Integral $\int_0^\infty x^ne^{-x} dx$

Question

I know that is true when $t = \frac{x-n}{\sqrt n}$ $$\begin{align} n! = \int_0^\infty x^ne^{-x} dx \stackrel{x = n + \sqrt{n}t}{=} \int_{-\sqrt{n}}^\infty \left( n + \sqrt{n}t \right)^n e^{-\left(n+\sqrt{n}t\right)} \sqrt{n} dt =\frac{n^n\sqrt{n}}{e^n} \int_{-\sqrt{n}}^\infty \left(1 + \frac{t}{\sqrt{n}}\right)^n e^{-\sqrt{n}t} dt.\end{align}$$
but I don't know why limits of my integral are different and why that appeared $\sqrt{n}$. Can someone please help me?

Answer 1

The Laplace transform is

so by setting $s=1$ we get $n!$ The proof is very easy. You just use

inegration by parts to show that $I_{n+1}=(n+1)I_{n}$ and that proves the result!!

Answer 2

This is the gamma function when the entry is a natural number : \begin{equation} \Gamma(n+1)=\int_0^{+\infty}x^ne^{-x}dx,\hspace{0.3cm}n\in\mathbb N. \end{equation} Two methods can be used. The main one is that you can use the integration by parts. Or proof by induction (That is, if you already know the result).

Answer 3

When you made the substitution

$$t=\frac {x-n}{\sqrt n}\qquad\implies\qquad x=n+t\sqrt n$$

Plug in $x=0$ into your substitution to find the value $t$ takes on. In this case,

$$t=-\frac n{\sqrt n}=-\sqrt n$$

Similarly, as $x\to\infty$, then $t\to\infty$. These are your bounds on the integral. Factoring out an $n$ from the $(n+t\sqrt n)^n$ term gives the final result in your question

$$\int\limits_{-\sqrt n}^{+\infty}(n+t\sqrt n)^n e^{-(n+t \sqrt n)} \,\mathrm dt=\frac {n^n\sqrt n}{e^n}\int\limits_{-\sqrt n}^{+\infty}\left(1+\frac t{\sqrt n}\right)^n e^{-t\sqrt n}\,\mathrm dt$$

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If you want to prove that

$$n!=\int\limits_0^{+\infty} x^n e^{-x}\,\mathrm dx$$

Then use repeated integration by parts, with $u=x^n$ and $\mathrm dv=e^{-x}\,\mathrm dx$. Integrating by parts once gives

$$\begin{align*}I_n\equiv\int\limits_0^{+\infty} x^n e^{-x}\,\mathrm dx & =-x^ne^{-x}\,\Biggr|_0^{+\infty}+n\int\limits_0^{+\infty} x^{n-1}e^{-x}\,\mathrm dx\\ & =nI_{n-1}\end{align*}$$

If we integrate by parts again, then we get

$$I_n=n(n-1)I_{n-2}$$

If we keep on integrating by parts a total of $n$ times, then the remaining integral simply becomes

$$\begin{align*}I_n & =n(n-1)(n-2)\cdots2\cdot 1I_0\\ & =n!\int\limits_0^{+\infty}e^{-x}\,\mathrm dx\\ & =n!\end{align*}$$