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

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

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Answer 1 · 2018-06-01T07:11:34

$I = \frac{7!}{m^{8}}$ $I=(7!)/m^8$

Explanation:

We want to solve

$I = \int_{0}^{\infty} x^{7} e^{- m x} d x$ $I=int_0^oo x^7e^(-mx)dx$

Consider a slightly different problem

$I_{0} = \int_{0}^{\infty} e^{- m x} d x = \frac{1}{m}$ $I_0=int_0^oo e^(-mx)dx=1/m$

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

$I_{1} = \int_{0}^{\infty} x e^{- m x} d x = \frac{1}{m^{2}}$ $I_1=int_0^oo xe^(-mx)dx=1/m^2$
$I_{2} = \int_{0}^{\infty} x^{2} e^{- m x} d x = \frac{1 \cdot 2}{m^{3}}$ $I_2=int_0^oo x^2e^(-mx)dx=(1*2)/m^3$
$I_{3} = \int_{0}^{\infty} x^{3} e^{- m x} d x = \frac{1 \cdot 2 \cdot 3}{m^{4}}$ $I_3=int_0^oo x^3e^(-mx)dx=(1*2*3)/m^4$
$I_{4} = \int_{0}^{\infty} x^{4} e^{- m x} d x = \frac{1 \cdot 2 \cdot 3 \cdot 4}{m^{5}}$ $I_4=int_0^oo x^4e^(-mx)dx=(1*2*3*4)/m^5$

We may have spotted the pattern, for the general case

$I_{n} = \int_{0}^{\infty} x^{a} e^{- m x} d x = \frac{a!}{m^{a + 1}}$ $I_n=int_0^oo x^ae^(-mx)dx=(a!)/m^(a+1)$

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

$I_{7} = \int_{0}^{\infty} x^{7} e^{- m x} d x = \frac{7!}{m^{8}}$ $I_7=int_0^oo x^7e^(-mx)dx=(7!)/m^8$

Bonus info

Consider the integral we found

$\int_{0}^{\infty} x^{a} e^{- m x} d x = \frac{a!}{m^{a + 1}}$ $int_0^oo x^ae^(-mx)dx=(a!)/m^(a+1)$

For the case $m = 1$ $color(blue)(m=1$ , 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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∫e^(-mx).x^(7)dx Limits zero to infinity ????

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