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I'm trying to evaluate the following definite integral $$ \int_{-\infty}^{\infty} dx\,\exp\left[-\beta (1+x^2)^2\right]\,, $$ for $\beta>0$, which Mathematica's Integrate[] evaluates as $$ \frac{\exp(-\beta/2)K_{1/4}(\beta/2)}{\sqrt{2}}\,, $$ where $K_{(\cdot)}(\cdot)$ is the modified Bessel function of the second kind.

I'm not able to see this result from the standard integral representations of the modified Bessel function of the second kind. The closest representation to Mathematica's results in the NIST library seems to be Eqs. 10.32.8 and 10.32.10, but I'm not able to turn my integral to either. Are there other integral representations of $K$ that I'm not aware of?

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By http://dlmf.nist.gov/12.5.E1 and http://dlmf.nist.gov/12.7.E10, we have \begin{align*} \int_{ - \infty }^{ + \infty } {\exp \left( { - \beta (1 + x^2 )^2 } \right)dx} & \mathop = \limits^{t = x^2 } e^{ - \beta } \int_0^{ + \infty } {t^{ - \frac{1}{2}} \exp \left( { - \beta t^2 - 2\beta t} \right)dt} \\& \mathop = \limits^{s = \sqrt {2\beta } t} e^{ - \beta } (2\beta )^{ - 1/4} \int_0^{ + \infty } {s^{ - \frac{1}{2}} \exp \left( { - \tfrac{1}{2}s^2 - \sqrt {2\beta } s} \right)ds} \\ & = e^{ - \beta /2} (2\beta )^{ - 1/4} \sqrt \pi U(0,\sqrt {2\beta } ) = \frac{{e^{ - \beta /2} }}{{\sqrt 2 }}K_{1/4} (\beta /2). \end{align*}

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  • $\begingroup$ I just discovered this myself. I didn't know about the relationship between the parabolic cylindrical function $D_{-1/2}$ and $K_{1/4}$, and I just saw it in Chapter 46 of An Atlas of Function, 2e, by Oldham et al. Nonetheless, thanks for the detailed answer! $\endgroup$ Jul 21 at 8:02
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    $\begingroup$ I recommed you to use the NIST Digital Library of Mathematical Functions or its printed version, the NIST Handbook of Mathematical Functions. These are more modern references. $\endgroup$
    – Gary
    Jul 21 at 8:11

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