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Integral Calculus
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any function f(x) can be expanded into
f(x) = f(0) + x(f'(0)/1!) + x^2(f''(0)/2!) + x^3(f'''(0)/3!) .... + x^n((nth derivative of f(x) at x=0)/n!)
so using this e^x can be written as
1 + x/1! + x^2/2! + x^3/3! ... + x^n/n! because all derivatives of e^x at x=0 is 1...
So the denominator is nothing but e^x!!!
Assuming the integration from zero to infinity
with n>0, integrating by part with u=x^n and dv=e^-x dx
with have du = n x^(n-1) dx and v = -e^-x, so
Integral[0-inf, (x^n) e^-x] = Evaluate[0-inf, -(x^n)(e^-x)] + n Integral[0-inf, (x^(n-1)) e^-x]
where Evaluate[a-b, f(x)] means f(b)-f(a) (or with the appropriate limit when there's an infinity).
Here Evaluate[0-inf, -(x^n)(e^-x)] = 0 because e^-x tends to 0 much quicker than x^n for a given n.
So Integral[0-inf, (x^n) e^-x] = n Integral[0-inf, (x^(n-1)) e^-x]
Similarly Integral[0-inf, (x^(n-1)) e^-x] = (n-1) Integral[0-inf, (x^(n-2)) e^-x]
and so on.
We then have:
Integral[0-inf, (x^n) e^-x] = n*(n-1)...3*2 Integral[0-inf, x e^-x]
Integral[0-inf, (x^n) e^-x] = n*(n-1)...3*2*1 Integral[0-inf, e^-x]
Integral[0-inf, (x^n) e^-x] = n*(n-1)...3*2*1 = n!
Sorry but the formula editor is not working, so had to post like this.
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