$E(1/(1+x^2)) $under a normal distribution 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
 A: 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
$$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} dx \frac{1}{1+x^2} \, e^{-x^2} = \sqrt{\frac{\pi}{2}} \, e \, \text{erfc}(1) \approx 0.535897$$
A: You can have a closed form solution for your integral in terms of the erf function

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

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{e^{-x^2}}{1+x^2} dx = \frac{\sqrt{\pi}}{\sqrt{2}} \,{{ e}}\, \left( 1-{{\rm erf}\left(1\right)} \right) = \frac{\sqrt{\pi}}{\sqrt{2}} \,{{ e}}\,\rm erfc(1)\sim 0.5358965410. $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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\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.
