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

2 Answers
Aug 28, 2016

#= 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)#

Aug 28, 2016

#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!#