TITLE: Considering $\int_0^\infty2^{-x}(1+x)^ndx$
QUESTION [5 upvotes]: $\displaystyle I_n=\int_0^\infty2^{-x}(1+x)^ndx$

By integral by parts, I can get $I_n=\frac{1}{\ln 2}+\frac{n}{\ln 2}I_{n-1}$ where $I_0=\frac{1}{\ln 2}$. How to proceed to conclude an explicit expression of $I_n=\sum\limits_{m=0}^n\cdots$? Thank you.

REPLY [4 votes]: Iterate. Since $I_n = \frac{1}{\ln 2}(1 + n I_{n-1})$, then $I_{n-1} = \frac{1}{\ln 2}[1 + (n-1) I_{n-2}]$, and so-on. Repeating down to $I_0$ yields
$$ I_n = \frac{1}{\ln 2}(1 + n I_{n-1}) = \frac{1}{\ln 2}(1 + n[\frac{1}{\ln 2}[1 + (n-1) I_{n-2}]]) \\ = \text{...} = \frac{1}{\ln2} + \frac{n}{(\ln2)^2} + \frac{n(n-1)}{(\ln2)^3} + \frac{n(n-1)(n-2)}{(\ln2)^4} + \text{...} = \sum_{k=0}^n \frac{P(n,k)}{(\ln2)^{k+1}}$$
just as Marvis found using the (very slick) Binomial expansion + Gamma function approach above.