Conjectured identity involving an elliptic integral and a Bessel function

Question

Let $K$ be the complete elliptic integral of the first kind and $K_\nu$ a modified Bessel function of the second kind. I'm wondering whether the following identity is true and if so, how to prove it: $$\int_{1}^\infty e^{-2 a u^2} u^{1/2}K\left(\frac{u-1}{2u}\right) du \overset{?}{=}\sqrt{\frac{\pi}{32a}} e^{-a}K_{1/4}(a)\tag{1}$$ for $a$ real and strictly positive. I've searched several integral tables but could not find anything that looks related.

The reason why I suspect this might be true is the following. Start from the integral $$Z(g)=\int_{-\infty}^{\infty} \frac{dx}{\sqrt{2\pi}}\, \exp{\left(-\frac{1}{2} x^2 -\frac{g}{4!}x^4\right)} \tag{2}$$ This can be expressed in terms of a Bessel function, for example as in this question, $$Z(g)=\sqrt{\frac{3}{2\pi g}} \exp{\left(\frac{3}{4g}\right)} K_{1/4}\left(\frac{3}{4 g}\right)$$ However, going back to $(2)$, expanding the second exponential and exchanging the order of integration and summation, one gets the asymptotic series $$Z(g)\sim \sum_{n=0}^\infty \frac{(4n)!}{2^{2n}(2n)!n!}\left(-\frac{g}{24}\right)^n$$ Its Borel transform sums to a complete elliptic integral of the first kind, as in this question: $$\mathcal{B}Z(g)= \sum_{n=0}^\infty \frac{(4n)!}{2^{2n}(2n)!(n!)^2}\left(-\frac{g}{24}\right)^n = \frac{2}{\pi} \frac{1}{\left(1+2g/3\right)^{1/4}}K\left(\frac{-1+\sqrt{1+2g/3}}{2\sqrt{1+2g/3}}\right)$$ The Borel transform converges for $|g| < 3/2$, however the right hand side makes sense for all $g > -3/2$. Now the statement $(1)$ is equivalent to the assertion that $Z$ is equal to its Borel sum, that is $Z(g) \overset{?}{=} \int_0^\infty e^{-t} \mathcal{B}Z(gt)$ after replacing $u^2 = 1+2gt/3$ and $a=3/(4g)$ in the integral for cleanliness.

Answer 1

With the change of integration variables $u = \sqrt {1 + t}$, your original identity reads $$ K_{1/4} (a) = \sqrt \pi \mathrm{e}^{ - a} \sqrt {2a} \int_0^{ + \infty } {\mathrm{e}^{ - 2at} \frac{2}{{\pi (1 + t)^{1/4} }}K\!\left( {\frac{{\sqrt {1 + t} - 1}}{{2\sqrt {1 + t} }}} \right)\mathrm{d}t} . $$ By Entry $19.2$ on page $195$ of F. Oberhettinger, L. Badii, Tables of Laplace Transforms, we have $$ K_\nu (a) = \sqrt \pi \mathrm{e}^{ - a} (2a)^{\lambda - 1/2} \int_0^{ + \infty } {\mathrm{e}^{ - 2at} t^{\lambda - 1} \frac{1}{{\Gamma (\lambda )}}{}_2F_1 \! \left( {\nu + \tfrac{1}{2}, - \nu + \tfrac{1}{2};\lambda ; - t} \right)\mathrm{d}t} $$ for any $\nu \in \mathbb C$, $\operatorname{Re}(a)>0$ and $\operatorname{Re}(\lambda) >0$. Here ${}_2 F_1$ stands for the Gauss hypergeometric function. Thus it is enough to show that $$ \frac{2}{{\pi (1 + t)^{1/4} }}K\!\left( {\frac{{\sqrt {1 + t} - 1}}{{2\sqrt {1 + t} }}} \right) = {}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - t} \right),\;\;t>0. $$ This is shown here.

Proof using the asymptotics of the Bessel function. We start with the expansion $$ F(a): = \sqrt {\frac{2}{\pi a}} \mathrm{e}^a K_{1/4} (a) = \sum\limits_{n = 0}^N {( - 1)^n \frac{{(4n)!}}{{128^n n!(2n)!}}\frac{1}{{a^{n + 1} }}} + R_{N} (a), $$ where $$\tag{1} \left| {R_{N} (a)} \right| \le \frac{{(4N + 4)!}}{{128^{N + 1} (N + 1)!(2N + 2)!}}\frac{1}{{\left| a \right|^{N + 2} }} $$ provided that $|\arg a|\leq\frac{\pi}{2}$ and $N\geq 0$, see for instance here. Now with an arbitrary $c>0$, let $$ f(t): = \frac{1}{{2\pi \mathrm{i}}}\int_{c - \mathrm{i}\infty }^{c + \mathrm{i}\infty } {\mathrm{e}^{at} F(a)\mathrm{d}a} = \sum\limits_{n = 1}^N {( - 1)^n \frac{{(4n)!}}{{128^n n!^2 (2n)!}}t^n } + \frac{1}{{2\pi \mathrm{i}}}\int_{c - \mathrm{i}\infty }^{c + \mathrm{i}\infty } {\mathrm{e}^{at} R_{N} (a)\mathrm{d}a} $$ for any $t>0$. Differentiating $N$ times gives $$\tag{2} f^{(N)} (t) = ( - 1)^N \frac{{(4N)!}}{{128^N N!(2N)!}} + \frac{1}{{2\pi \mathrm{i}}}\int_{c - \mathrm{i}\infty }^{c + \mathrm{i}\infty } {\mathrm{e}^{at} a^N R_{N} (a)\mathrm{d}a} $$ where, by $(1)$, $$\tag{3} \left| {\frac{1}{{2\pi \mathrm{i}}}\int_{c - \mathrm{i}\infty }^{c + \mathrm{i}\infty } {\mathrm{e}^{at} a^N R_{N} (a)\mathrm{d}a} } \right| \le \frac{{(4N + 4)!}}{{128^{N + 1} (N + 1)!(2N + 2)!}}\frac{{\mathrm{e}^{ct} }}{{2c}}. $$ Stirling's formula then implies that $$ \frac{{\left| {f^{(N)} (t)} \right|}}{{N!}} \le C\frac{N}{{2^N }}\frac{{\mathrm{e}^{ct} }}{c} $$ with an absolute $C>0$. Thus, by Taylor's formula and the Cauchy–Hadamard theorem, $f(t)$ is analytic in the tubular neighbourhood $U:=\left\{ {t:\text{dist}(t,\mathbb R^ + ) < 2} \right\}$ of the positive axis. Employing $(2)$ and $(3)$ with $t=0$ and letting $c\to +\infty$ shows that $$ \frac{{f^{(N)} (0)}}{{N!}} = ( - 1)^N \frac{{(4N)!}}{{128^N N!^2 (2N)!}}, $$ i.e., the Maclaurin series of $f(t)$ and ${}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - \frac{t}{2}} \right)$ are the same. By analytic continuation, $f(t)={}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - \frac{t}{2}} \right)$ for any $t\in U$. Accordingly, $$ F(a) = \int_0^{ + \infty } {\mathrm{e}^{ - at} {}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - \tfrac{t}{2}} \right)\mathrm{d}t} = 2\int_0^{ + \infty } {\mathrm{e}^{ - 2at} {}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - t} \right)\mathrm{d}t} $$ provided $\operatorname{Re}(a) >0$.