13

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.

Ivan
  • 2,324

2 Answers2

1

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

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.

Xoff
  • 10,310
  • Sorry, I should probably change $n$ to $\alpha$. $n$ is not integer, therefore integration by parts leads nowhere, there is still need to evaluate an integral with $0<n<1$. – Ivan Aug 31 '12 at 17:07
  • If you're looking for a bound, you can consider that for $n\in(0,1)$, you are between $I_0$ and $I_1$, then you get your polynomial bound by the recursive formula. – Xoff Aug 31 '12 at 17:09
  • I don't think you are between $I_0$ and $I_1$: $1>x^n>x$ for $x<1$, $1<x^n<x$ for $x>1$. – Ivan Aug 31 '12 at 17:15
  • 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. – Sebastien B Nov 30 '12 at 15:32
1

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.

Nathaniel Bubis
  • 33,018