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

Jun 1, 2018

I=(7!)/m^8

#### Explanation:

We want to solve

$I = {\int}_{0}^{\infty} {x}^{7} {e}^{- m x} \mathrm{dx}$

Consider a slightly different problem

${I}_{0} = {\int}_{0}^{\infty} {e}^{- m x} \mathrm{dx} = \frac{1}{m}$

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

${I}_{1} = {\int}_{0}^{\infty} x {e}^{- m x} \mathrm{dx} = \frac{1}{m} ^ 2$
${I}_{2} = {\int}_{0}^{\infty} {x}^{2} {e}^{- m x} \mathrm{dx} = \frac{1 \cdot 2}{m} ^ 3$
${I}_{3} = {\int}_{0}^{\infty} {x}^{3} {e}^{- m x} \mathrm{dx} = \frac{1 \cdot 2 \cdot 3}{m} ^ 4$
${I}_{4} = {\int}_{0}^{\infty} {x}^{4} {e}^{- m x} \mathrm{dx} = \frac{1 \cdot 2 \cdot 3 \cdot 4}{m} ^ 5$

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