Question

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

Answer 1

`= 1000!`

Explanation:

`int_0^oo x^1000e^-x dx`

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

`mathcal (L) { x^1000}_(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

`color(blue)(int_0^oo e^(- alpha x) dx)`

`= [- 1/ alpha e^(- alpha x)]_0^oo color(blue)(= 1/alpha)`

`d/(d alpha) int_0^oo e^(- alpha x) dx = d/(dalpha) (1/ alpha)`

`implies int_0^oo (-x) e^(- alpha x) dx = - 1/ alpha^2` or `color(blue)( int_0^oo x e^(- alpha x) dx = 1/ alpha^2)`

Again `d/(d alpha) int_0^oo x e^(- alpha x) dx = d/(d alpha)( 1/ alpha^2)`

`implies int_0^oo (-x) x e^(- alpha x) dx = - 2/ alpha^3` or `color(blue)(int_0^oo x^2 e^(- alpha x) dx = 2/ alpha^3)`

Such that

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

Answer 2

`1000!`

Explanation:

Supposing `n in NN`

`d/(dx)(x^n e^(-x)) = nx^(n-1)e^(-x)-x^n e^(-x)`

Calling `I_n = int x^n e^(-x)dx` we have

`I_n-nI_(n-1)=-x^n e^(-x)`

Here `I_0 = int e^(-x)dx = -e^(-x)`

so we have

`e^xI_n - n e^xI_(n-1)=-x^n`

Considering now `J_n = e^xI_n`

the recurrence equation reads

`J_n-nJ_(n-1)=-x^n`

developping

`J_1 = J_0-x`
`J_2=2J_1-x^2`
`J_3=3J_2-x^3`
`cdots`
`J_n = nJ_(n-1)-x^n`

or

`J_2 = 2(J_0-x)-x^2`
`J_3=3(2(J_0-x)-x^2)-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!`