n-1}e^{-x} \bigg\vert_{\lambda}^{\infty} - \int_{\lambda}^{\infty} - (n-1) \ x^{n-2} dx \right) \bigg]$$ $$= \frac{1}{n!} \bigg[ x^{n} e^{-\lambda} + n \Big( \cancelto{0}{- \frac{x^{n-1} }{e^{\infty} }} + x^{n-1}e^{-\lambda} + \int_{\lambda}^{\infty} (n-1) \ x^{n-2} dx \Big) \bigg]$$ $$= \frac{1}{n!} \bigg[ x^{n} e^{\lambda} + nx^{n-1}e^{\lambda} + \int_{\lambda}^{\infty} (n-1) \ x^{n-2} dx \ \bigg]$$

$$= \frac{1}{n!} \bigg[ \frac{n!}{(n)!} \ x^{n} e^{\lambda} + \frac{n!}{(n-1)!} \ x^{n-1}e^{\lambda} + \int_{\lambda}^{\infty} (n-1) \ x^{n-2} dx \ \bigg]$$

By taking this pattern to its conclusion, we introduce a summation:

$$= \frac{1}{n!} \bigg[ \sum_{i=0}^{n} \frac{n!}{(n-i)!} x^{n-i} e^{-x} \bigg]_{\lambda}^{\infty}$$ $$= \bigg[ \sum_{i=0}^{n} \frac{ x^{n-i} \ e^{-x} }{ (n-i)! } \bigg]_{\lambda}^{\infty}$$

At this point I get stuck and don't know how to reach:

$$\sum_{i=0}^{n} e^{-\lambda} \frac{\lambda^{i}}{i!}$$

I would appreciate it if someone could give me a hint as to where I should be heading now. I can't seem to go any further.

There are some errors in the integrations by parts. We have $$\frac{1}{n!}\int_{\lambda}^{\infty}x^{n}e^{-x}dx=\frac{1}{n!}\left(\left.-e^{-x}x^{n}\right|_{\lambda}^{\infty}+n\int_{\lambda}^{\infty}e^{-x}x^{n-1}dx\right)$$ and so on, then $$\frac{1}{n!}\int_{\lambda}^{\infty}x^{n}e^{-x}dx=\sum_{i=0}^{n}\frac{e^{-\lambda}\lambda^{n-i}}{\left(n-i\right)!}$$ and now use the fact that, for every function $f$, holds $$\sum_{k\leq n}f\left(k\right)=\sum_{k\leq n}f\left(n-k\right).$$
• $x$ is the variable of the integral. So you have to consider $x$ at infinity and $x=\lambda$. – Marco Cantarini Nov 10 '15 at 6:58