How do you evaluate the integral int x^1000e^-x from 0 to oo?

Aug 28, 2016

= 1000!

Explanation:

${\int}_{0}^{\infty} {x}^{1000} {e}^{-} x \mathrm{dx}$

you can do this very quickly by noting that it is the equivalent of

$m a t h c a l \left(L\right) {\left\{{x}^{1000}\right\}}_{s = 1}$

and as mathcal(L) {t ^n} = (n!)/(s^(n+1))

int_0^oo x^1000e^-x = (1000!)/(1)^1001 = 1000!

Or in terms of the gamma function:

Gamma(n+1) = int_0^oo x^(n) e^(-x) dx = n!

To generate your own solution you could start as per the factorial function with this

$\textcolor{b l u e}{{\int}_{0}^{\infty} {e}^{- \alpha x} \mathrm{dx}}$

$= {\left[- \frac{1}{\alpha} {e}^{- \alpha x}\right]}_{0}^{\infty} \textcolor{b l u e}{= \frac{1}{\alpha}}$

$\frac{d}{d \alpha} {\int}_{0}^{\infty} {e}^{- \alpha x} \mathrm{dx} = \frac{d}{\mathrm{da} l p h a} \left(\frac{1}{\alpha}\right)$

$\implies {\int}_{0}^{\infty} \left(- x\right) {e}^{- \alpha x} \mathrm{dx} = - \frac{1}{\alpha} ^ 2$ or $\textcolor{b l u e}{{\int}_{0}^{\infty} x {e}^{- \alpha x} \mathrm{dx} = \frac{1}{\alpha} ^ 2}$

Again $\frac{d}{d \alpha} {\int}_{0}^{\infty} x {e}^{- \alpha x} \mathrm{dx} = \frac{d}{d \alpha} \left(\frac{1}{\alpha} ^ 2\right)$

$\implies {\int}_{0}^{\infty} \left(- x\right) x {e}^{- \alpha x} \mathrm{dx} = - \frac{2}{\alpha} ^ 3$ or $\textcolor{b l u e}{{\int}_{0}^{\infty} {x}^{2} {e}^{- \alpha x} \mathrm{dx} = \frac{2}{\alpha} ^ 3}$

Such that

 int_0^oo x^n e^(- alpha x) dx = (n!)/ alpha^(n+1)

Aug 28, 2016

1000!

Explanation:

Supposing $n \in \mathbb{N}$

$\frac{d}{\mathrm{dx}} \left({x}^{n} {e}^{- x}\right) = n {x}^{n - 1} {e}^{- x} - {x}^{n} {e}^{- x}$

Calling ${I}_{n} = \int {x}^{n} {e}^{- x} \mathrm{dx}$ we have

${I}_{n} - n {I}_{n - 1} = - {x}^{n} {e}^{- x}$

Here ${I}_{0} = \int {e}^{- x} \mathrm{dx} = - {e}^{- x}$

so we have

${e}^{x} {I}_{n} - n {e}^{x} {I}_{n - 1} = - {x}^{n}$

Considering now ${J}_{n} = {e}^{x} {I}_{n}$

${J}_{n} - n {J}_{n - 1} = - {x}^{n}$

developping

${J}_{1} = {J}_{0} - x$
${J}_{2} = 2 {J}_{1} - {x}^{2}$
${J}_{3} = 3 {J}_{2} - {x}^{3}$
$\cdots$
${J}_{n} = n {J}_{n - 1} - {x}^{n}$

or

${J}_{2} = 2 \left({J}_{0} - x\right) - {x}^{2}$
${J}_{3} = 3 \left(2 \left({J}_{0} - x\right) - {x}^{2}\right) - {x}^{3}$
$\cdots$
J_n = n!J_0-sum_(k=1)^n ((n!)/(k!))x^k

so

I_n = n!I_0-e^(-x)(sum_(k=1)^n ((n!)/(k!))x^k)

or

I_n = n!e^(-x)-e^(-x)(sum_(k=1)^n ((n!)/(k!))x^k)

Finally

int_0^oo x^n e^(-x)dx = n!

so

int_0^oo x^(1000) e^(-x)dx = 1000!