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Is there any closed form, asymptotics, and/or approximations for the following integral:

$$f(c) := \frac{1}{\sqrt{2\sigma^2\pi}}\int e^{-x^2/2\sigma^2}\frac{x^2}{1-cx^2}dx,$$ where $\sigma^2$ is real valued and $c$ is complex? I am integrating over $\mathbb{R}$.

Even in the case where $c$ is real would be interesting! Pointers to relevant references are appreciated!

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  • $\begingroup$ Are you integrating over $\mathbb R$? $\endgroup$ Dec 20, 2016 at 0:37
  • $\begingroup$ That's correct, I just updated the question to clarify that. $\endgroup$
    – user19346
    Dec 20, 2016 at 0:40
  • $\begingroup$ If you integrate over $x\in\mathbb{R}$ and $c$ is real, then $c\leq 0$; else, you have poles. $\endgroup$ Dec 20, 2016 at 1:10
  • $\begingroup$ @user19346: I've posted an alternative approach to the solution. $\endgroup$ Dec 21, 2016 at 0:29

3 Answers 3

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Obviously you need $\sigma^2 > 0$ (and I'll assume $\sigma > 0$), and $c$ should not be a positive real to avoid having singularities at $x = \pm \sqrt{1/c}$. According to Maple, the result is

$$ -{\frac {\sqrt {\sigma}}{c}}- \frac{\sqrt{-\pi c}}{ c^2 \sqrt{2\sigma}} \exp\left(-\frac{1}{2c\sigma^2}\right) + \frac{\sqrt{\pi}}{\sqrt{2\sigma} c^{3/2}} \exp\left(-\frac{1}{2c\sigma^2}\right) \text{erfi}\left(\frac{1}{\sigma \sqrt{2c}}\right)$$

This appears to check out numerically.

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  • $\begingroup$ Thanks! Is there an analogous expression when c is complex (instead of real)? $\endgroup$
    – user19346
    Dec 20, 2016 at 7:37
  • $\begingroup$ @user, "...$c$ should not be a positive real..." (emphasis mine). $\endgroup$ Dec 20, 2016 at 13:48
  • $\begingroup$ Yes, this is for complex $c$. $\endgroup$ Dec 20, 2016 at 20:30
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Here is a direct computation. For $\beta>0$, we need the following fact $$\int_{\mathbb{R}}e^{-x^2\beta^2}dx=\frac{\sqrt{\pi}}{\beta},$$ and the so-called Error function $$\pmb{\text{erf}}(s)=\frac2{\sqrt{\pi}}\int_0^se^{-y^2}dy.$$ Then, the complementary error function is defined as $\pmb{\text{erfc}}(s)=1-\pmb{\text{erf}}(s)$. Let $b^2:=-c, b>0$.

Now, write $x^2=\frac{1+b^2x^2-1}{b^2}$ so that \begin{align} \int_{\mathbb{R}}e^{-x^2/2\sigma^2}\frac{x^2dx}{1+b^2x^2} &=\frac1{b^2}\int_{\mathbb{R}}e^{-x^2/2\sigma^2}dx-\frac1{b^2}\int_{\mathbb{R}} e^{-x^2/2\sigma^2}\frac{dx}{1+b^2x^2} \\ &=\frac{\sigma\sqrt{2\pi}}{b^2}-\frac1{b^3}\int_{\mathbb{R}} e^{-u^2/2b^2\sigma^2}\frac{du}{1+u^2}. \end{align} Let $a:=1/2b^2\sigma^2$ and rewrite $$\int_{\mathbb{R}}e^{-u^2a}\frac{du}{1+u^2}=e^a\int_{\mathbb{R}}e^{-(1+u^2)a}\frac{du}{1+u^2}:=e^ag(a).$$ Computing the derivative $\frac{d}{da}$, we find that $$g'(a)=-\int_{\mathbb{R}}e^{-(1+u^2)a}du=-e^{-a}\int_{\mathbb{R}}e^{-u^2a}du =-\sqrt{\pi}\frac{e^{-a}}a.$$ This differential equation and initial value $g(0)=\pi$ lead to \begin{align} g(a) &=-\sqrt{\pi}\int_0^a\frac{e^{-t}}tdt+\pi =-2\sqrt{\pi}\int_0^{\sqrt{a}}e^{-y^2}dy+\pi \\ &=-\pi\left(\frac2{\sqrt{\pi}}\int_0^{\sqrt{a}}e^{-y^2}dy\right)+\pi =-\pi\cdot\pmb{\text{erf}}(\sqrt{a})+\pi \\ &=\pi\cdot\pmb{\text{erfc}}(\sqrt{a}). \end{align} Finally, we combine all that we calculated so far to obtain \begin{align} \frac1{\sqrt{2\sigma^2\pi}}\int_{\mathbb{R}}e^{-x^2/2\sigma^2}\frac{x^2dx}{1+b^2x^2} &=\frac1{\sqrt{2\sigma^2\pi}}\left[\frac{\sigma\sqrt{2\pi}}{b^2} -\frac{\pi}{b^3}e^{1/2b^2\sigma^2}\cdot\pmb{\text{erfc}}\left(\frac1{b\sigma\sqrt{2}}\right)\right] \\ &=\frac1{b^2}-\frac{\sqrt{\pi}}{b^3\sigma\sqrt{2}}\cdot e^{1/2b^2\sigma^2}\cdot\pmb{\text{erfc}}\left(\frac1{b\sigma\sqrt{2}}\right). \end{align} We made the assumption $\sigma>0$.

REMARK. The integral in the OP's question can be regarded, up to scaling, as the Weierstrass transform $$H(y)=\frac1{\sqrt{4\pi}}\int_{\mathbb{R}}h(x)e^{-(y-x)^2/4}dx$$ of the function $h(x)=\frac1{1+b^2x^2}$, but evaluated at $y=0$; that is, $H(0)$.

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Here is a beautiful analogy of Gaussian integral which I solved recently check it out.. It might be of help to u. https://mymathware.blogspot.nl/2016/12/gaussian-or-euler-poisson-integral.html?m=1

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  • $\begingroup$ You seem to have only calculated the Gaussian integral itself, not an analogue. $\endgroup$
    – Ben McKay
    Dec 29, 2016 at 17:13

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