How to do this integral?

**jfz23** · September 26th 2009, 03:36 PM

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:

$\int_0^{\infty}x^2e^{-x(y+1)}dx$

**galactus** · September 26th 2009, 04:11 PM

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.

**jfz23** · September 26th 2009, 04:18 PM

You sir are brilliant! Thanks! Does this technique have a name?

**galactus** · September 26th 2009, 04:31 PM

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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