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I need to solve two integrals:

1.) $$ \int_0^\infty {xne^{n\theta-nx}}dx $$

2.) $$ \int_0^\infty {x^2ne^{n\theta-nx}}dx $$

n and $\theta$ are constants.

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What have you tried so far? –  draks ... Oct 11 '12 at 13:41
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Have a look at partial integration methods. –  Lord_Farin Oct 11 '12 at 13:43
    
You don't solve integrals, you evaluate them. –  Stefan Smith Oct 13 '12 at 1:52
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4 Answers 4

up vote 1 down vote accepted

1) using integration by parts. $$\int_0^\infty {xne^{n\theta-nx}}dx=\left [ -xe^{n\theta-nx} \right ]_0^\infty + \int_0^\infty e^{n\theta-nx} dx$$ Here, $e^{-x}$ goes to zero quicker than $x$ to $\infty$. So you are just left with the integral, which is: $$\left [ -\frac{e^{n\theta-nx}}{n} \right ]_0^\infty=\frac{e^{n\theta}}{n}$$

2) left as an exercise ;) but it's exactly the same! Just do integration by parts twice.

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Or guess that the (indefinite) integral will be $p(x)e^{n\theta-nx}$ for some polynomial $p$ with $\text{deg }p \leq 2$. Find the derivative of $p(x)e^{n\theta-nx}$, and determine the polynomials coeffiecients by comparing to your integrand.

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HINT: Use partial integration ($\int_a^b uv' dx=[uv]_a^b-\int_a^bu'v dx$) and the fact that $x'=1$ (at least for problem 1)

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... but first of all: Take the factor that don't depened on $x$ out of the integral: $\int a f(x) dx=a\int f(x)dx$ –  draks ... Oct 11 '12 at 13:57
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You could do that but there's no need to. –  Alex Oct 11 '12 at 14:05
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No, but it makes things clearer... –  draks ... Oct 11 '12 at 14:07
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It is useful to remember that

$$ \Gamma(n+1) = \int_{0}^{+\infty} x^n\,e^{-x}\,dx = n!, $$

so, through the substitution $x=y/n$, we get that the first integral is equal to $e^{n\theta}/n$ and the second integral is equal to $2e^{n\theta}/n^2$.

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