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Is there an analytic solution or approximation for the following Gaussian-like integration? $\frac{1}{\eta^{2n}} \frac{1}{\sqrt{2 \pi}} \int_{-\eta}^{+\eta} e^{-x^2/2} x^{2n} dx$? The numerical plot suggests that it initially decrease faster, but reach a steady decrease of $(2n)^{-1.06}$ numerically when $2n > 100$ for all $\eta$.

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  • $\begingroup$ The integral is asympotically $Θ(1/n)$, so the $1.06$ is actually $1$. Are you asking for the coefficient in the $Θ$? $\endgroup$
    – LeechLattice
    Commented Dec 27, 2019 at 3:34
  • $\begingroup$ Write $x^{2n} = e^{2n \ln x} $. When $n$ becomes large, the integrand is strongly peaked around the maximum of the exponent, i.e., at $x=\sqrt{2n} $. If that's within the range of integration, saddle point approximation should be good. $\endgroup$
    – Michael Engelhardt
    Commented Dec 27, 2019 at 3:34
  • $\begingroup$ @MichaelEngelhardt I understand the question as $n→+\infty$ for fixed $η$, so $x=\sqrt{2n}$ is outside the range of integration. $\endgroup$
    – LeechLattice
    Commented Dec 27, 2019 at 3:37
  • $\begingroup$ @LeechLattice Ok, then the dominant contribution comes from the endpoints, $x=\pm \eta $. $\endgroup$
    – Michael Engelhardt
    Commented Dec 27, 2019 at 3:39

3 Answers 3

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An exact result (in terms of the incomplete Gamma function) and the large-$n$ asymptotics are as follows: $$\frac{1}{\eta^{2n}} \frac{1}{\sqrt{2 \pi}} \int_{-\eta}^{+\eta} e^{-x^2/2} x^{2n} dx=\pi^{-1/2}2^{n} \eta^{-2 n} \left[\Gamma \left(n+\tfrac{1}{2}\right)-\Gamma \left(n+\tfrac{1}{2},\tfrac{1}{2}\eta^2\right)\right]$$ $$\rightarrow \frac{e^{-\eta^2/2} \eta}{\sqrt{2 \pi } n},\;\;\text{for}\;\;n\gg 1.$$ The convergence to the large-$n$ result is shown in the plot for $\eta=5$ (blue is the integral, gold the large-$n$ value):

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  • $\begingroup$ Thanks! Can you send me your address so that I can acknowledge you properly? It still looks like that the trend is less than 1/n $\endgroup$
    – CPW
    Commented Dec 27, 2019 at 12:26
  • $\begingroup$ Can you give me a link to the 2nd gamma function which has two arguments? I can only find the definition for the first one at en.wikipedia.org/wiki/Gamma_function $\endgroup$
    – CPW
    Commented Dec 27, 2019 at 12:32
  • $\begingroup$ en.wikipedia.org/wiki/Incomplete_gamma_function ; no need for an acknowledgment, but for completeness, the proper way to refer to MO postings is explained here: meta.mathoverflow.net/a/4352/11260 $\endgroup$
    – Carlo Beenakker
    Commented Dec 27, 2019 at 14:11
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See paragraph 3.8 in https://people.sc.fsu.edu/~%20jburkardt/presentations/truncated_normal.pdf and https://people.smp.uq.edu.au/YoniNazarathy/teaching_projects/studentWork/EricOrjebin_TruncatedNormalMoments.pdf where recursive Formulas are given for the moments of a truncated normal distribution.

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Mathematica 12.0 answers

AsymptoticIntegrate[Exp[-x^2/2]*x^(2*n), {x, -\[Eta], \[Eta]}, {n, Infinity, 1},
Assumptions -> n>0&&n\[Element] Integers && \[Eta] > 0]/  Sqrt[2*Pi]/\[Eta]^(2*n)//Simplify

$$\frac{e^{-\frac{\eta ^2}{2}} \eta ^{-2 n} \left(\eta ^2\right)^{n+\frac{1}{2}}}{\sqrt{2 \pi } n} $$

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  • $\begingroup$ Thanks you so much! Can you send me your address so that I can acknowledge you properly? $\endgroup$
    – CPW
    Commented Dec 27, 2019 at 12:13
  • $\begingroup$ The answer seems can be simplified further as $ \frac{e^{-\frac{\eta^2}{2}} \eta}{\sqrt{2 \pi} \frac{1}{n}$ $\endgroup$
    – CPW
    Commented Dec 27, 2019 at 12:22
  • $\begingroup$ Yes, the command of Mathematica Simplify[%, Assumptions -> n > 0 && n \ [Element] Integers && \ [Eta] > 0] performs $$\frac{e^{-\frac{\eta ^2}{2}} \eta }{\sqrt{2 \pi } n} .$$ $\endgroup$
    – user64494
    Commented Dec 27, 2019 at 18:52

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