Is this equation correct? And if so, is this famous?

Question

$$\int_1^{\infty}x^n e^{-x}dx=\frac{1}{e} \sum_{k=0}^n\frac{n!}{(n-k)!}$$

My proof.¹

\begin{align*} &\int_{1}^{\infty} e^{-\alpha x} \, \mathrm{d}x = \frac{e^{-\alpha}}{\alpha}, \\ &\Rightarrow \qquad \int_{1}^{\infty} (-x)^n e^{-\alpha x} \, \mathrm{d}x = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} (-1)^k k! \frac{1}{\alpha^k}(-1)^{n-k}e^{-\alpha} \\ &\Rightarrow \qquad \int_{1}^{\infty} x^n e^{-\alpha x} \, \mathrm{d}x = \sum_{k=0}^{n} \frac{n!}{(n-k)!} \frac{e^{-\alpha}}{\alpha^k}. \end{align*} $\alpha = 1$とすれば、 \begin{align*} \int_{1}^{\infty} x^n e^{-x} \, \mathrm{d}x = \frac{1}{e} \sum_{k=0}^{n} \frac{n!}{(n-k)!}. \end{align*} が成り立つ。 $\blacksquare$

Original images: (1, 2)

@StarlightRoad: I won't believe you unless you show that proof. — , Jun 12 '19 at 11:49
Certainly $\Gamma (z) = \int_0^{\infty} t^{z-1} e^{-t} dt$ is a famous function, and yours is just slightly removed. https://en.wikipedia.org/wiki/Gamma_function — user317176, Jun 12 '19 at 12:00
A slightly more general form of this integral is number 3.351.2 (with $u=\mu=1$) on page 310 of Table of Integrals, Series, and Products by Gradshteyn and Ryzhik. The indefinite integral (from which yours can be derived) is number 521 on page 379 of the CRC Standard Mathematical Tables, 24th edition. — David K, Jun 12 '19 at 12:52

score 11 · Answer 1 · answered Jun 12 '19 at 12:28

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

In:
Gradshteyn and Rhyzik, Table of Integrals, Series, and Products
formula 3.351.2 is $$ \int_u^\infty x^n e^{-\mu x} dx = e^{-u \mu} \sum_{k=0}^n \frac{n!}{k!}\; \frac{u^k}{\mu^{n-k+1}} $$ Now if you take $u=\mu=1$ you get your formula.

score 8 · Answer 2 · edited Jun 12 '20 at 10:38

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. $\text{Exp}(1)$ random variables. Then

$X$ has the PDF $f_X(x) = (x^n e^{-x} / n!) \mathbf{1}(x > 0)$, and so,

$$ \int_{1}^{\infty} x^n e^{-x} \, \mathrm{d}x = n! \,\mathbf{P}(X > 1). $$
$X$ can be realized as the $(n+1)$-th arrival time of the Poisson process $N = (N_t)_{t\geq 0}$ with unit rate, and so,

\begin{align*} \mathbf{P}(X > 1) &= \mathbf{P}(\text{$(n+1)$-th arrival has not occurred by time $1$}) \\ &= \mathbf{P}(\text{there are at most $n$ arrivals by time $1$}) \\ &= \mathbf{P}(N(1) \leq n) = \sum_{k=0}^{n} \frac{1}{k!}e^{-1}. \end{align*}

Combining two observations proves the desired identity.

score 2 · Answer 3 · edited Jun 12 '20 at 10:38

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

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

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

case n = 0:

$\int_{1}^{\infty} x^0 e^{-x}dx = [-e^{-x}]_1^\infty = 0 + \frac{1}{e} = \frac{1}{e} \frac{0!}{0!}$

step $n \rightarrow n+1$:

$\int_{1}^{\infty} x^{n+1} e^{-x}dx = [-x^{n+1}e^{-x}]_1^\infty + (n+1)\int_{1}^{\infty} x^{n} e^{-x}dx = \frac{1}{e} + (n+1)\left(\frac{1}{e}\sum_{k = 0}^{n} \frac{n!}{k!} \right) = \frac{1}{e} \left(1 + \sum_{k = 0}^{n} \frac{(n+1)!}{k!}\right) = \frac{1}{e}\sum_{k = 0}^{n+1} \frac{(n+1)!}{k!}$

Is this equation correct? And if so, is this famous?

3 Answers3