Xjyuk

$$int_1^inftyx^n e^-xdx=frac1e sum_k=0^nfracn!(n-k)!$$

My proof.¹

beginalign*
&int_1^infty e^-alpha x , mathrmdx = frace^-alphaalpha, \
&Rightarrow qquad int_1^infty (-x)^n e^-alpha x , mathrmdx = sum_k=0^n fracn!k!(n-k)! (-1)^k k! frac1alpha^k(-1)^n-ke^-alpha \
&Rightarrow qquad int_1^infty x^n e^-alpha x , mathrmdx = sum_k=0^n fracn!(n-k)! frace^-alphaalpha^k.
endalign*
$alpha = 1$とすれば、
beginalign*
int_1^infty x^n e^-x , mathrmdx = frac1e sum_k=0^n fracn!(n-k)!.
endalign*
が成り立つ。 $blacksquare$

Original images: (1, 2)

Is it famous? probalby not. Is it known? Yes.

In:

Gradshteyn and Rhyzik, Table of Integrals, Series, and Products

formula 3.351.2 is
$$
int_u^infty x^n e^-mu x dx = e^-u mu sum_k=0^n fracn!k!;
fracu^kmu^n-k+1
$$
Now if you take $u=mu=1$ you get your formula.

It depends on what it means to be famous, but it is a quite well-known formula in probability theory. Indeed, let $X$ be the sum of $(n+1)$ i.i.d. $textExp(1)$ random variables. Then

1. 
   

   $X$ has the PDF $f_X(x) = (x^n e^-x / n!) mathbf1(x > 0)$, and so,

   
   
   

   $$ int_1^infty x^n e^-x , mathrmdx = n! ,mathbfP(X > 1). $$

   
   


2. 
   

   $X$ can be realized as the $(n+1)$-th arrival time of the Poisson process $N = (N_t)_tgeq 0$ with unit rate, and so,

   
   
   

   beginalign*
   mathbfP(X > 1)
   &= mathbfP(text$(n+1)$-th arrival has not occurred by time $1$) \
   &= mathbfP(textthere are at most $n$ arrivals by time $1$) \
   &= mathbfP(N(1) leq n) = sum_k=0^n frac1k!e^-1. endalign*

   
   

Combining two observations proves the desired identity.

It is true and changing the summation order results in a slightly different expression, which I find easier to prove:

$int_1^infty x^n e^-xdx = frac1esum_k = 0^n fracn!k!$

You can prove it by using induction and integration by parts:

case n = 0:

$int_1^infty x^0 e^-xdx = [-e^-x]_1^infty = 0 + frac1e = frac1e frac0!0!$

step $n rightarrow n+1$:

$int_1^infty x^n+1 e^-xdx = [-x^n+1e^-x]_1^infty + (n+1)int_1^infty x^n e^-xdx = frac1e + (n+1)left(frac1esum_k = 0^n fracn!k! right) = frac1e left(1 + sum_k = 0^n frac(n+1)!k!right) = frac1esum_k = 0^n+1 frac(n+1)!k!$

This formula can be proven by simple induction.
Note, that we can also start with $n=0$ since both sides evaluate to $frac1e$.
[At first I stated this to be wrong for n=0.]

For $n=1$ we conclude $int_1^infty x e^-x, dx=frac2e=frac1esum_k=0^1 frac1!(1-k)!$ with partial integration.

The inductive step is done similary.

We have:

$int_1^infty x^n+1e^-x, dx =-e^xx^n|_1^infty+(n+1)int_1^infty e^-xx^n, dx$

$=frac1e+(n+1)sum_k=0^n fracn!(n-k)!$. By the inductive claim.

$=frac1e(1+sum_k=0^n frac(n+1)!(n-k)!)$

With an index shift, we get:

$=frac1e(1+sum_k=1^n+1 frac(n+1)!n-(k-1)!)$

$=frac1esum_k=0^n+1frac(n+1)!(n+1-k)!$