Best way to integrate a moment generating function?

trojansc82
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Homework Statement



∫etxx2e-x

Homework Equations



M(t) = etx f(x) dx

The Attempt at a Solution



I know the solution is -1/(t-3)3, however I'm having difficulty integrating the function. UV - ∫ V DU is extremely long and challenging, I'm wondering if there is a shortcut (i.e. quotient rule?)

Also, there is a process used here but I'm unable to understand it:

Untitled-1.jpg
 
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trojansc82 said:

Homework Statement



∫etxx2e-x

Homework Equations



M(t) = etx f(x) dx

The Attempt at a Solution



I know the solution is -1/(t-3)3, however I'm having difficulty integrating the function. UV - ∫ V DU is extremely long and challenging, I'm wondering if there is a shortcut (i.e. quotient rule?)

Unfortunately, integration by parts is the antiderivative method generated by the product or quotient rules. I don't know of any shortcut. Let u = x2 the first round than u = x the second round and you should be able to integrate the result directly.
 
It looks like that example integrated by parts multiple times. It is easy if you practice. Another method could be differentiation by the parameter.
Let In(t) be integrals like yours where n is the power of x

Then notice
In(t)=DnI0(t)=Dn(-1/(t-1))

where
D is differentiation with respect to t
I0(t)=(-1/(t-1))
 
LCKurtz said:
Unfortunately, integration by parts is the antiderivative method generated by the product or quotient rules. I don't know of any shortcut. Let u = x2 the first round than u = x the second round and you should be able to integrate the result directly.

I did the integration by parts, and I ended up with 1/(1-t)3...is that incorrect?

The mean and variance were still the same as the book's answers (μ = 3, σ2 = 3)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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