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How to prove this integral as following is finite?

$$ \int_{-\infty}^{+\infty}\left|x\right|^ke^{-(x-3)^2/2}dx $$

k is a positive integer

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  • $\begingroup$ $k \in \Bbb{Z}$ I guess? $\endgroup$ – MathematicianByMistake May 9 '16 at 9:37
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    $\begingroup$ k is a positive integer. $\endgroup$ – hliu May 9 '16 at 9:37
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    $\begingroup$ For any k if $x$ is large enough we have $e^x>x^k$ use this to estimate integral. $\endgroup$ – user175968 May 9 '16 at 9:43
  • $\begingroup$ $|x|^k e^{-x^2/4}$ is bounded by a constant and $\exp\left(x^2/4-(x-3)^2/2\right)$ is integrable. $\endgroup$ – Jack D'Aurizio May 9 '16 at 11:09
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Fix $n \in \mathbb{N}$, and consider $f_n (x)= \chi_{[-n,n]}(x) |x|^k \exp( - (x-3)^2 /2)$.
Then we have that $f_n$ increases pointwise to $f$ and thus we have by MCT $$\lim_{n \to \infty} \int_{-\infty}^\infty f_n(x) \textrm{d}x = \int_{-\infty}^\infty \lim_{n \to \infty} f_n(x) \textrm{d}x = \int_{-\infty}^\infty f(x) \textrm{d}x.$$ Furthermore, we can find $\int_{-\infty}^\infty f_n(x) \textrm{d}x $ by using the power series of $\exp(- (x-3)^2 /2)$, and denote $g_{n,m}$ to be the partial sum till the $m^{\text{th}}$ term on $[-n,n]$ and $0$ outside $[-n,n]$.
As the $g_{n,m}$ converges pointwise to $f_n$ and the series $g_{n,m}$ is bounded by the integrable function $g_n(x) = \Lambda_n \chi_{[-n,n]}(x)$ where $\Lambda_n$ is a finite upper bound of $\{g_{n,m} ,f_{n}\}$, we have by DCT $$ \lim_{m \to \infty} \int_{-\infty}^\infty g_{n,m}(x) \textrm{d}x = \int_{- \infty}^\infty f_n(x) \textrm{d}x.$$ And thus $$ \int_{-\infty}^\infty f(x) \textrm{d}x = \lim_{n \to \infty} \lim_{m \to \infty} \int_{-n}^n g_{n,m}(x) \textrm{d}x.$$

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We have: $$ \int_{\mathbb{R}}\exp\left(\rho x^2+(x-3)^2/2\right)\,dx = \sqrt{\frac{2\pi}{1-2\rho}}\,\exp\left(\frac{9\rho}{1-2\rho}\right)$$ for any $\rho<\frac{1}{2}$, while $|x|^k e^{-\rho x^2}$ has its maxima at $x=\pm\sqrt{\frac{k}{2\rho}}$, hence the original integral is bounded by

$$ \left(\frac{k}{2\rho}\right)^{k/2}\sqrt{\frac{2\pi}{1-2\rho}}\,\exp\left(\frac{9\rho}{1-2\rho}-\frac{k}{2}\right)$$ for any $\rho\in\left(0,\frac{1}{2}\right)$, and by choosing $\rho$ as $\frac{k}{2(9+k)}$ we get the upper bound:

$$\int_{\mathbb{R}}|x|^k e^{-(x-3)^2/2}\,dx \leq \color{red}{\frac{\sqrt{2\pi}}{3}\, (9+k)^{(k+1)/2}}.$$

As an alternative, you may also show that $$ f(\rho) = \int_{\mathbb{R}}\exp\left(-\rho|x|-(x-3)^2/2\right)\,dx $$ is an analytic function in a neighbourhood of $\rho=0$. $$ M_k = (-1)^k f^{(k)}(0) = \int_\mathbb{R}|x|^k e^{-(x-3)^2/2}\,dx $$ is a log-convex sequence by Cauchy-Schwarz inequality.

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You can use Taylor series representation for exponential function. Then interchange integration and summation with Fubinis theorem.

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  • $\begingroup$ How can we use Fubini? The power series is not non-negative, thus we need absolute integrability in some sort already. $\endgroup$ – Hetebrij May 9 '16 at 10:02

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