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

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

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. – Yves Daoust Jun 12 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. en.wikipedia.org/wiki/Gamma_function – Doug M Jun 12 at 12:00
• How do you justify the first implication ? – Yves Daoust Jun 12 at 12:09
• How do you justify the first implication ? – Yves Daoust Jun 12 at 12:14
• 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 at 12:52

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.

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

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

$$\int_{1}^{\infty} x^n e^{-x} \, \mathrm{d}x = n! \,\mathbf{P}(X > 1).$$

2. $$X$$ 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.

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 $$n \rightarrow n+1$$:

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

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 $$\int_1^\infty x e^{-x}\, dx=\frac2e=\frac1e\sum_{k=0}^1 \frac{1!}{(1-k)!}$$ with partial integration.

The inductive step is done similary.

We have:

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

$$=\frac1e+(n+1)\sum_{k=0}^n \frac{n!}{(n-k)!}$$. By the inductive claim.

$$=\frac1e(1+\sum_{k=0}^n \frac{(n+1)!}{(n-k)!})$$

With an index shift, we get:

$$=\frac1e(1+\sum_{k=1}^{n+1} \frac{(n+1)!}{n-(k-1)!})$$

$$=\frac{1}{e}\sum_{k=0}^{n+1}\frac{(n+1)!}{(n+1-k)!}$$

• The equation still holds when $n=0$ – Julian Mejia Jun 12 at 14:49
• @JulianMejia You are right. I edit that. – Cornman Jun 12 at 17:53