Closed-form for $\int_0^\infty \frac{a}{x}\sqrt{x^2+\frac{a^2}{x^2}} \exp\left(-x^2-\frac{a^2}{x^2}\right)dx$

Question

Let $a > 0$. Is there a closed-form for $$ \int_0^\infty \frac{a}{x}\sqrt{x^2+\frac{a^2}{x^2}} e^{-x^2-\frac{a^2}{x^2}} dx \ ? $$ I have searched a lot on this website and the literature, but the closest results I could find are the following integrals: \begin{align*} &\int_0^\infty \left(x^2+\frac{a^2}{x^2}\right) e^{-x^2-\frac{a^2}{x^2}} dx = \frac{\sqrt{\pi}}{4}(1+4a) e^{-2a}, \\ &\int_0^\infty \frac{a}{x}\left(x^2+\frac{a^2}{x^2}\right) e^{-x^2-\frac{a^2}{x^2}} dx = 2 a^2 \,\mathrm{K}_1(2a), \end{align*} where $\mathrm{K}$ is the modified Bessel function of the second kind.

The motivation is to calculate the level crossing rate of the product of two Rayleigh fading models.

Answer 1

By doing a simplification of the integral, and then using the help of WolframAlpha, I could find a closed form which looks similar to the ones you mentioned on the question.

\begin{align*} I(a) = & \int_0^{\infty} \frac{a}{x}\sqrt{x^2+\frac{a^2}{x^2}}e^{-(x^2+\frac{a^2}{x^2})} dx \\ = & a\sqrt{a}\int_0^{\infty} \frac{1}{x}\sqrt{x^2+\frac{1}{x^2}}e^{-a(x^2+\frac{1}{x^2})} dx \quad (\text{use} \; x \mapsto \sqrt{a}x) \\ = & a\sqrt{a}\left( \int_0^{1} \frac{1}{x}\sqrt{x^2+\frac{1}{x^2}}e^{-a(x^2+\frac{1}{x^2})} dx + \int_1^{\infty} \frac{1}{x}\sqrt{x^2+\frac{1}{x^2}}e^{-a(x^2+\frac{1}{x^2})} dx \right) \\ &\text{for the first integral use} \; x \mapsto \frac{1}{x} \\ = & 2a\sqrt{a}\int_1^{\infty} \frac{1}{x}\sqrt{x^2+\frac{1}{x^2}}e^{-a(x^2+\frac{1}{x^2})} dx \\ & \text{use} \; x^2 = t^2+\sqrt{t^2-1} \; \text{or} \; t = \frac12\left(x^2+\dfrac{1}{x^2}\right) \\ & = \sqrt{2a^3} \int_1^{\infty} \sqrt{\frac{t}{t^2-1}}e^{-2at} dt \\ & = \sqrt{\frac{2}{\pi}} a^2 \mathrm{K}_{-\frac34}(a) \mathrm{K}_{-\frac14}(a) \end{align*}

The last step, i.e., solving $\int_1^{\infty} \sqrt{\frac{t}{t^2-1}}e^{-2at} dt$ is done by WolframAlpha. I have not solved it yet.

However, using the fact that $\; \mathrm{K}_{\alpha}(z) = \mathrm{K}_{-\alpha}(z)$ and $$\mathrm{K}_\alpha(z) \sim \sqrt{\frac{\pi}{2 z}} e^{-z}\left(1+\frac{4 \alpha^2-1}{8 z}+\frac{\left(4 \alpha^2-1\right)\left(4 \alpha^2-9\right)}{2 !(8 z)^2}+\frac{\left(4 \alpha^2-1\right)\left(4 \alpha^2-9\right)\left(4 \alpha^2-25\right)}{3 !(8 z)^3}+\cdots\right) \quad \text { for }|\arg z|<\frac{3 \pi}{2}$$

We get that $$I(a) \sim \sqrt {\frac{\pi }{2}} ae^{-2a} \left( {1 + \frac{1}{{16a}} - \frac{{15}}{{512a^2 }} + \cdots } \right)$$

Hope this helps you.

Answer 2

This answer gives only an asymptotics for large $a$. Denote your integral by $I(a)$. Performing the change of variables from $x$ to $t$ via $x = \sqrt {a{\rm e}^t }$ and using the evenness of the resulting integrand yields $$\tag{1} I(a) = \sqrt 2 a^{3/2} {\rm e}^{ - 2a} \int_0^{ + \infty } {\sqrt {\cosh t} \exp ( - 2a(\cosh t - 1))} {\rm d}t. $$ Performing the change of variables from $t$ to $s$ via $t = \cosh ^{ - 1} (1 + s) $ then gives $$ I(a) = a^{3/2} {\rm e}^{ - 2a} \int_0^{ + \infty } {\sqrt {\frac{{2(1 + s)}}{{s(2 + s)}}} {\rm e}^{ - 2as}} {\rm d}s. $$ Since $$ \sqrt {\frac{{2(1 + s)}}{{s(2 + s)}}} = s^{ - 1/2} + \frac{1}{{4 }}s^{1/2} - \frac{5}{{32 }}s^{3/2} + \ldots $$ as $s\to 0^+$, Watson's lemma gives the asymptotic expansion $$ I(a) \sim \sqrt {\frac{\pi }{2}} a{\rm e}^{ - 2a} \!\left( {1 + \frac{1}{{16a}} - \frac{{15}}{{512a^2 }} + \ldots } \right) $$ as $a\to+\infty$.

Addendum. By $(10.39.2)$ and $(10.32.17)$, $(1)$ becomes $$ I(a) = 2\sqrt {\frac{2}{\pi }} a^2 \int_0^{ + \infty } {K_{1/2} (2a\cosh t)\cosh t\,{\rm d}t} = \sqrt {\frac{2}{\pi }} a^2 K_{3/4} (a)K_{1/4} (a) $$ for $\operatorname{Re}(a)>0$. For $(10.32.17)$, see page $440$ of G. N. Watson's book A Treatise on the Theory of Bessel Functions.