∫e^(-mx).x^(7)dx Limits zero to infinity ????

1 Answer
Øko
Jun 1, 2018

#I=(7!)/m^8#

Explanation:

We want to solve

#I=int_0^oo x^7e^(-mx)dx#

Consider a slightly different problem

#I_0=int_0^oo e^(-mx)dx=1/m#

By repeated differentiation of both sides w.r.t. m
(Notice how the negative sign cancels out each time)

#I_1=int_0^oo xe^(-mx)dx=1/m^2#
#I_2=int_0^oo x^2e^(-mx)dx=(1*2)/m^3#
#I_3=int_0^oo x^3e^(-mx)dx=(1*2*3)/m^4#
#I_4=int_0^oo x^4e^(-mx)dx=(1*2*3*4)/m^5#

We may have spotted the pattern, for the general case

#I_n=int_0^oo x^ae^(-mx)dx=(a!)/m^(a+1)#

Or for your case #color(blue)(a=7#

#I_7=int_0^oo x^7e^(-mx)dx=(7!)/m^8#

Bonus info

Consider the integral we found

#int_0^oo x^ae^(-mx)dx=(a!)/m^(a+1)#

For the case #color(blue)(m=1#, this is what we call the gamma function

#Gamma(a+1)=int_0^oo x^ae^(-x)dx=a!#

This is an extended definition of the factorial

From this definition, results which seems odd by our usually definition can be derived
As example

#(0!)=1# and #(1/2!)=sqrt(pi)/2#

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