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I need lower and upper bounds (as tight as possible) on the following integral: $\int_0^\infty x^n\exp\left(-\frac{(x-x_0)^2}2\right)\, dx$

$n$ is a real number greater than $0$, and $x_0>0$. I am guessing the bounds will be of the form $B(n)x_0^n+C(n)$, but I don't have a way of showing it (it might not be true either).

EDIT: I guess for the cases $n>1$ and $n<1$, one can use the convexity/concavity of the $x^n$ and make estimates of the form $x_0^n+nxx_0^{n-1}\le(x+x_0)^n\le2^{n-1}(x^n+x_0^n)$.

EDIT: $n$ is not an integer. More importantly, I would like an actual bound on the integral in terms of elementary functions.

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    @Ivan - what's wrong with gamma functions?2013-01-12

2 Answers 2

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$f(x)=\exp\left(-\frac{(x-x_0)^2}{2}\right)$ $f'(x)=(x_0-x)f(x)$ $xf(x)=x_0f(x)-f'(x)$

Hence

$I_n(x_0)=\int_0^{\infty}x^nf(x)=x_0I_{n-1}(x_0)-\int_0^{\infty}x^{n-1}f'(x)= x_0I_{n-1}(x_0)+(n-1)I_{n-2}(x_0)$

And $I_0(x_0)$ can be computed from the usual error function. So you get a quick way to compute the results.

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    For the n>1 here is a bound (not very tight). With $n\leq\alpha\leq n+1$, $\int_0^1 x^\alpha\exp(-(x-x_0)^2/2)dx\leq \int_0^1 x^n\exp(-(x-x_0)^2/2)dx$ and $\int_1^\infty x^\alpha\exp(-(x-x_0)^2/2)dx\leq \int_1^\infty x^{n+1}\exp(-(x-x_0)^2/2)dx$ and thus $I_\alpha(x_0)\leq I_n(x_0)+I_{n+1}(x_0)$ and compute $I_n(x_0)$ and $I_{n+1}(x_0)$$ as above.2012-11-30
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If I am not mistaken, for $n>0$, your integral is bounded by: $2^{1/2 - n/2} e^{-x_0^2/2} \Gamma(1 + n)$ This follows by actually calculating the integral which involves hypergeometric functions.