So, I've got an integral in the following form:

$$\int_{-\infty}^{\infty} \frac{x^2 e^{-x^2/2}}{a+bx^2}dx$$

where $b<0$ and $a\in\mathbb{R}$.

I've tried substituting $y=x^2$ (after changing changing lower limit to 0 and multiplying by 2 of course) and $z=y+a$ but there is that pesky square root in the denominator...

Anyone with better ideas? Is this thing even soluble?

  • $\begingroup$ Is this in a course of Fourier analysis? In order to help you it would be good to know a bit of your knowledge. $\endgroup$ – AD. Oct 10 '11 at 5:02
  • $\begingroup$ @AD. Sadly, this is not in a course (otherwise there'd be a solution :) .) This is from a research project. I have taken multivariate calculus in college (long time ago). Unfortunately I wasn't smart enough back then to take anything beyond linear algebra. I've taken a few math-related courses in grad school (probability theory, signal theory, information theory). I know almost nothing about Fourier analysis (besides properties of characteristic functions). I've had to do lots of integration for the recent research projects and am quite familiar with Gradshtein & Ryzhyk. That about sums it up. $\endgroup$ – M.B.M. Oct 10 '11 at 5:19
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    $\begingroup$ Note that to find your desired integral if suffices to be able to evaluate $\displaystyle \int^{\infty}_{-\infty} \frac{\exp(-x^2)}{ a^2+x^2} dx = \frac{ \pi e^{a^2} (1-\text{erf} (a))}{a} $. Mathematicia gave me that result by the way. So at least now we know what the final form we need is. If this is not for a math paper, because you don't need a derivation and simply having the answer is enough. $\endgroup$ – Ragib Zaman Oct 10 '11 at 5:23
  • $\begingroup$ @Ragib Zaman: I guess you assume $b >0$. $\endgroup$ – AD. Oct 10 '11 at 5:41
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    $\begingroup$ The integrand is real, so the value of the integral is real. But if $b$ is negative and $a$ positive, the integrand will have simple poles and therefore the integral will be divergent (unless you interpret it as a Cauchy principal value). $\endgroup$ – Hans Lundmark Oct 10 '11 at 6:51

Consider the function $$\mathcal{I}(a)=\int_{-\infty}^{+\infty} \frac{a}{a^2+x^2}e^{-(a^2+x^2)}dx.$$ Integration by parts gives $$\mathcal{I}(a)=\left[\tan^{-1}\left(\frac{x}{a}\right) e^{-(a^2+x^2)}\right]_{-\infty}^{+\infty}-\int_{-\infty}^{+\infty}\tan^{-1}\left(\frac{x}{a}\right)(-2x)e^{-(a^2+x^2)}dx$$ $$=\int_{-\infty}^{+\infty}\tan^{-1}\left(\frac{x}{a}\right)2xe^{-(a^2+x^2)}dx.$$ Now differentiate $\mathcal{I}$ with respect to $a$ and obtain $$\frac{d\,\mathcal{I}}{da}=\int_{-\infty}^{+\infty}\left[-\frac{x}{a^2+x^2}\right]2xe^{-(a^2+x^2)}+\tan^{-1}\left(\frac{x}{a}\right)\left[(-2a)2xe^{-(a^2+x^2)}\right]dx$$ $$=-\int_{-\infty}^{+\infty}\frac{2x^2}{a^2+x^2}e^{-(a^2+x^2)}dx-2a\mathcal{I}(a) $$ $$=-\int_{-\infty}^{+\infty}\left(\frac{2x^2}{a^2+x^2}+2a\frac{a}{a^2+x^2}\right)e^{-(a^2+x^2)}dx $$ $$=-2\int_{-\infty}^{+\infty}e^{-(x^2+a^2)}dx=-2\sqrt{\pi}e^{-a^2}.$$ Equipped with the fact $\lim\limits_{a\to\infty}\mathcal{I}(a)=0$, we arrive at $$\mathcal{I}(a)=\int_{+\infty}^a -2\sqrt{\pi}e^{-u^2}du= \pi \,\mathrm{erfc}(a),$$ where $\mathrm{erfc}$ is the complementary error function. Note this agrees as $a\to0$ because of the distributional fact that $a/(a^2+x^2)\to\delta(x)$. This implies $$\int_{-\infty}^{+\infty}\frac{1}{x^2+a}e^{-x^2}dx=\pi e^a\frac{\mathrm{erfc}\left(\sqrt{a}\right)}{\sqrt{a}}.$$

Finally, observe that $$\int_{-\infty}^{+\infty}\frac{x^2}{a+bx^2}e^{-x^2/2}dx=\frac{1}{b}\int_{-\infty}^{+\infty}\left(1-\frac{a}{a+bx^2}\right)e^{-x^2/2}dx$$ $$=\frac{1}{b}\left(\sqrt{2\pi}-\frac{a}{\sqrt{2}b}\int_{-\infty}^{+\infty}\frac{1}{\frac{a}{2b}+x^2}e^{-x^2}dx\right)$$ $$=\frac{1}{b}\left(\sqrt{2\pi}-\sqrt{\frac{a}{b}}\pi\exp\left(\frac{a}{2b}\right)\mathrm{erfc}\left(\sqrt{\frac{a}{2b}}\right)\right).$$

  • $\begingroup$ (+1). Excellent solution, this should have more up votes. Now that I see it can be done in such an elementary manner, I sorely regret not having a try at it myself! $\endgroup$ – Ragib Zaman Oct 10 '11 at 10:16
  • $\begingroup$ Wow, what an elegant solution! Thanks, @anon! This tells me that $a$ has to be negative... $\endgroup$ – M.B.M. Oct 10 '11 at 15:16
  • $\begingroup$ @anon: One thing I don't understand is how do you get the implication that: $$\int_{-\infty}^{+\infty}\frac{1}{x^2+a}e^{-x^2}dx=\pi e^a\frac{\mathrm{erfc}\left(\sqrt{a}\right)}{\sqrt{a}}$$ I'm probably missing something trivial, but could you please elaborate? Again, thanks for the solution! $\endgroup$ – M.B.M. Oct 10 '11 at 17:58
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    $\begingroup$ @Bullmoose: Multiply the definition of $\mathcal{I}(a)$ through by $e^{a^2}/a$, then replace $a$ with its square root. $\endgroup$ – anon Oct 10 '11 at 18:01
  • $\begingroup$ @anon Aha! Thanks again! $\endgroup$ – M.B.M. Oct 10 '11 at 18:11

For positive $a$ and negative $b$ the integral is divergent. However, it is possible to consider the principal value, for which mathematica gives $$ \frac{\sqrt{\pi } \left(2 \sqrt{-a b} F\left(\sqrt{-\frac{a}{2b}}\right)+\sqrt{2} b\right)}{b^2} $$ where $F\;$ is the Dawson integral.


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