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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

Mhenni Benghorbal
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sachinruk
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3 Answers3

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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

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

Ron Gordon
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  • I'm guessing you simply took the Fourier Transform in step 1 and then transforming it from the Fourier Domain to the real domain by the integral. But why is there no $e^{ikt}$ term?

    Also forgive my ignorance but where did the constant terms (eg. $\sqrt{\pi}$) come from?

     – sachinruk Apr 15 '13 at 05:53
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    @user72388: not necessary in using Parseval's theorem, which just states that the integral of a product of a pair of functions is equal to the integral of the product of the Fourier transforms of that pair of functions. In the first step, I replaced the $1/(1+x^2)$ by its FT ($\pi e^{-|k|}$) and $e^{-x^2}$ by its FT ($\sqrt{\pi} e^{-k^2/4}$). These are well-known transform pairs and I did not feel the need to derive them. – Ron Gordon Apr 15 '13 at 05:58
  • @user72388: the constant terms are a result of the FT definition I used. Note the factor $1/(2 \pi)$ I used - all of the constants are the result of a consistent application of the definition of a FT:

    $$\hat{f}(k) = \int_{-\infty}^{\infty} dx : f(x) e^{i k x}$$

     – Ron Gordon Apr 15 '13 at 06:03
  • Just a thing, that the OP didn't see, it seems: a $N(0,1)$ has density $\frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$. – Jean-Claude Arbaut Mar 28 '16 at 21:24
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You can have a closed form solution for your integral in terms of the erf function

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

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

Mhenni Benghorbal
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  • @Downvoter: I forgot to multiply the integral by $\frac{1}{\sqrt{2\pi}}$ and it's been corrected. – Mhenni Benghorbal Apr 15 '13 at 13:35
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    Mhenni: I did not and would not downvote you, but I suspect that, whoever did, might have done so because you gave no insights as to where your answer came from. (I do not agree with that as a reason, but it might be possible.) I did, however, downvote the other answer because it does the OP a disservice and shows that (s)he didn't bother reading anything else in the page. – Ron Gordon Apr 15 '13 at 14:00
  • @RonGordon: I just wanted to inform the OP that he could find a closed form for his integral as a first step. Then if he is interested in techniques of solving it we may give more details. Notice that, many of the problems come from applications and they only need the final solution. If you read his question, you can see that he is wondering if there is an analytic solution or no. – Mhenni Benghorbal Apr 15 '13 at 14:05
  • Just a thing, that the OP didn't see, it seems: a $N(0,1)$ has density $\frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$. – Jean-Claude Arbaut Mar 28 '16 at 21:25
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

Felix Marin
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