How do we determine the closed for $\int_{-\infty}^{\infty}x^ne^{-(x-\pi)^2\over \pi}dx=F(n)?$

Question

$$\int_{-\infty}^{\infty}e^{-(x-\pi)^2\over \pi}dx=\pi$$

$$\int_{-\infty}^{\infty}xe^{-(x-\pi)^2\over \pi}dx=\pi\left({0\over 2}+\pi\right)$$

$$\int_{-\infty}^{\infty}x^2e^{-(x-\pi)^2\over \pi}dx=\pi^2\left({1\over 2}+\pi\right)$$

$$\int_{-\infty}^{\infty}x^3e^{-(x-\pi)^2\over \pi}dx=\pi^3\left({3\over 2}+\pi\right)$$

What is the general formula for this integral

$$I=\int_{-\infty}^{\infty}x^ne^{-(x-\pi)^2\over \pi}dx=F(n)?$$

Let try an apply integration by parts

$u=x^n$ then $du=nx^{n-1}dx$

$dv=e^{{-1\over \pi}\left(x-\pi\right)^2}dx$ Then

$v=\int_{-\infty}^{\infty}e^{{-1\over \pi}\left(x-\pi\right)^2}dx=\pi$

$I=x^n\pi-n\pi\int_{-\infty}^{\infty}x^{n-1}dx$

$I=x^n\pi-n\pi[x^{\infty}-x^{-\infty}]$

This part seem meaningless, so integration by parts doesn't work.

Answer 1

If $X$ is a normal random variable with mean $\mu=\pi$ and variance $\sigma^2=\pi/2$, then $F(n)/\pi$ are the moments of $X$.

You can find formulae for the non-central moments here for example (see the table). Plugging in the values of $\mu$ and $\sigma$ above give the following values for $F(n)/\pi$. \begin{align} \pi && n=1\\ \pi^2 + \pi/2 && n=2\\ \pi^3 + 3\pi^2/2 && n=3\\ \pi^4 + 3 \pi^3 + 3 \pi^2/2 && n=4\\ &&\vdots \end{align}

A sketch of the derivation of the non-central moments as it applies to your problem appears below.

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The hairy part is due to the non-central nature (the $x-\pi$ term in the exponent).

If we consider $Z$ being standard normal, then the moments are much easier to compute.

For $E[Z^n] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty z^n e^{-z^2/2} \mathop{dz}$ for $n$ odd, one simply notes the integrand is odd, so the integral $\int_{-\infty}^0$ cancels with the integral $\int_0^\infty$ to yield $E[Z^n]=0$.

For $n$ even, one uses integration by parts and induction. \begin{align} \sqrt{2\pi}E[Z^n] &= \int_{-\infty}^\infty z^n e^{-z^2/2} \mathop{dz}\\ &= \left[-z^{n-1} e^{-z^2/2}\right]_{-\infty}^\infty + (n-1) \int_{-\infty}^\infty z^{n-2} e^{-z^2/2} \mathop{dz}\\ &= (n-1)\sqrt{2\pi} E[Z^{n-2}]. \end{align} With the base case $E[Z^0]=1$, we have $E[Z^n] = 1 \cdot 3 \cdot 5 \cdots (n-1) =: (n-1)!!$ for $n$ even.

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Armed with the central moments $E[Z^n]$, we can move to the moments of $X$.

\begin{align} E[X^n] &= E[(\sqrt{\pi/2} \cdot Z+ \pi)^n]\\ &= \sum_{k=0}^n \binom{n}{k} \pi^{n-k/2} 2^{-k/2} E[Z^k]\\ &= \sum_{j=0}^{\lfloor n/2\rfloor} \binom{n}{2j} \pi^{n-j} 2^{-j} (2j-1)!! \end{align}

Answer 2

That $\displaystyle\int_{-\infty}^\infty e^{-x^2/2}\,dx = \sqrt{2\pi}$ has been the subject of postings here many times. Some rescaling gives you the value of $\displaystyle\int_{-\infty}^\infty e^{-(x-\pi)^2/\pi} \,dx.$

Now consider $\displaystyle \int_{-\infty}^\infty x^n e^{-x^2/2} \,dx.$

When $n$ is odd, this is clearly $0.$ When $n$ is even, you can start by observing that it's equal to $\displaystyle2\int_0^\infty x^n e^{-x^2/2} \,dx.$ Then let $u=x^2/2$ so that $x\,dx = du.$ Then you have $$ 2\int_0^\infty \sqrt{2u}^{n-1} e^{-u}\,du = \sqrt 2^{n+1} \int_0^\infty u^{(n-1)/2} e^{-u}\,du = \sqrt 2^{n+1} \Gamma\left( \frac{n+1} 2 \right). $$ This can be expressed in terms of factorials and $\sqrt\pi$ via standard Gamma function identities.

But you have $(x-\pi)^2$. Let $w=x-\pi$, so that $dw=dx$, and then $x^n=(w+\pi)^n$. Apply the binomial theorem.

This may be be a "closed form" since $n$ will be the upper bound of summation: $\sum\limits_{k=0}^n.$ Whether further identities can reduce it to a closed form might be of interest.