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?
Best Answer
By $(12.5.1)$ and $(12.7.10)$, we have \begin{align*} \int_{ - \infty }^{ + \infty } {\exp \left( { - \beta (1 + x^2 )^2 } \right)\,\mathrm{d}x} & \mathop = \limits^{t = x^2 } \mathrm{e}^{ - \beta } \int_0^{ + \infty } {t^{ - \frac{1}{2}} \exp \left( { - \beta t^2 - 2\beta t} \right)\,\mathrm{d}t} \\& \!\!\!\mathop = \limits^{s = \sqrt {2\beta } t} \mathrm{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)\,\mathrm{d}s} \\ &\; = \mathrm{e}^{ - \beta /2} (2\beta )^{ - 1/4} \sqrt \pi U(0,\sqrt {2\beta } ) = \frac{{\mathrm{e}^{ - \beta /2} }}{{\sqrt 2 }}K_{1/4} (\beta /2). \end{align*}