$E(1/(1+x^2)) $under a normal distribution

Question

I want to know as mentioned in topic $E(1/(1+x^2))$ under a normal distribution $N(0,1)$. If it's not analytical, are there any bounds that are possible?

So basically, $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{e^{-x^2}}{1+x^2} dx $$

Thanks in advance, Sachin

Answer 1

To evaluate this integral, use Parseval's Theorem:

$$\begin{align}\int_{-\infty}^{\infty} dx \frac{1}{1+x^2} \, e^{-x^2} &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \:\pi e^{- |k|} \, \sqrt{\pi} e^{-k^2/4}\\ &= \frac{\sqrt{\pi}}{2}\left[\int_{-\infty}^{0} dk \:e^{k} \, e^{-k^2/4} + \int_{-\infty}^{\infty} dk \: e^{-k} \, e^{-k^2/4} \right ]\\ &= \frac{\sqrt{\pi}}{2} e \left[\int_{-\infty}^{0} dk \: \:e^{-(k-2)^2/4}+\int_{0}^{\infty} dk \: \:e^{-(k+2)^2/4} \right ]\\ &= \frac{\sqrt{\pi}}{2} e \left[\int_{-\infty}^{-2} dk \: \:e^{-k^2/4}+\int_{2}^{\infty} dk \: \:e^{-k^2/4} \right ]\\ &= \sqrt{\pi} e \left[\int_{-\infty}^{-1} dk \: \:e^{-k^2}+\int_{1}^{\infty} dk \: \:e^{-k^2} \right ]\\ &=2 \sqrt{\pi} e \left [\frac{\sqrt{\pi}}{2}\text{erfc}(1)+\frac{\sqrt{\pi}}{2} \text{erfc}(1) \right ]\\ &= \pi\, e\, \text{erfc}(1) \end{align}$$

where

$$\text{erfc}(y) = \frac{2}{\sqrt{\pi}} \int_y^{\infty} dt \: e^{-t^2}$$

Therefore

$$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} dx \frac{1}{1+x^2} \, e^{-x^2} = \sqrt{\frac{\pi}{2}} \, e \, \text{erfc}(1) \approx 0.535897$$

Answer 2

You can have a closed form solution for your integral in terms of the erf function

$$ \rm erf(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2} dt.$$

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{e^{-x^2}}{1+x^2} dx = \frac{\sqrt{\pi}}{\sqrt{2}} \,{{ e}}\, \left( 1-{{\rm erf}\left(1\right)} \right) = \frac{\sqrt{\pi}}{\sqrt{2}} \,{{ e}}\,\rm erfc(1)\sim 0.5358965410. $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{{1 \over \root{2\pi}} \int_{-\infty}^{\infty}{\expo{-x^{2}} \over 1 + x^{2}}\,\dd x} = {\expo{} \over \root{2\pi}} \int_{-\infty}^{\infty}{\expo{-\pars{x^{2} + 1}} \over 1 + x^{2}}\,\dd x \\[5mm] = &\ {\expo{} \over \root{2\pi}} \int_{-\infty}^{\infty} \int_{1}^{\infty}\expo{-\pars{x^{2} + 1}y}\,\,\dd y\,\dd x \\[5mm] = &\ {\expo{} \over \root{2\pi}} \int_{1}^{\infty}\expo{-y}\int_{-\infty}^{\infty} \expo{-yx^{2}}\,\,\dd x\,\dd y = {\expo{} \over \root{2\pi}}\int_{1}^{\infty}\expo{-y} \pars{\root{\pi}y^{-1/2}}\dd y \\[5mm] = &\ \root{2}\expo{}\int_{1}^{\infty}\expo{-y^{2}}\,\dd y = \root{2}\expo{}\bracks{{\root{\pi} \over 2}\on{erfc}\pars{1}} \\[5mm] & = \bbx{{\root{2\pi} \over 2}\,\expo{}\on{erfc}\pars{1}} \approx 0.5359\\ & \end{align} See this DLMF link.