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Math Help - How to do this integral?

  1. #1
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    How to do this integral?

    I tried using integration by parts but it just seems to get more complicated regardless of which function I set for f(x) or g(x). Here is the integral:

    \int_0^{\infty}x^2e^{-x(y+1)}dx
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  2. #2
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    Here is one way to tackle it.

    \int x^{2}e^{-x(y+1)}dx

    Since the e term is a product of a polynomial, x^2, we can try this:

    \frac{d}{dx}[ax^{2}+bx+c]e^{-x(y+1)}=x^{2}e^{-x(y+1)}..............[1]

    Use the product rule on the left side and get:

    (-ay-a)x^{2}+x(-by+2a-b)-cy+b-c=x^{2}e^{-x(y+1)}

    Equate coefficients:

    -ay-a=1

    -by+2a-b=0

    -cy+b-c=0

    Solving the system results in:

    a=\frac{-1}{y+1}, \;\ b=\frac{-2}{(y+1)^{2}}, \;\ c=\frac{-2}{(y+1)^{3}}

    Sub them back into [1]:

    \frac{d}{dx}\left[\frac{-x^{2}}{y+1}-\frac{2x}{(y+1)^{2}}-\frac{2}{(y+1)^{3}}\right]e^{-x(y+1)}=x^{2}e^{-x(y+1)}

    Integrate both sides and get:

    \left(\frac{-x^{2}}{y+1}-\frac{2x}{(y+1)^{2}}-\frac{2}{(y+1)^{3}}\right)e^{-x(y+1)}=\int x^{2}e^{-x(y+1)}dx

    There she is.
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  3. #3
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    You sir are brilliant! Thanks! Does this technique have a name?
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  4. #4
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    I don't know. It probably does. But it works well when one has a polynomial and e product to integrate.
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